Linear Algebra Examples

$[\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}]$

Step 1

Set up the formula to find the characteristic equation

$p (λ)$ .

$p (λ) = determinant (A - λ I_{4})$

Step 2

The identity matrix or unit matrix of size

$4$ is the

$4 \times 4$ square matrix with ones on the main diagonal and zeros elsewhere.

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$

Step 3

Substitute the known values into

$p (λ) = determinant (A - λ I_{4})$ .

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Step 3.1

Substitute

$[\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}]$ for

$A$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] - λ I_{4})$

Step 3.2

Substitute

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$ for

$I_{4}$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply

$- λ$ by each element of the matrix.

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.5.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.9.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.10

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.10.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.10.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.11

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.12

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.12.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.12.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.13

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.13.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.13.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.14

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.14.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.14.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.15

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.15.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.15.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.16

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} - 2 & 1 & - 2 & - 4 \\ 12 & 1 & 4 & 9 \\ 6 & 5 & - 2 & - 4 \\ 3 & - 4 & 5 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 + 0 & - 2 + 0 & - 4 + 0 \\ 12 + 0 & 1 - λ & 4 + 0 & 9 + 0 \\ 6 + 0 & 5 + 0 & - 2 - λ & - 4 + 0 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add

$1$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 + 0 & - 4 + 0 \\ 12 + 0 & 1 - λ & 4 + 0 & 9 + 0 \\ 6 + 0 & 5 + 0 & - 2 - λ & - 4 + 0 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.2

Add

$- 2$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 + 0 \\ 12 + 0 & 1 - λ & 4 + 0 & 9 + 0 \\ 6 + 0 & 5 + 0 & - 2 - λ & - 4 + 0 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.3

Add

$- 4$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 + 0 & 1 - λ & 4 + 0 & 9 + 0 \\ 6 + 0 & 5 + 0 & - 2 - λ & - 4 + 0 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.4

Add

$12$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 & 1 - λ & 4 + 0 & 9 + 0 \\ 6 + 0 & 5 + 0 & - 2 - λ & - 4 + 0 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.5

Add

$4$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 & 1 - λ & 4 & 9 + 0 \\ 6 + 0 & 5 + 0 & - 2 - λ & - 4 + 0 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.6

Add

$9$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 & 1 - λ & 4 & 9 \\ 6 + 0 & 5 + 0 & - 2 - λ & - 4 + 0 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.7

Add

$6$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 & 1 - λ & 4 & 9 \\ 6 & 5 + 0 & - 2 - λ & - 4 + 0 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.8

Add

$5$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 & 1 - λ & 4 & 9 \\ 6 & 5 & - 2 - λ & - 4 + 0 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.9

Add

$- 4$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 & 1 - λ & 4 & 9 \\ 6 & 5 & - 2 - λ & - 4 \\ 3 + 0 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.10

Add

$3$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 & 1 - λ & 4 & 9 \\ 6 & 5 & - 2 - λ & - 4 \\ 3 & - 4 + 0 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.11

Add

$- 4$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 & 1 - λ & 4 & 9 \\ 6 & 5 & - 2 - λ & - 4 \\ 3 & - 4 & 5 + 0 & 10 - λ \end{array}]$

Step 4.3.12

Add

$5$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 2 - λ & 1 & - 2 & - 4 \\ 12 & 1 - λ & 4 & 9 \\ 6 & 5 & - 2 - λ & - 4 \\ 3 & - 4 & 5 & 10 - λ \end{array}]$

Step 5

Find the determinant.

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Step 5.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$1$ by its cofactor and add.

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Step 5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{array} |$

Step 5.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 5.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} 1 - λ & 4 & 9 \\ 5 & - 2 - λ & - 4 \\ - 4 & 5 & 10 - λ \end{array} |$

Step 5.1.4

Multiply element

$a_{11}$ by its cofactor.

$(- 2 - λ) | \begin{array}{c} 1 - λ & 4 & 9 \\ 5 & - 2 - λ & - 4 \\ - 4 & 5 & 10 - λ \end{array} |$

Step 5.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} |$

Step 5.1.6

Multiply element

$a_{12}$ by its cofactor.

$- 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} |$

Step 5.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} |$

Step 5.1.8

Multiply element

$a_{13}$ by its cofactor.

$- 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} |$

Step 5.1.9

The minor for

$a_{14}$ is the determinant with row

$1$ and column

$4$ deleted.

$| \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.1.10

Multiply element

$a_{14}$ by its cofactor.

$4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.1.11

Add the terms together.

$p (λ) = (- 2 - λ) | \begin{array}{c} 1 - λ & 4 & 9 \\ 5 & - 2 - λ & - 4 \\ - 4 & 5 & 10 - λ \end{array} | - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2

Evaluate

$| \begin{array}{c} 1 - λ & 4 & 9 \\ 5 & - 2 - λ & - 4 \\ - 4 & 5 & 10 - λ \end{array} |$ .

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Step 5.2.1

Choose the row or column with the most

$0$ elements. If there are no

$0$ elements choose any row or column. Multiply every element in row

$1$ by its cofactor and add.

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Step 5.2.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.2.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 5.2.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} - 2 - λ & - 4 \\ 5 & 10 - λ \end{array} |$

Step 5.2.1.4

Multiply element

$a_{11}$ by its cofactor.

$(1 - λ) | \begin{array}{c} - 2 - λ & - 4 \\ 5 & 10 - λ \end{array} |$

Step 5.2.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} |$

Step 5.2.1.6

Multiply element

$a_{12}$ by its cofactor.

$- 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} |$

Step 5.2.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |$

Step 5.2.1.8

Multiply element

$a_{13}$ by its cofactor.

$9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |$

Step 5.2.1.9

Add the terms together.

$p (λ) = (- 2 - λ) ((1 - λ) | \begin{array}{c} - 2 - λ & - 4 \\ 5 & 10 - λ \end{array} | - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2

Evaluate

$| \begin{array}{c} - 2 - λ & - 4 \\ 5 & 10 - λ \end{array} |$ .

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Step 5.2.2.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) ((1 - λ) ((- 2 - λ) (10 - λ) - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2

Simplify the determinant.

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Step 5.2.2.2.1

Simplify each term.

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Step 5.2.2.2.1.1

Expand

$(- 2 - λ) (10 - λ)$ using the FOIL Method.

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Step 5.2.2.2.1.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) ((1 - λ) (- 2 (10 - λ) - λ (10 - λ) - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.1.2

Apply the distributive property.

$p (λ) = (- 2 - λ) ((1 - λ) (- 2 \cdot 10 - 2 (- λ) - λ (10 - λ) - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.1.3

Apply the distributive property.

$p (λ) = (- 2 - λ) ((1 - λ) (- 2 \cdot 10 - 2 (- λ) - λ \cdot 10 - λ (- λ) - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.2

Simplify and combine like terms.

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Step 5.2.2.2.1.2.1

Simplify each term.

