Linear Algebra Examples

$[\begin{array}{c} 2 & 3 & - 1 & 0 \\ 1 & 2 & 4 & 3 \\ - 2 & 1 & 3 & 2 \\ - 1 & - 2 & - 3 & 0 \end{array}]$

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 $[\begin{array}{c} 2 & 3 & - 1 & 0 \\ 1 & 2 & 4 & 3 \\ - 2 & 1 & 3 & 2 \\ - 1 & - 2 & - 3 & 0 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 3 & - 1 & 0 \\ 1 & 2 & 4 & 3 \\ - 2 & 1 & 3 & 2 \\ - 1 & - 2 & - 3 & 0 \end{array}] - λ I_{4})$

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 $I_{4}$ .

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

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.

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.5

Multiply $- λ \cdot 0$ .

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Step 4.1.2.5.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.5.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.6

Multiply $- 1$ by $1$ .

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

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.9

Multiply $- λ \cdot 0$ .

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Step 4.1.2.9.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.9.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.10

Multiply $- λ \cdot 0$ .

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Step 4.1.2.10.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.10.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.11

Multiply $- 1$ by $1$ .

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

Step 4.1.2.12

Multiply $- λ \cdot 0$ .

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Step 4.1.2.12.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.12.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.13

Multiply $- λ \cdot 0$ .

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Step 4.1.2.13.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.13.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.14

Multiply $- λ \cdot 0$ .

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Step 4.1.2.14.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.14.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.15

Multiply $- λ \cdot 0$ .

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Step 4.1.2.15.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.15.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.16

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add $3$ and $0$ .

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

Step 4.3.2

Add $- 1$ and $0$ .

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

Step 4.3.3

Add $0$ and $0$ .

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

Step 4.3.4

Add $1$ and $0$ .

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

Step 4.3.5

Add $4$ and $0$ .

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

Step 4.3.6

Add $3$ and $0$ .

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

Step 4.3.7

Add $- 2$ and $0$ .

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

Step 4.3.8

Add $1$ and $0$ .

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

Step 4.3.9

Add $2$ and $0$ .

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

Step 4.3.10

Add $- 1$ and $0$ .

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

Step 4.3.11

Add $- 2$ and $0$ .

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

Step 4.3.12

Add $- 3$ and $0$ .

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

Step 4.3.13

Subtract $λ$ from $0$ .

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

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.

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

Step 5.1.4

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

$(2 - λ) | \begin{array}{c} 2 - λ & 4 & 3 \\ 1 & 3 - λ & 2 \\ - 2 & - 3 & - λ \end{array} |$

Step 5.1.5

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

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

Step 5.1.6

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

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

Step 5.1.7

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

$| \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} |$

Step 5.1.8

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

$- 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} |$

Step 5.1.9

The minor for $a_{14}$ is the determinant with row $1$ and column $4$ deleted.

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

Step 5.1.10

Multiply element $a_{14}$ by its cofactor.

$0 | \begin{array}{c} 1 & 2 - λ & 4 \\ - 2 & 1 & 3 - λ \\ - 1 & - 2 & - 3 \end{array} |$

Step 5.1.11

Add the terms together.

$p (λ) = (2 - λ) | \begin{array}{c} 2 - λ & 4 & 3 \\ 1 & 3 - λ & 2 \\ - 2 & - 3 & - λ \end{array} | - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0 | \begin{array}{c} 1 & 2 - λ & 4 \\ - 2 & 1 & 3 - λ \\ - 1 & - 2 & - 3 \end{array} |$

Step 5.2

Multiply $0$ by $| \begin{array}{c} 1 & 2 - λ & 4 \\ - 2 & 1 & 3 - λ \\ - 1 & - 2 & - 3 \end{array} |$ .

Step 5.3

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

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

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

Step 5.3.1.4

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

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

Step 5.3.1.5

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

$| \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} |$

Step 5.3.1.6

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

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

Step 5.3.1.7

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

$| \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |$

Step 5.3.1.8

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

$3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |$

Step 5.3.1.9

Add the terms together.

$p (λ) = (2 - λ) ((2 - λ) | \begin{array}{c} 3 - λ & 2 \\ - 3 & - λ \end{array} | - 4 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | + 3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.2

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

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Step 5.3.2.1

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

$p (λ) = (2 - λ) ((2 - λ) ((3 - λ) (- λ) - (- 3 \cdot 2)) - 4 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | + 3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

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.