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Step 5.2.2.2.1.2.1.1

Multiply

$- 2$ by

$10$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 - 2 (- λ) - λ \cdot 10 - λ (- λ) - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.2.1.2

Multiply

$- 1$ by

$- 2$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 + 2 λ - λ \cdot 10 - λ (- λ) - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.2.1.3

Multiply

$10$ by

$- 1$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 + 2 λ - 10 λ - λ (- λ) - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 + 2 λ - 10 λ - 1 \cdot - 1 λ \cdot λ - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.2.1.5

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 5.2.2.2.1.2.1.5.1

Move

$λ$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 + 2 λ - 10 λ - 1 \cdot - 1 (λ \cdot λ) - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.2.1.5.2

Multiply

$λ$ by

$λ$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 + 2 λ - 10 λ - 1 \cdot - 1 λ^{2} - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.2.1.6

Multiply

$- 1$ by

$- 1$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 + 2 λ - 10 λ + 1 λ^{2} - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.2.1.7

Multiply

$λ^{2}$ by

$1$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 + 2 λ - 10 λ + λ^{2} - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.2.2

Subtract

$10 λ$ from

$2 λ$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 - 8 λ + λ^{2} - 5 \cdot - 4) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.1.3

Multiply

$- 5$ by

$- 4$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 20 - 8 λ + λ^{2} + 20) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.2

Combine the opposite terms in

$- 20 - 8 λ + λ^{2} + 20$ .

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Step 5.2.2.2.2.1

Add

$- 20$ and

$20$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 8 λ + λ^{2} + 0) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.2.2

Add

$- 8 λ + λ^{2}$ and

$0$ .

$p (λ) = (- 2 - λ) ((1 - λ) (- 8 λ + λ^{2}) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.2.2.3

Reorder

$- 8 λ$ and

$λ^{2}$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.3

Evaluate

$| \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} |$ .

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Step 5.2.3.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (5 (10 - λ) - (- 4 \cdot - 4)) + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.3.2

Simplify the determinant.

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Step 5.2.3.2.1

Simplify each term.

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Step 5.2.3.2.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (5 \cdot 10 + 5 (- λ) - (- 4 \cdot - 4)) + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.3.2.1.2

Multiply

$5$ by

$10$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (50 + 5 (- λ) - (- 4 \cdot - 4)) + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.3.2.1.3

Multiply

$- 1$ by

$5$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (50 - 5 λ - (- 4 \cdot - 4)) + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.3.2.1.4

Multiply

$- (- 4 \cdot - 4)$ .

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Step 5.2.3.2.1.4.1

Multiply

$- 4$ by

$- 4$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (50 - 5 λ - 1 \cdot 16) + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.3.2.1.4.2

Multiply

$- 1$ by

$16$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (50 - 5 λ - 16) + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.3.2.2

Subtract

$16$ from

$50$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4

Evaluate

$| \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |$ .

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Step 5.2.4.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (5 \cdot 5 - (- 4 (- 2 - λ)))) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4.2

Simplify the determinant.

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Step 5.2.4.2.1

Simplify each term.

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Step 5.2.4.2.1.1

Multiply

$5$ by

$5$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (25 - (- 4 (- 2 - λ)))) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4.2.1.2

Apply the distributive property.

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (25 - (- 4 \cdot - 2 - 4 (- λ)))) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4.2.1.3

Multiply

$- 4$ by

$- 2$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (25 - (8 - 4 (- λ)))) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4.2.1.4

Multiply

$- 1$ by

$- 4$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (25 - (8 + 4 λ))) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4.2.1.5

Apply the distributive property.

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (25 - 1 \cdot 8 - (4 λ))) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4.2.1.6

Multiply

$- 1$ by

$8$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (25 - 8 - (4 λ))) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4.2.1.7

Multiply

$4$ by

$- 1$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (25 - 8 - 4 λ)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4.2.2

Subtract

$8$ from

$25$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (17 - 4 λ)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.4.2.3

Reorder

$17$ and

$- 4 λ$ .

$p (λ) = (- 2 - λ) ((1 - λ) (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5

Simplify the determinant.

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Step 5.2.5.1

Simplify each term.

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Step 5.2.5.1.1

Expand

$(1 - λ) (λ^{2} - 8 λ)$ using the FOIL Method.

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Step 5.2.5.1.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (1 (λ^{2} - 8 λ) - λ (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.1.2

Apply the distributive property.

$p (λ) = (- 2 - λ) (1 λ^{2} + 1 (- 8 λ) - λ (λ^{2} - 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.1.3

Apply the distributive property.

$p (λ) = (- 2 - λ) (1 λ^{2} + 1 (- 8 λ) - λ \cdot λ^{2} - λ (- 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2

Simplify and combine like terms.

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Step 5.2.5.1.2.1

Simplify each term.

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Step 5.2.5.1.2.1.1

Multiply

$λ^{2}$ by

$1$ .

$p (λ) = (- 2 - λ) (λ^{2} + 1 (- 8 λ) - λ \cdot λ^{2} - λ (- 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.1.2

Multiply

$- 8 λ$ by

$1$ .

$p (λ) = (- 2 - λ) (λ^{2} - 8 λ - λ \cdot λ^{2} - λ (- 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.1.3

Multiply

$λ$ by

$λ^{2}$ by adding the exponents.

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Step 5.2.5.1.2.1.3.1

Move

$λ^{2}$ .

$p (λ) = (- 2 - λ) (λ^{2} - 8 λ - (λ^{2} λ) - λ (- 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.1.3.2

Multiply

$λ^{2}$ by

$λ$ .

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Step 5.2.5.1.2.1.3.2.1

Raise

$λ$ to the power of

$1$ .

$p (λ) = (- 2 - λ) (λ^{2} - 8 λ - (λ^{2} λ^{1}) - λ (- 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.1.3.2.2

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = (- 2 - λ) (λ^{2} - 8 λ - λ^{2 + 1} - λ (- 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.1.3.3

Add

$2$ and

$1$ .

$p (λ) = (- 2 - λ) (λ^{2} - 8 λ - λ^{3} - λ (- 8 λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (- 2 - λ) (λ^{2} - 8 λ - λ^{3} - 1 \cdot - 8 λ \cdot λ - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.1.5

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 5.2.5.1.2.1.5.1

Move

$λ$ .

$p (λ) = (- 2 - λ) (λ^{2} - 8 λ - λ^{3} - 1 \cdot - 8 (λ \cdot λ) - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.1.5.2

Multiply

$λ$ by

$λ$ .

$p (λ) = (- 2 - λ) (λ^{2} - 8 λ - λ^{3} - 1 \cdot - 8 λ^{2} - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.1.6

Multiply

$- 1$ by

$- 8$ .

$p (λ) = (- 2 - λ) (λ^{2} - 8 λ - λ^{3} + 8 λ^{2} - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.2.2

Add

$λ^{2}$ and

$8 λ^{2}$ .

$p (λ) = (- 2 - λ) (9 λ^{2} - 8 λ - λ^{3} - 4 (- 5 λ + 34) + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.3

Apply the distributive property.