$p (λ) = (2 - λ) ((2 - λ) (3 (- λ) - λ (- λ) - (- 3 \cdot 2)) - 4 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | + 3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.2.2.1.2

Multiply $- 1$ by $3$ .

$p (λ) = (2 - λ) ((2 - λ) (- 3 λ - λ (- λ) - (- 3 \cdot 2)) - 4 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | + 3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.2.2.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (2 - λ) ((2 - λ) (- 3 λ - 1 \cdot - 1 λ \cdot λ - (- 3 \cdot 2)) - 4 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | + 3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.2.2.1.4

Simplify each term.

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Step 5.3.2.2.1.4.1

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

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Step 5.3.2.2.1.4.1.1

Move $λ$ .

$p (λ) = (2 - λ) ((2 - λ) (- 3 λ - 1 \cdot - 1 (λ \cdot λ) - (- 3 \cdot 2)) - 4 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | + 3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.2.2.1.4.1.2

Multiply $λ$ by $λ$ .

$p (λ) = (2 - λ) ((2 - λ) (- 3 λ - 1 \cdot - 1 λ^{2} - (- 3 \cdot 2)) - 4 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | + 3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.2.2.1.4.2

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) ((2 - λ) (- 3 λ + 1 λ^{2} - (- 3 \cdot 2)) - 4 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | + 3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.2.2.1.4.3

Multiply $λ^{2}$ by $1$ .

$p (λ) = (2 - λ) ((2 - λ) (- 3 λ + λ^{2} - (- 3 \cdot 2)) - 4 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | + 3 | \begin{array}{c} 1 & 3 - λ \\ - 2 & - 3 \end{array} |) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.2.2.1.5

Multiply $- (- 3 \cdot 2)$ .

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Step 5.3.2.2.1.5.1

Multiply $- 3$ by $2$ .

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

Step 5.3.2.2.1.5.2

Multiply $- 1$ by $- 6$ .

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

Step 5.3.2.2.2

Reorder $- 3 λ$ and $λ^{2}$ .

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

Step 5.3.3

Evaluate $| \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} |$ .

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Step 5.3.3.1

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

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

Step 5.3.3.2

Simplify each term.

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Step 5.3.3.2.1

Multiply $- λ$ by $1$ .

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

Step 5.3.3.2.2

Multiply $- (- 2 \cdot 2)$ .

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Step 5.3.3.2.2.1

Multiply $- 2$ by $2$ .

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

Step 5.3.3.2.2.2

Multiply $- 1$ by $- 4$ .

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

Step 5.3.4

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

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Step 5.3.4.1

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

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

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 $- 3$ by $1$ .

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

Step 5.3.4.2.1.2

Apply the distributive property.

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

Step 5.3.4.2.1.3

Multiply $- 2$ by $3$ .

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

Step 5.3.4.2.1.4

Multiply $- 1$ by $- 2$ .

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

Step 5.3.4.2.1.5

Apply the distributive property.

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

Step 5.3.4.2.1.6

Multiply $- 1$ by $- 6$ .

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

Step 5.3.4.2.1.7

Multiply $2$ by $- 1$ .

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

Step 5.3.4.2.2

Add $- 3$ and $6$ .

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

Step 5.3.4.2.3

Reorder $3$ and $- 2 λ$ .

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

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

Expand $(2 - λ) (λ^{2} - 3 λ + 6)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 5.3.5.1.2

Simplify each term.

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Step 5.3.5.1.2.1

Multiply $- 3$ by $2$ .

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

Step 5.3.5.1.2.2

Multiply $2$ by $6$ .

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

Step 5.3.5.1.2.3

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

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Step 5.3.5.1.2.3.1

Move $λ^{2}$ .

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

Step 5.3.5.1.2.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.3.5.1.2.3.2.1

Raise $λ$ to the power of $1$ .

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

Step 5.3.5.1.2.3.2.2

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

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

Step 5.3.5.1.2.3.3

Add $2$ and $1$ .

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

Step 5.3.5.1.2.4

Rewrite using the commutative property of multiplication.

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

Step 5.3.5.1.2.5

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

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Step 5.3.5.1.2.5.1

Move $λ$ .

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

Step 5.3.5.1.2.5.2

Multiply $λ$ by $λ$ .

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

Step 5.3.5.1.2.6

Multiply $- 1$ by $- 3$ .

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

Step 5.3.5.1.2.7

Multiply $6$ by $- 1$ .

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

Step 5.3.5.1.3

Add $2 λ^{2}$ and $3 λ^{2}$ .

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

Step 5.3.5.1.4

Subtract $6 λ$ from $- 6 λ$ .

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

Step 5.3.5.1.5

Apply the distributive property.

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

Step 5.3.5.1.6

Multiply $- 1$ by $- 4$ .

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

Step 5.3.5.1.7

Multiply $- 4$ by $4$ .

$p (λ) = (2 - λ) (5 λ^{2} - 12 λ + 12 - λ^{3} + 4 λ - 16 + 3 (- 2 λ + 3)) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.5.1.8

Apply the distributive property.

$p (λ) = (2 - λ) (5 λ^{2} - 12 λ + 12 - λ^{3} + 4 λ - 16 + 3 (- 2 λ) + 3 \cdot 3) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.5.1.9

Multiply $- 2$ by $3$ .

$p (λ) = (2 - λ) (5 λ^{2} - 12 λ + 12 - λ^{3} + 4 λ - 16 - 6 λ + 3 \cdot 3) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.5.1.10

Multiply $3$ by $3$ .

$p (λ) = (2 - λ) (5 λ^{2} - 12 λ + 12 - λ^{3} + 4 λ - 16 - 6 λ + 9) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.5.2

Add $- 12 λ$ and $4 λ$ .

$p (λ) = (2 - λ) (5 λ^{2} - 8 λ + 12 - λ^{3} - 16 - 6 λ + 9) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.5.3

Subtract $6 λ$ from $- 8 λ$ .

$p (λ) = (2 - λ) (5 λ^{2} - 14 λ + 12 - λ^{3} - 16 + 9) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.5.4

Subtract $16$ from $12$ .

$p (λ) = (2 - λ) (5 λ^{2} - 14 λ - λ^{3} - 4 + 9) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.5.5

Add $- 4$ and $9$ .

$p (λ) = (2 - λ) (5 λ^{2} - 14 λ - λ^{3} + 5) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.5.6

Move $- 14 λ$ .

$p (λ) = (2 - λ) (5 λ^{2} - λ^{3} - 14 λ + 5) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.3.5.7

Reorder $5 λ^{2}$ and $- λ^{3}$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 | \begin{array}{c} 1 & 4 & 3 \\ - 2 & 3 - λ & 2 \\ - 1 & - 3 & - λ \end{array} | - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4

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

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

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

Step 5.4.1.4

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

$2 | \begin{array}{c} 4 & 3 \\ - 3 & - λ \end{array} |$

Step 5.4.1.5

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

$| \begin{array}{c} 1 & 3 \\ - 1 & - λ \end{array} |$

Step 5.4.1.6

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

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

Step 5.4.1.7

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

$| \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |$

Step 5.4.1.8

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

$- 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |$

Step 5.4.1.9

Add the terms together.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 | \begin{array}{c} 4 & 3 \\ - 3 & - λ \end{array} | + (3 - λ) | \begin{array}{c} 1 & 3 \\ - 1 & - λ \end{array} | - 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.2

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

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Step 5.4.2.1

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

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (4 (- λ) - (- 3 \cdot 3)) + (3 - λ) | \begin{array}{c} 1 & 3 \\ - 1 & - λ \end{array} | - 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.2.2

Simplify each term.

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Step 5.4.2.2.1

Multiply $- 1$ by $4$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ - (- 3 \cdot 3)) + (3 - λ) | \begin{array}{c} 1 & 3 \\ - 1 & - λ \end{array} | - 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.2.2.2

Multiply $- (- 3 \cdot 3)$ .

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Step 5.4.2.2.2.1

Multiply $- 3$ by $3$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ - - 9) + (3 - λ) | \begin{array}{c} 1 & 3 \\ - 1 & - λ \end{array} | - 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.2.2.2.2

Multiply $- 1$ by $- 9$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) | \begin{array}{c} 1 & 3 \\ - 1 & - λ \end{array} | - 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.3

Evaluate $| \begin{array}{c} 1 & 3 \\ - 1 & - λ \end{array} |$ .

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Step 5.4.3.1

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

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) (1 (- λ) - (- 1 \cdot 3)) - 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.3.2

Simplify each term.

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Step 5.4.3.2.1

Multiply $- λ$ by $1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) (- λ - (- 1 \cdot 3)) - 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.3.2.2

Multiply $- (- 1 \cdot 3)$ .