$p (λ) = (- 2 - λ) (9 λ^{2} - 8 λ - λ^{3} - 4 (- 5 λ) - 4 \cdot 34 + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.4

Multiply $- 5$ by $- 4$ .

$p (λ) = (- 2 - λ) (9 λ^{2} - 8 λ - λ^{3} + 20 λ - 4 \cdot 34 + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.5

Multiply $- 4$ by $34$ .

$p (λ) = (- 2 - λ) (9 λ^{2} - 8 λ - λ^{3} + 20 λ - 136 + 9 (- 4 λ + 17)) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.6

Apply the distributive property.

$p (λ) = (- 2 - λ) (9 λ^{2} - 8 λ - λ^{3} + 20 λ - 136 + 9 (- 4 λ) + 9 \cdot 17) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.7

Multiply $- 4$ by $9$ .

$p (λ) = (- 2 - λ) (9 λ^{2} - 8 λ - λ^{3} + 20 λ - 136 - 36 λ + 9 \cdot 17) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.1.8

Multiply $9$ by $17$ .

$p (λ) = (- 2 - λ) (9 λ^{2} - 8 λ - λ^{3} + 20 λ - 136 - 36 λ + 153) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.2

Add $- 8 λ$ and $20 λ$ .

$p (λ) = (- 2 - λ) (9 λ^{2} - λ^{3} + 12 λ - 136 - 36 λ + 153) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.3

Subtract $36 λ$ from $12 λ$ .

$p (λ) = (- 2 - λ) (9 λ^{2} - λ^{3} - 24 λ - 136 + 153) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.4

Add $- 136$ and $153$ .

$p (λ) = (- 2 - λ) (9 λ^{2} - λ^{3} - 24 λ + 17) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.2.5.5

Reorder $9 λ^{2}$ and $- λ^{3}$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 | \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} | - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3

Evaluate $| \begin{array}{c} 12 & 4 & 9 \\ 6 & - 2 - λ & - 4 \\ 3 & 5 & 10 - λ \end{array} |$ .

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Step 5.3.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $2$ by its cofactor and add.

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Step 5.3.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.3.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 5.3.1.3

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

$| \begin{array}{c} 4 & 9 \\ 5 & 10 - λ \end{array} |$

Step 5.3.1.4

Multiply element $a_{21}$ by its cofactor.

$- 6 | \begin{array}{c} 4 & 9 \\ 5 & 10 - λ \end{array} |$

Step 5.3.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

$| \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} |$

Step 5.3.1.6

Multiply element $a_{22}$ by its cofactor.

$(- 2 - λ) | \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} |$

Step 5.3.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

$| \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |$

Step 5.3.1.8

Multiply element $a_{23}$ by its cofactor.

$4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |$

Step 5.3.1.9

Add the terms together.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 | \begin{array}{c} 4 & 9 \\ 5 & 10 - λ \end{array} | + (- 2 - λ) | \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.2

Evaluate $| \begin{array}{c} 4 & 9 \\ 5 & 10 - λ \end{array} |$ .

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Step 5.3.2.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (4 (10 - λ) - 5 \cdot 9) + (- 2 - λ) | \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.2.2

Simplify the determinant.

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Step 5.3.2.2.1

Simplify each term.

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Step 5.3.2.2.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (4 \cdot 10 + 4 (- λ) - 5 \cdot 9) + (- 2 - λ) | \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.2.2.1.2

Multiply $4$ by $10$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (40 + 4 (- λ) - 5 \cdot 9) + (- 2 - λ) | \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.2.2.1.3

Multiply $- 1$ by $4$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (40 - 4 λ - 5 \cdot 9) + (- 2 - λ) | \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.2.2.1.4

Multiply $- 5$ by $9$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (40 - 4 λ - 45) + (- 2 - λ) | \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.2.2.2

Subtract $45$ from $40$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) | \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.3

Evaluate $| \begin{array}{c} 12 & 9 \\ 3 & 10 - λ \end{array} |$ .

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Step 5.3.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (12 (10 - λ) - 3 \cdot 9) + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.3.2

Simplify the determinant.

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Step 5.3.3.2.1

Simplify each term.

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Step 5.3.3.2.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (12 \cdot 10 + 12 (- λ) - 3 \cdot 9) + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.3.2.1.2

Multiply $12$ by $10$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (120 + 12 (- λ) - 3 \cdot 9) + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.3.2.1.3

Multiply $- 1$ by $12$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (120 - 12 λ - 3 \cdot 9) + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.3.2.1.4

Multiply $- 3$ by $9$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (120 - 12 λ - 27) + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.3.2.2

Subtract $27$ from $120$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (- 12 λ + 93) + 4 | \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.4

Evaluate $| \begin{array}{c} 12 & 4 \\ 3 & 5 \end{array} |$ .

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Step 5.3.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (- 12 λ + 93) + 4 (12 \cdot 5 - 3 \cdot 4)) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.4.2

Simplify the determinant.

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Step 5.3.4.2.1

Simplify each term.

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Step 5.3.4.2.1.1

Multiply $12$ by $5$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (- 12 λ + 93) + 4 (60 - 3 \cdot 4)) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.4.2.1.2

Multiply $- 3$ by $4$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (- 12 λ + 93) + 4 (60 - 12)) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.4.2.2

Subtract $12$ from $60$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ - 5) + (- 2 - λ) (- 12 λ + 93) + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5

Simplify the determinant.

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Step 5.3.5.1

Simplify each term.

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Step 5.3.5.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 6 (- 4 λ) - 6 \cdot - 5 + (- 2 - λ) (- 12 λ + 93) + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.2

Multiply $- 4$ by $- 6$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ - 6 \cdot - 5 + (- 2 - λ) (- 12 λ + 93) + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.3

Multiply $- 6$ by $- 5$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 + (- 2 - λ) (- 12 λ + 93) + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.4

Expand $(- 2 - λ) (- 12 λ + 93)$ using the FOIL Method.

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Step 5.3.5.1.4.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 - 2 (- 12 λ + 93) - λ (- 12 λ + 93) + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.4.2

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 - 2 (- 12 λ) - 2 \cdot 93 - λ (- 12 λ + 93) + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.4.3

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 - 2 (- 12 λ) - 2 \cdot 93 - λ (- 12 λ) - λ \cdot 93 + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.5

Simplify and combine like terms.

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Step 5.3.5.1.5.1

Simplify each term.

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Step 5.3.5.1.5.1.1

Multiply $- 12$ by $- 2$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 + 24 λ - 2 \cdot 93 - λ (- 12 λ) - λ \cdot 93 + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.5.1.2

Multiply $- 2$ by $93$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 + 24 λ - 186 - λ (- 12 λ) - λ \cdot 93 + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.5.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 + 24 λ - 186 - 1 \cdot - 12 λ \cdot λ - λ \cdot 93 + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.5.1.4

Multiply $λ$ by $λ$ by adding the exponents.

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Step 5.3.5.1.5.1.4.1

Move $λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 + 24 λ - 186 - 1 \cdot - 12 (λ \cdot λ) - λ \cdot 93 + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.5.1.4.2

Multiply $λ$ by $λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 + 24 λ - 186 - 1 \cdot - 12 λ^{2} - λ \cdot 93 + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.5.1.5

Multiply $- 1$ by $- 12$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 + 24 λ - 186 + 12 λ^{2} - λ \cdot 93 + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.5.1.6

Multiply $93$ by $- 1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 + 24 λ - 186 + 12 λ^{2} - 93 λ + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.5.2

Subtract $93 λ$ from $24 λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 - 69 λ - 186 + 12 λ^{2} + 4 \cdot 48) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.1.6

Multiply $4$ by $48$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (24 λ + 30 - 69 λ - 186 + 12 λ^{2} + 192) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.2

Subtract $69 λ$ from $24 λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 45 λ + 30 - 186 + 12 λ^{2} + 192) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.3

Subtract $186$ from $30$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 45 λ - 156 + 12 λ^{2} + 192) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.4

Add $- 156$ and $192$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (- 45 λ + 12 λ^{2} + 36) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.3.5.5

Reorder $- 45 λ$ and $12 λ^{2}$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 | \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} | + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4

Evaluate $| \begin{array}{c} 12 & 1 - λ & 9 \\ 6 & 5 & - 4 \\ 3 & - 4 & 10 - λ \end{array} |$ .

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Step 5.4.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $1$ by its cofactor and add.

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Step 5.4.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.4.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 5.4.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} |$

Step 5.4.1.4

Multiply element $a_{11}$ by its cofactor.

$12 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} |$

Step 5.4.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} |$

Step 5.4.1.6

Multiply element $a_{12}$ by its cofactor.

$- (1 - λ) | \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} |$

Step 5.4.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |$

Step 5.4.1.8

Multiply element $a_{13}$ by its cofactor.

$9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |$

Step 5.4.1.9

Add the terms together.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 | \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} | - (1 - λ) | \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} | + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.2

Evaluate $| \begin{array}{c} 5 & - 4 \\ - 4 & 10 - λ \end{array} |$ .

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Step 5.4.2.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (5 (10 - λ) - (- 4 \cdot - 4)) - (1 - λ) | \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} | + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.2.2

Simplify the determinant.

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Step 5.4.2.2.1

Simplify each term.

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Step 5.4.2.2.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (5 \cdot 10 + 5 (- λ) - (- 4 \cdot - 4)) - (1 - λ) | \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} | + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.2.2.1.2

Multiply $5$ by $10$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (50 + 5 (- λ) - (- 4 \cdot - 4)) - (1 - λ) | \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} | + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.2.2.1.3

Multiply $- 1$ by $5$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (50 - 5 λ - (- 4 \cdot - 4)) - (1 - λ) | \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} | + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.2.2.1.4

Multiply $- (- 4 \cdot - 4)$ .

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Step 5.4.2.2.1.4.1

Multiply $- 4$ by $- 4$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (50 - 5 λ - 1 \cdot 16) - (1 - λ) | \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} | + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.2.2.1.4.2

Multiply $- 1$ by $16$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (50 - 5 λ - 16) - (1 - λ) | \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} | + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.2.2.2

Subtract $16$ from $50$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) | \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} | + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.3

Evaluate $| \begin{array}{c} 6 & - 4 \\ 3 & 10 - λ \end{array} |$ .

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Step 5.4.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (6 (10 - λ) - 3 \cdot - 4) + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.3.2

Simplify the determinant.

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Step 5.4.3.2.1

Simplify each term.

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Step 5.4.3.2.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (6 \cdot 10 + 6 (- λ) - 3 \cdot - 4) + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.3.2.1.2

Multiply $6$ by $10$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (60 + 6 (- λ) - 3 \cdot - 4) + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.3.2.1.3

Multiply $- 1$ by $6$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (60 - 6 λ - 3 \cdot - 4) + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.3.2.1.4

Multiply $- 3$ by $- 4$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (60 - 6 λ + 12) + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.3.2.2

Add $60$ and $12$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (- 6 λ + 72) + 9 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.4

Evaluate $| \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |$ .

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Step 5.4.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (- 6 λ + 72) + 9 (6 \cdot - 4 - 3 \cdot 5)) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.4.2

Simplify the determinant.

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Step 5.4.4.2.1

Simplify each term.

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Step 5.4.4.2.1.1

Multiply $6$ by $- 4$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (- 6 λ + 72) + 9 (- 24 - 3 \cdot 5)) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.4.2.1.2

Multiply $- 3$ by $5$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (- 6 λ + 72) + 9 (- 24 - 15)) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.4.2.2

Subtract $15$ from $- 24$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ + 34) - (1 - λ) (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5

Simplify the determinant.

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Step 5.4.5.1

Simplify each term.

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Step 5.4.5.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (12 (- 5 λ) + 12 \cdot 34 - (1 - λ) (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.2

Multiply $- 5$ by $12$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 12 \cdot 34 - (1 - λ) (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.3

Multiply $12$ by $34$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 - (1 - λ) (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.4

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + (- 1 \cdot 1 - - λ) (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.5

Multiply $- 1$ by $1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + (- 1 - - λ) (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.6

Multiply $- - λ$ .

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Step 5.4.5.1.6.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + (- 1 + 1 λ) (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.6.2

Multiply $λ$ by $1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + (- 1 + λ) (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.7

Expand $(- 1 + λ) (- 6 λ + 72)$ using the FOIL Method.

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Step 5.4.5.1.7.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 - 1 (- 6 λ + 72) + λ (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.7.2

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 - 1 (- 6 λ) - 1 \cdot 72 + λ (- 6 λ + 72) + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.7.3

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 - 1 (- 6 λ) - 1 \cdot 72 + λ (- 6 λ) + λ \cdot 72 + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.8

Simplify and combine like terms.

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Step 5.4.5.1.8.1

Simplify each term.

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Step 5.4.5.1.8.1.1

Multiply $- 6$ by $- 1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + 6 λ - 1 \cdot 72 + λ (- 6 λ) + λ \cdot 72 + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.8.1.2

Multiply $- 1$ by $72$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + 6 λ - 72 + λ (- 6 λ) + λ \cdot 72 + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.8.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + 6 λ - 72 - 6 λ \cdot λ + λ \cdot 72 + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.8.1.4

Multiply $λ$ by $λ$ by adding the exponents.