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Step 5.4.3.2.2.1

Multiply $- 1$ by $3$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) (- λ - - 3) - 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.3.2.2.2

Multiply $- 1$ by $- 3$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) (- λ + 3) - 2 | \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.4

Evaluate $| \begin{array}{c} 1 & 4 \\ - 1 & - 3 \end{array} |$ .

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Step 5.4.4.1

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

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) (- λ + 3) - 2 (1 \cdot - 3 - (- 1 \cdot 4))) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

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 $- 3$ by $1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) (- λ + 3) - 2 (- 3 - (- 1 \cdot 4))) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.4.2.1.2

Multiply $- (- 1 \cdot 4)$ .

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Step 5.4.4.2.1.2.1

Multiply $- 1$ by $4$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) (- λ + 3) - 2 (- 3 - - 4)) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.4.2.1.2.2

Multiply $- 1$ by $- 4$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) (- λ + 3) - 2 (- 3 + 4)) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.4.2.2

Add $- 3$ and $4$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ + 9) + (3 - λ) (- λ + 3) - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

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.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (2 (- 4 λ) + 2 \cdot 9 + (3 - λ) (- λ + 3) - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.2

Multiply $- 4$ by $2$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 2 \cdot 9 + (3 - λ) (- λ + 3) - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.3

Multiply $2$ by $9$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 + (3 - λ) (- λ + 3) - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.4

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

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Step 5.4.5.1.4.1

Apply the distributive property.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 + 3 (- λ + 3) - λ (- λ + 3) - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.4.2

Apply the distributive property.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 + 3 (- λ) + 3 \cdot 3 - λ (- λ + 3) - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.4.3

Apply the distributive property.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 + 3 (- λ) + 3 \cdot 3 - λ (- λ) - λ \cdot 3 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.5

Simplify and combine like terms.

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Step 5.4.5.1.5.1

Simplify each term.

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Step 5.4.5.1.5.1.1

Multiply $- 1$ by $3$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 3 λ + 3 \cdot 3 - λ (- λ) - λ \cdot 3 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.5.1.2

Multiply $3$ by $3$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 3 λ + 9 - λ (- λ) - λ \cdot 3 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.5.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 3 λ + 9 - 1 \cdot - 1 λ \cdot λ - λ \cdot 3 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.5.1.4

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

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Step 5.4.5.1.5.1.4.1

Move $λ$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 3 λ + 9 - 1 \cdot - 1 (λ \cdot λ) - λ \cdot 3 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.5.1.4.2

Multiply $λ$ by $λ$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 3 λ + 9 - 1 \cdot - 1 λ^{2} - λ \cdot 3 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.5.1.5

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 3 λ + 9 + 1 λ^{2} - λ \cdot 3 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.5.1.6

Multiply $λ^{2}$ by $1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 3 λ + 9 + λ^{2} - λ \cdot 3 - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.5.1.7

Multiply $3$ by $- 1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 3 λ + 9 + λ^{2} - 3 λ - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.5.2

Subtract $3 λ$ from $- 3 λ$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 6 λ + 9 + λ^{2} - 2 \cdot 1) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.1.6

Multiply $- 2$ by $1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 8 λ + 18 - 6 λ + 9 + λ^{2} - 2) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.2

Subtract $6 λ$ from $- 8 λ$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 14 λ + 18 + 9 + λ^{2} - 2) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.3

Add $18$ and $9$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 14 λ + 27 + λ^{2} - 2) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.4

Subtract $2$ from $27$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (- 14 λ + λ^{2} + 25) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.4.5.5

Reorder $- 14 λ$ and $λ^{2}$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 | \begin{array}{c} 1 & 2 - λ & 3 \\ - 2 & 1 & 2 \\ - 1 & - 2 & - λ \end{array} | + 0$

Step 5.5

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

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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.

$| \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} |$

Step 5.5.1.4

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

$1 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} |$

Step 5.5.1.5

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

$| \begin{array}{c} - 2 & 2 \\ - 1 & - λ \end{array} |$

Step 5.5.1.6

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

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

Step 5.5.1.7

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

$| \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |$

Step 5.5.1.8

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

$3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |$

Step 5.5.1.9

Add the terms together.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 | \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} | - (2 - λ) | \begin{array}{c} - 2 & 2 \\ - 1 & - λ \end{array} | + 3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |) + 0$

Step 5.5.2

Evaluate $| \begin{array}{c} 1 & 2 \\ - 2 & - λ \end{array} |$ .