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Step 5.4.5.1.8.1.4.1

Move $λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + 6 λ - 72 - 6 (λ \cdot λ) + λ \cdot 72 + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.8.1.4.2

Multiply $λ$ by $λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + 6 λ - 72 - 6 λ^{2} + λ \cdot 72 + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.8.1.5

Move $72$ to the left of $λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + 6 λ - 72 - 6 λ^{2} + 72 λ + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.8.2

Add $6 λ$ and $72 λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + 78 λ - 72 - 6 λ^{2} + 9 \cdot - 39) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.1.9

Multiply $9$ by $- 39$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 60 λ + 408 + 78 λ - 72 - 6 λ^{2} - 351) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.2

Add $- 60 λ$ and $78 λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (18 λ + 408 - 72 - 6 λ^{2} - 351) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.3

Subtract $72$ from $408$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (18 λ + 336 - 6 λ^{2} - 351) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.4

Subtract $351$ from $336$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (18 λ - 6 λ^{2} - 15) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.4.5.5

Reorder $18 λ$ and $- 6 λ^{2}$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 | \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$

Step 5.5

Evaluate $| \begin{array}{c} 12 & 1 - λ & 4 \\ 6 & 5 & - 2 - λ \\ 3 & - 4 & 5 \end{array} |$ .

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Step 5.5.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $1$ by its cofactor and add.

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Step 5.5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 5.5.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 5.5.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |$

Step 5.5.1.4

Multiply element $a_{11}$ by its cofactor.

$12 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |$

Step 5.5.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} |$

Step 5.5.1.6

Multiply element $a_{12}$ by its cofactor.

$- (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} |$

Step 5.5.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

$| \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |$

Step 5.5.1.8

Multiply element $a_{13}$ by its cofactor.

$4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |$

Step 5.5.1.9

Add the terms together.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 | \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} | - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2

Evaluate $| \begin{array}{c} 5 & - 2 - λ \\ - 4 & 5 \end{array} |$ .

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Step 5.5.2.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (5 \cdot 5 - (- 4 (- 2 - λ))) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2.2

Simplify the determinant.

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Step 5.5.2.2.1

Simplify each term.

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Step 5.5.2.2.1.1

Multiply $5$ by $5$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (25 - (- 4 (- 2 - λ))) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2.2.1.2

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (25 - (- 4 \cdot - 2 - 4 (- λ))) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2.2.1.3

Multiply $- 4$ by $- 2$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (25 - (8 - 4 (- λ))) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2.2.1.4

Multiply $- 1$ by $- 4$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (25 - (8 + 4 λ)) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2.2.1.5

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (25 - 1 \cdot 8 - (4 λ)) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2.2.1.6

Multiply $- 1$ by $8$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (25 - 8 - (4 λ)) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2.2.1.7

Multiply $4$ by $- 1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (25 - 8 - 4 λ) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2.2.2

Subtract $8$ from $25$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (17 - 4 λ) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.2.2.3

Reorder $17$ and $- 4 λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) | \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} | + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.3

Evaluate $| \begin{array}{c} 6 & - 2 - λ \\ 3 & 5 \end{array} |$ .

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Step 5.5.3.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (6 \cdot 5 - 3 (- 2 - λ)) + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.3.2

Simplify the determinant.

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Step 5.5.3.2.1

Simplify each term.

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Step 5.5.3.2.1.1

Multiply $6$ by $5$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (30 - 3 (- 2 - λ)) + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.3.2.1.2

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (30 - 3 \cdot - 2 - 3 (- λ)) + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.3.2.1.3

Multiply $- 3$ by $- 2$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (30 + 6 - 3 (- λ)) + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.3.2.1.4

Multiply $- 1$ by $- 3$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (30 + 6 + 3 λ) + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.3.2.2

Add $30$ and $6$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (36 + 3 λ) + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.3.2.3

Reorder $36$ and $3 λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (3 λ + 36) + 4 | \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |)$

Step 5.5.4

Evaluate $| \begin{array}{c} 6 & 5 \\ 3 & - 4 \end{array} |$ .

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Step 5.5.4.1

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (3 λ + 36) + 4 (6 \cdot - 4 - 3 \cdot 5))$

Step 5.5.4.2

Simplify the determinant.

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Step 5.5.4.2.1

Simplify each term.

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Step 5.5.4.2.1.1

Multiply $6$ by $- 4$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (3 λ + 36) + 4 (- 24 - 3 \cdot 5))$

Step 5.5.4.2.1.2

Multiply $- 3$ by $5$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (3 λ + 36) + 4 (- 24 - 15))$

Step 5.5.4.2.2

Subtract $15$ from $- 24$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ + 17) - (1 - λ) (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5

Simplify the determinant.

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Step 5.5.5.1

Simplify each term.

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Step 5.5.5.1.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (12 (- 4 λ) + 12 \cdot 17 - (1 - λ) (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5.1.2

Multiply $- 4$ by $12$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 12 \cdot 17 - (1 - λ) (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5.1.3

Multiply $12$ by $17$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - (1 - λ) (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5.1.4

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 + (- 1 \cdot 1 - - λ) (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5.1.5

Multiply $- 1$ by $1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 + (- 1 - - λ) (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5.1.6

Multiply $- - λ$ .

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Step 5.5.5.1.6.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 + (- 1 + 1 λ) (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5.1.6.2

Multiply $λ$ by $1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 + (- 1 + λ) (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5.1.7

Expand $(- 1 + λ) (3 λ + 36)$ using the FOIL Method.

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Step 5.5.5.1.7.1

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - 1 (3 λ + 36) + λ (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5.1.7.2

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - 1 (3 λ) - 1 \cdot 36 + λ (3 λ + 36) + 4 \cdot - 39)$

Step 5.5.5.1.7.3

Apply the distributive property.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - 1 (3 λ) - 1 \cdot 36 + λ (3 λ) + λ \cdot 36 + 4 \cdot - 39)$

Step 5.5.5.1.8

Simplify and combine like terms.

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Step 5.5.5.1.8.1

Simplify each term.

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Step 5.5.5.1.8.1.1

Multiply $3$ by $- 1$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - 3 λ - 1 \cdot 36 + λ (3 λ) + λ \cdot 36 + 4 \cdot - 39)$

Step 5.5.5.1.8.1.2

Multiply $- 1$ by $36$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - 3 λ - 36 + λ (3 λ) + λ \cdot 36 + 4 \cdot - 39)$

Step 5.5.5.1.8.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - 3 λ - 36 + 3 λ \cdot λ + λ \cdot 36 + 4 \cdot - 39)$

Step 5.5.5.1.8.1.4

Multiply $λ$ by $λ$ by adding the exponents.

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Step 5.5.5.1.8.1.4.1

Move $λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - 3 λ - 36 + 3 (λ \cdot λ) + λ \cdot 36 + 4 \cdot - 39)$

Step 5.5.5.1.8.1.4.2

Multiply $λ$ by $λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - 3 λ - 36 + 3 λ^{2} + λ \cdot 36 + 4 \cdot - 39)$

Step 5.5.5.1.8.1.5

Move $36$ to the left of $λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 - 3 λ - 36 + 3 λ^{2} + 36 λ + 4 \cdot - 39)$

Step 5.5.5.1.8.2

Add $- 3 λ$ and $36 λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 + 33 λ - 36 + 3 λ^{2} + 4 \cdot - 39)$

Step 5.5.5.1.9

Multiply $4$ by $- 39$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 48 λ + 204 + 33 λ - 36 + 3 λ^{2} - 156)$

Step 5.5.5.2

Add $- 48 λ$ and $33 λ$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 15 λ + 204 - 36 + 3 λ^{2} - 156)$

Step 5.5.5.3

Subtract $36$ from $204$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 15 λ + 168 + 3 λ^{2} - 156)$

Step 5.5.5.4

Subtract $156$ from $168$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (- 15 λ + 3 λ^{2} + 12)$

Step 5.5.5.5

Reorder $- 15 λ$ and $3 λ^{2}$ .

$p (λ) = (- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17) - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6

Simplify the determinant.

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Step 5.6.1

Simplify each term.

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Step 5.6.1.1

Expand $(- 2 - λ) (- λ^{3} + 9 λ^{2} - 24 λ + 17)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = - 2 (- λ^{3}) - 2 (9 λ^{2}) - 2 (- 24 λ) - 2 \cdot 17 - λ (- λ^{3}) - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2

Simplify each term.

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Step 5.6.1.2.1

Multiply $- 1$ by $- 2$ .

$p (λ) = 2 λ^{3} - 2 (9 λ^{2}) - 2 (- 24 λ) - 2 \cdot 17 - λ (- λ^{3}) - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.2

Multiply $9$ by $- 2$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} - 2 (- 24 λ) - 2 \cdot 17 - λ (- λ^{3}) - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.3

Multiply $- 24$ by $- 2$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 2 \cdot 17 - λ (- λ^{3}) - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.4

Multiply $- 2$ by $17$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 - λ (- λ^{3}) - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.5

Rewrite using the commutative property of multiplication.

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 - 1 \cdot - 1 λ \cdot λ^{3} - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.6

Multiply $λ$ by $λ^{3}$ by adding the exponents.

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Step 5.6.1.2.6.1

Move $λ^{3}$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 - 1 \cdot - 1 (λ^{3} λ) - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.6.2

Multiply $λ^{3}$ by $λ$ .

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Step 5.6.1.2.6.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 - 1 \cdot - 1 (λ^{3} λ^{1}) - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.6.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 - 1 \cdot - 1 λ^{3 + 1} - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.6.3

Add $3$ and $1$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 - 1 \cdot - 1 λ^{4} - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.7

Multiply $- 1$ by $- 1$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + 1 λ^{4} - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.8

Multiply $λ^{4}$ by $1$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - λ (9 λ^{2}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.9

Rewrite using the commutative property of multiplication.

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 1 \cdot 9 λ \cdot λ^{2} - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.10

Multiply $λ$ by $λ^{2}$ by adding the exponents.

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Step 5.6.1.2.10.1

Move $λ^{2}$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 1 \cdot 9 (λ^{2} λ) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.10.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.6.1.2.10.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 1 \cdot 9 (λ^{2} λ^{1}) - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.10.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 1 \cdot 9 λ^{2 + 1} - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.10.3

Add $2$ and $1$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 1 \cdot 9 λ^{3} - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.11

Multiply $- 1$ by $9$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 9 λ^{3} - λ (- 24 λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.12

Rewrite using the commutative property of multiplication.

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 9 λ^{3} - 1 \cdot - 24 λ \cdot λ - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.13

Multiply $λ$ by $λ$ by adding the exponents.

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Step 5.6.1.2.13.1

Move $λ$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 9 λ^{3} - 1 \cdot - 24 (λ \cdot λ) - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.13.2

Multiply $λ$ by $λ$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 9 λ^{3} - 1 \cdot - 24 λ^{2} - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.14

Multiply $- 1$ by $- 24$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 9 λ^{3} + 24 λ^{2} - λ \cdot 17 - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.2.15

Multiply $17$ by $- 1$ .

$p (λ) = 2 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} - 9 λ^{3} + 24 λ^{2} - 17 λ - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.3

Subtract $9 λ^{3}$ from $2 λ^{3}$ .

$p (λ) = - 7 λ^{3} - 18 λ^{2} + 48 λ - 34 + λ^{4} + 24 λ^{2} - 17 λ - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.4

Add $- 18 λ^{2}$ and $24 λ^{2}$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 48 λ - 34 + λ^{4} - 17 λ - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.5

Subtract $17 λ$ from $48 λ$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 1 (12 λ^{2} - 45 λ + 36) - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.6

Apply the distributive property.

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 1 (12 λ^{2}) - 1 (- 45 λ) - 1 \cdot 36 - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.7

Simplify.

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Step 5.6.1.7.1

Multiply $12$ by $- 1$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} - 1 (- 45 λ) - 1 \cdot 36 - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.7.2

Multiply $- 45$ by $- 1$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 1 \cdot 36 - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.7.3

Multiply $- 1$ by $36$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 - 2 (- 6 λ^{2} + 18 λ - 15) + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.8

Apply the distributive property.

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 - 2 (- 6 λ^{2}) - 2 (18 λ) - 2 \cdot - 15 + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.9

Simplify.

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Step 5.6.1.9.1

Multiply $- 6$ by $- 2$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 + 12 λ^{2} - 2 (18 λ) - 2 \cdot - 15 + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.9.2

Multiply $18$ by $- 2$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 + 12 λ^{2} - 36 λ - 2 \cdot - 15 + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.9.3

Multiply $- 2$ by $- 15$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 + 12 λ^{2} - 36 λ + 30 + 4 (3 λ^{2} - 15 λ + 12)$

Step 5.6.1.10

Apply the distributive property.

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 + 12 λ^{2} - 36 λ + 30 + 4 (3 λ^{2}) + 4 (- 15 λ) + 4 \cdot 12$

Step 5.6.1.11

Simplify.

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Step 5.6.1.11.1

Multiply $3$ by $4$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 + 12 λ^{2} - 36 λ + 30 + 12 λ^{2} + 4 (- 15 λ) + 4 \cdot 12$

Step 5.6.1.11.2

Multiply $- 15$ by $4$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 + 12 λ^{2} - 36 λ + 30 + 12 λ^{2} - 60 λ + 4 \cdot 12$

Step 5.6.1.11.3

Multiply $4$ by $12$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 + 12 λ^{2} - 36 λ + 30 + 12 λ^{2} - 60 λ + 48$

Step 5.6.2

Combine the opposite terms in $- 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} - 12 λ^{2} + 45 λ - 36 + 12 λ^{2} - 36 λ + 30 + 12 λ^{2} - 60 λ + 48$ .

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Step 5.6.2.1

Add $- 12 λ^{2}$ and $12 λ^{2}$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} + 45 λ - 36 + 0 - 36 λ + 30 + 12 λ^{2} - 60 λ + 48$

Step 5.6.2.2

Add $- 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} + 45 λ - 36$ and $0$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} + 31 λ - 34 + λ^{4} + 45 λ - 36 - 36 λ + 30 + 12 λ^{2} - 60 λ + 48$

Step 5.6.3

Add $6 λ^{2}$ and $12 λ^{2}$ .