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Step 5.5.2.1

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

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (1 (- λ) - (- 2 \cdot 2)) - (2 - λ) | \begin{array}{c} - 2 & 2 \\ - 1 & - λ \end{array} | + 3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |) + 0$

Step 5.5.2.2

Simplify each term.

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Step 5.5.2.2.1

Multiply $- λ$ by $1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ - (- 2 \cdot 2)) - (2 - λ) | \begin{array}{c} - 2 & 2 \\ - 1 & - λ \end{array} | + 3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |) + 0$

Step 5.5.2.2.2

Multiply $- (- 2 \cdot 2)$ .

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Step 5.5.2.2.2.1

Multiply $- 2$ by $2$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ - - 4) - (2 - λ) | \begin{array}{c} - 2 & 2 \\ - 1 & - λ \end{array} | + 3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |) + 0$

Step 5.5.2.2.2.2

Multiply $- 1$ by $- 4$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) | \begin{array}{c} - 2 & 2 \\ - 1 & - λ \end{array} | + 3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |) + 0$

Step 5.5.3

Evaluate $| \begin{array}{c} - 2 & 2 \\ - 1 & - λ \end{array} |$ .

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Step 5.5.3.1

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

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) (- 2 (- λ) - (- 1 \cdot 2)) + 3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |) + 0$

Step 5.5.3.2

Simplify each term.

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Step 5.5.3.2.1

Multiply $- 1$ by $- 2$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) (2 λ - (- 1 \cdot 2)) + 3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |) + 0$

Step 5.5.3.2.2

Multiply $- (- 1 \cdot 2)$ .

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Step 5.5.3.2.2.1

Multiply $- 1$ by $2$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) (2 λ - - 2) + 3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |) + 0$

Step 5.5.3.2.2.2

Multiply $- 1$ by $- 2$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) (2 λ + 2) + 3 | \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |) + 0$

Step 5.5.4

Evaluate $| \begin{array}{c} - 2 & 1 \\ - 1 & - 2 \end{array} |$ .

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Step 5.5.4.1

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

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) (2 λ + 2) + 3 (- 2 \cdot - 2 - (- 1 \cdot 1))) + 0$

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 $- 2$ by $- 2$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) (2 λ + 2) + 3 (4 - (- 1 \cdot 1))) + 0$

Step 5.5.4.2.1.2

Multiply $- (- 1 \cdot 1)$ .

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Step 5.5.4.2.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) (2 λ + 2) + 3 (4 - - 1)) + 0$

Step 5.5.4.2.1.2.2

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) (2 λ + 2) + 3 (4 + 1)) + 0$

Step 5.5.4.2.2

Add $4$ and $1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (1 (- λ + 4) - (2 - λ) (2 λ + 2) + 3 \cdot 5) + 0$

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

Multiply $- λ + 4$ by $1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - (2 - λ) (2 λ + 2) + 3 \cdot 5) + 0$

Step 5.5.5.1.2

Apply the distributive property.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 + (- 1 \cdot 2 - - λ) (2 λ + 2) + 3 \cdot 5) + 0$

Step 5.5.5.1.3

Multiply $- 1$ by $2$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 + (- 2 - - λ) (2 λ + 2) + 3 \cdot 5) + 0$

Step 5.5.5.1.4

Multiply $- - λ$ .

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Step 5.5.5.1.4.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 + (- 2 + 1 λ) (2 λ + 2) + 3 \cdot 5) + 0$

Step 5.5.5.1.4.2

Multiply $λ$ by $1$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 + (- 2 + λ) (2 λ + 2) + 3 \cdot 5) + 0$

Step 5.5.5.1.5

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

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Step 5.5.5.1.5.1

Apply the distributive property.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 2 (2 λ + 2) + λ (2 λ + 2) + 3 \cdot 5) + 0$

Step 5.5.5.1.5.2

Apply the distributive property.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 2 (2 λ) - 2 \cdot 2 + λ (2 λ + 2) + 3 \cdot 5) + 0$

Step 5.5.5.1.5.3

Apply the distributive property.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 2 (2 λ) - 2 \cdot 2 + λ (2 λ) + λ \cdot 2 + 3 \cdot 5) + 0$

Step 5.5.5.1.6

Simplify and combine like terms.

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Step 5.5.5.1.6.1

Simplify each term.