$p (λ) = - 7 λ^{3} + 18 λ^{2} + 31 λ - 34 + λ^{4} + 45 λ - 36 - 36 λ + 30 - 60 λ + 48$

Step 5.6.4

Add $31 λ$ and $45 λ$ .

$p (λ) = - 7 λ^{3} + 18 λ^{2} + 76 λ - 34 + λ^{4} - 36 - 36 λ + 30 - 60 λ + 48$

Step 5.6.5

Subtract $36 λ$ from $76 λ$ .

$p (λ) = - 7 λ^{3} + 18 λ^{2} + 40 λ - 34 + λ^{4} - 36 + 30 - 60 λ + 48$

Step 5.6.6

Subtract $60 λ$ from $40 λ$ .

$p (λ) = - 7 λ^{3} + 18 λ^{2} - 20 λ - 34 + λ^{4} - 36 + 30 + 48$

Step 5.6.7

Subtract $36$ from $- 34$ .

$p (λ) = - 7 λ^{3} + 18 λ^{2} - 20 λ + λ^{4} - 70 + 30 + 48$

Step 5.6.8

Add $- 70$ and $30$ .

$p (λ) = - 7 λ^{3} + 18 λ^{2} - 20 λ + λ^{4} - 40 + 48$

Step 5.6.9

Add $- 40$ and $48$ .

$p (λ) = - 7 λ^{3} + 18 λ^{2} - 20 λ + λ^{4} + 8$

Step 5.6.10

Move $- 20 λ$ .

$p (λ) = - 7 λ^{3} + 18 λ^{2} + λ^{4} - 20 λ + 8$

Step 5.6.11

Move $18 λ^{2}$ .

$p (λ) = - 7 λ^{3} + λ^{4} + 18 λ^{2} - 20 λ + 8$

Step 5.6.12

Reorder $- 7 λ^{3}$ and $λ^{4}$ .

$p (λ) = λ^{4} - 7 λ^{3} + 18 λ^{2} - 20 λ + 8$

Step 6

Set the characteristic polynomial equal to $0$ to find the eigenvalues $λ$ .

$λ^{4} - 7 λ^{3} + 18 λ^{2} - 20 λ + 8 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Factor the left side of the equation.

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Step 7.1.1

Factor $λ^{4} - 7 λ^{3} + 18 λ^{2} - 20 λ + 8$ using the rational roots test.

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Step 7.1.1.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 8, \pm 2, \pm 4$

$q = \pm 1$

Step 7.1.1.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 8, \pm 2, \pm 4$

Step 7.1.1.3

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

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Step 7.1.1.3.1

Substitute $1$ into the polynomial.

$1^{4} - 7 \cdot 1^{3} + 18 \cdot 1^{2} - 20 \cdot 1 + 8$

Step 7.1.1.3.2

Raise $1$ to the power of $4$ .

$1 - 7 \cdot 1^{3} + 18 \cdot 1^{2} - 20 \cdot 1 + 8$

Step 7.1.1.3.3

Raise $1$ to the power of $3$ .

$1 - 7 \cdot 1 + 18 \cdot 1^{2} - 20 \cdot 1 + 8$

Step 7.1.1.3.4

Multiply $- 7$ by $1$ .

$1 - 7 + 18 \cdot 1^{2} - 20 \cdot 1 + 8$

Step 7.1.1.3.5

Subtract $7$ from $1$ .

$- 6 + 18 \cdot 1^{2} - 20 \cdot 1 + 8$

Step 7.1.1.3.6

Raise $1$ to the power of $2$ .

$- 6 + 18 \cdot 1 - 20 \cdot 1 + 8$

Step 7.1.1.3.7

Multiply $18$ by $1$ .

$- 6 + 18 - 20 \cdot 1 + 8$

Step 7.1.1.3.8

Add $- 6$ and $18$ .

$12 - 20 \cdot 1 + 8$

Step 7.1.1.3.9

Multiply $- 20$ by $1$ .

$12 - 20 + 8$

Step 7.1.1.3.10

Subtract $20$ from $12$ .

$- 8 + 8$

Step 7.1.1.3.11

Add $- 8$ and $8$ .

$0$

Step 7.1.1.4

Since $1$ is a known root, divide the polynomial by $λ - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{λ^{4} - 7 λ^{3} + 18 λ^{2} - 20 λ + 8}{λ - 1}$

Step 7.1.1.5

Divide $λ^{4} - 7 λ^{3} + 18 λ^{2} - 20 λ + 8$ by $λ - 1$ .

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Step 7.1.1.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

Step 7.1.1.5.2

Divide the highest order term in the dividend $λ^{4}$ by the highest order term in divisor $λ$ .

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

Step 7.1.1.5.3

Multiply the new quotient term by the divisor.

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

Step 7.1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $λ^{4} - λ^{3}$

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

Step 7.1.1.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

Step 7.1.1.5.6

Pull the next terms from the original dividend down into the current dividend.

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

Step 7.1.1.5.7

Divide the highest order term in the dividend $- 6 λ^{3}$ by the highest order term in divisor $λ$ .

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

Step 7.1.1.5.8

Multiply the new quotient term by the divisor.

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

Step 7.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 λ^{3} + 6 λ^{2}$

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

Step 7.1.1.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

Step 7.1.1.5.11

Pull the next terms from the original dividend down into the current dividend.

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

Step 7.1.1.5.12

Divide the highest order term in the dividend $12 λ^{2}$ by the highest order term in divisor $λ$ .

$λ^{3}$

$6 λ^{2}$

$12 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

Step 7.1.1.5.13

Multiply the new quotient term by the divisor.

$λ^{3}$

$6 λ^{2}$

$12 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

$12 λ^{2}$

$12 λ$

Step 7.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $12 λ^{2} - 12 λ$

$λ^{3}$

$6 λ^{2}$

$12 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

$12 λ^{2}$

$12 λ$

Step 7.1.1.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$λ^{3}$

$6 λ^{2}$

$12 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

$12 λ^{2}$

$12 λ$

$8 λ$

Step 7.1.1.5.16

Pull the next terms from the original dividend down into the current dividend.

$λ^{3}$

$6 λ^{2}$

$12 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

$12 λ^{2}$

$12 λ$

$8 λ$

$8$

Step 7.1.1.5.17

Divide the highest order term in the dividend $- 8 λ$ by the highest order term in divisor $λ$ .

$λ^{3}$

$6 λ^{2}$

$12 λ$

$8$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

$12 λ^{2}$

$12 λ$

$8 λ$

$8$

Step 7.1.1.5.18

Multiply the new quotient term by the divisor.

$λ^{3}$

$6 λ^{2}$

$12 λ$

$8$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

$12 λ^{2}$

$12 λ$

$8 λ$

$8$

$8 λ$

$8$

Step 7.1.1.5.19

The expression needs to be subtracted from the dividend, so change all the signs in $- 8 λ + 8$

$λ^{3}$

$6 λ^{2}$

$12 λ$

$8$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

$12 λ^{2}$

$12 λ$

$8 λ$

$8$

$8 λ$

$8$

Step 7.1.1.5.20

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$λ^{3}$

$6 λ^{2}$

$12 λ$

$8$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$18 λ^{2}$

$20 λ$

$8$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$18 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$12 λ^{2}$

$20 λ$

$12 λ^{2}$

$12 λ$

$8 λ$

$8$

$8 λ$

$8$

$0$

Step 7.1.1.5.21

Since the remander is $0$ , the final answer is the quotient.

$λ^{3} - 6 λ^{2} + 12 λ - 8$

Step 7.1.1.6

Write $λ^{4} - 7 λ^{3} + 18 λ^{2} - 20 λ + 8$ as a set of factors.

$(λ - 1) (λ^{3} - 6 λ^{2} + 12 λ - 8) = 0$

Step 7.1.2

Factor $λ^{3} - 6 λ^{2} + 12 λ - 8$ using the rational roots test.

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Step 7.1.2.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 8, \pm 2, \pm 4$

$q = \pm 1$

Step 7.1.2.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 8, \pm 2, \pm 4$

Step 7.1.2.3

Substitute $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

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Step 7.1.2.3.1

Substitute $2$ into the polynomial.

$2^{3} - 6 \cdot 2^{2} + 12 \cdot 2 - 8$

Step 7.1.2.3.2

Raise $2$ to the power of $3$ .

$8 - 6 \cdot 2^{2} + 12 \cdot 2 - 8$

Step 7.1.2.3.3

Raise $2$ to the power of $2$ .

$8 - 6 \cdot 4 + 12 \cdot 2 - 8$

Step 7.1.2.3.4

Multiply $- 6$ by $4$ .

$8 - 24 + 12 \cdot 2 - 8$

Step 7.1.2.3.5

Subtract $24$ from $8$ .

$- 16 + 12 \cdot 2 - 8$

Step 7.1.2.3.6

Multiply $12$ by $2$ .

$- 16 + 24 - 8$

Step 7.1.2.3.7

Add $- 16$ and $24$ .

$8 - 8$

Step 7.1.2.3.8

Subtract $8$ from $8$ .

$0$

Step 7.1.2.4

Since $2$ is a known root, divide the polynomial by $λ - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{λ^{3} - 6 λ^{2} + 12 λ - 8}{λ - 2}$

Step 7.1.2.5

Divide $λ^{3} - 6 λ^{2} + 12 λ - 8$ by $λ - 2$ .

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Step 7.1.2.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$λ$

$2$

$λ^{3}$

$6 λ^{2}$

$12 λ$

$8$

Step 7.1.2.5.2

Divide the highest order term in the dividend $λ^{3}$ by the highest order term in divisor $λ$ .

					$λ^{2}$
	$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$

Step 7.1.2.5.3

Multiply the new quotient term by the divisor.

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			+	$λ^{3}$	-	$2 λ^{2}$

Step 7.1.2.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $λ^{3} - 2 λ^{2}$

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$

Step 7.1.2.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$

Step 7.1.2.5.6

Pull the next terms from the original dividend down into the current dividend.

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$

Step 7.1.2.5.7

Divide the highest order term in the dividend $- 4 λ^{2}$ by the highest order term in divisor $λ$ .

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$

Step 7.1.2.5.8

Multiply the new quotient term by the divisor.

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$
					-	$4 λ^{2}$	+	$8 λ$

Step 7.1.2.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 λ^{2} + 8 λ$

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$
					+	$4 λ^{2}$	-	$8 λ$

Step 7.1.2.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$4 λ$

Step 7.1.2.5.11

Pull the next terms from the original dividend down into the current dividend.

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$4 λ$	-	$8$

Step 7.1.2.5.12

Divide the highest order term in the dividend $4 λ$ by the highest order term in divisor $λ$ .

				$λ^{2}$	-	$4 λ$	+	$4$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$4 λ$	-	$8$

Step 7.1.2.5.13

Multiply the new quotient term by the divisor.

				$λ^{2}$	-	$4 λ$	+	$4$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$4 λ$	-	$8$
							+	$4 λ$	-	$8$

Step 7.1.2.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $4 λ - 8$

				$λ^{2}$	-	$4 λ$	+	$4$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$4 λ$	-	$8$
							-	$4 λ$	+	$8$

Step 7.1.2.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$λ^{2}$	-	$4 λ$	+	$4$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$12 λ$	-	$8$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$12 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$4 λ$	-	$8$
							-	$4 λ$	+	$8$
										$0$

Step 7.1.2.5.16

Since the remander is $0$ , the final answer is the quotient.

$λ^{2} - 4 λ + 4$

Step 7.1.2.6

Write $λ^{3} - 6 λ^{2} + 12 λ - 8$ as a set of factors.

$(λ - 1) ((λ - 2) (λ^{2} - 4 λ + 4)) = 0$

Step 7.1.3

Factor using the perfect square rule.

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Step 7.1.3.1

Rewrite $4$ as $2^{2}$ .

$(λ - 1) ((λ - 2) (λ^{2} - 4 λ + 2^{2})) = 0$

Step 7.1.3.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 λ = 2 \cdot λ \cdot 2$

Step 7.1.3.3

Rewrite the polynomial.

$(λ - 1) ((λ - 2) (λ^{2} - 2 \cdot λ \cdot 2 + 2^{2})) = 0$

Step 7.1.3.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = λ$ and $b = 2$ .

$(λ - 1) ((λ - 2) {(λ - 2)}^{2}) = 0$

Step 7.1.4

Combine like factors.

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Step 7.1.4.1

Raise $λ - 2$ to the power of $1$ .

$(λ - 1) ((λ - 2) {(λ - 2)}^{2}) = 0$

Step 7.1.4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$(λ - 1) {(λ - 2)}^{1 + 2} = 0$

Step 7.1.4.3

Add $1$ and $2$ .

$(λ - 1) {(λ - 2)}^{3} = 0$

Step 7.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$λ - 1 = 0$

${(λ - 2)}^{3} = 0$

Step 7.3

Set $λ - 1$ equal to $0$ and solve for $λ$ .

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Step 7.3.1

Set $λ - 1$ equal to $0$ .

$λ - 1 = 0$

Step 7.3.2

Add $1$ to both sides of the equation.

$λ = 1$

Step 7.4

Set ${(λ - 2)}^{3}$ equal to $0$ and solve for $λ$ .

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Step 7.4.1

Set ${(λ - 2)}^{3}$ equal to $0$ .

${(λ - 2)}^{3} = 0$

Step 7.4.2

Solve ${(λ - 2)}^{3} = 0$ for $λ$ .

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Step 7.4.2.1

Set the $λ - 2$ equal to $0$ .

$λ - 2 = 0$

Step 7.4.2.2

Add $2$ to both sides of the equation.

$λ = 2$

Step 7.5

The final solution is all the values that make $(λ - 1) {(λ - 2)}^{3} = 0$ true.

$λ = 1, 2$