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Step 5.5.5.1.6.1.1

Multiply $2$ by $- 2$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 4 λ - 2 \cdot 2 + λ (2 λ) + λ \cdot 2 + 3 \cdot 5) + 0$

Step 5.5.5.1.6.1.2

Multiply $- 2$ by $2$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 4 λ - 4 + λ (2 λ) + λ \cdot 2 + 3 \cdot 5) + 0$

Step 5.5.5.1.6.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 4 λ - 4 + 2 λ \cdot λ + λ \cdot 2 + 3 \cdot 5) + 0$

Step 5.5.5.1.6.1.4

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

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Step 5.5.5.1.6.1.4.1

Move $λ$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 4 λ - 4 + 2 (λ \cdot λ) + λ \cdot 2 + 3 \cdot 5) + 0$

Step 5.5.5.1.6.1.4.2

Multiply $λ$ by $λ$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 4 λ - 4 + 2 λ^{2} + λ \cdot 2 + 3 \cdot 5) + 0$

Step 5.5.5.1.6.1.5

Move $2$ to the left of $λ$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 4 λ - 4 + 2 λ^{2} + 2 λ + 3 \cdot 5) + 0$

Step 5.5.5.1.6.2

Add $- 4 λ$ and $2 λ$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 2 λ - 4 + 2 λ^{2} + 3 \cdot 5) + 0$

Step 5.5.5.1.7

Multiply $3$ by $5$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ + 4 - 2 λ - 4 + 2 λ^{2} + 15) + 0$

Step 5.5.5.2

Combine the opposite terms in $- λ + 4 - 2 λ - 4 + 2 λ^{2} + 15$ .

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Step 5.5.5.2.1

Subtract $4$ from $4$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ - 2 λ + 0 + 2 λ^{2} + 15) + 0$

Step 5.5.5.2.2

Add $- λ - 2 λ$ and $0$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- λ - 2 λ + 2 λ^{2} + 15) + 0$

Step 5.5.5.3

Subtract $2 λ$ from $- λ$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (- 3 λ + 2 λ^{2} + 15) + 0$

Step 5.5.5.4

Reorder $- 3 λ$ and $2 λ^{2}$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15) + 0$

Step 5.6

Simplify the determinant.

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Step 5.6.1

Add $(2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$ and $0$ .

$p (λ) = (2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5) - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2

Simplify each term.

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Step 5.6.2.1

Expand $(2 - λ) (- λ^{3} + 5 λ^{2} - 14 λ + 5)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 2 (- λ^{3}) + 2 (5 λ^{2}) + 2 (- 14 λ) + 2 \cdot 5 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2

Simplify each term.

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Step 5.6.2.2.1

Multiply $- 1$ by $2$ .

$p (λ) = - 2 λ^{3} + 2 (5 λ^{2}) + 2 (- 14 λ) + 2 \cdot 5 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.2

Multiply $5$ by $2$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} + 2 (- 14 λ) + 2 \cdot 5 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.3

Multiply $- 14$ by $2$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 2 \cdot 5 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.4

Multiply $2$ by $5$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.5

Rewrite using the commutative property of multiplication.

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 - 1 \cdot - 1 λ \cdot λ^{3} - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.6

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

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Step 5.6.2.2.6.1

Move $λ^{3}$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 - 1 \cdot - 1 (λ^{3} λ) - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.6.2

Multiply $λ^{3}$ by $λ$ .

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Step 5.6.2.2.6.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 - 1 \cdot - 1 (λ^{3} λ^{1}) - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.6.2.2

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

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 - 1 \cdot - 1 λ^{3 + 1} - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.6.3

Add $3$ and $1$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 - 1 \cdot - 1 λ^{4} - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.7

Multiply $- 1$ by $- 1$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + 1 λ^{4} - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.8

Multiply $λ^{4}$ by $1$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - λ (5 λ^{2}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.9

Rewrite using the commutative property of multiplication.

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 1 \cdot 5 λ \cdot λ^{2} - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.10

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

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Step 5.6.2.2.10.1

Move $λ^{2}$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 1 \cdot 5 (λ^{2} λ) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.10.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.6.2.2.10.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 1 \cdot 5 (λ^{2} λ^{1}) - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.10.2.2

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

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 1 \cdot 5 λ^{2 + 1} - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.10.3

Add $2$ and $1$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 1 \cdot 5 λ^{3} - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.11

Multiply $- 1$ by $5$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 5 λ^{3} - λ (- 14 λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.12

Rewrite using the commutative property of multiplication.

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 5 λ^{3} - 1 \cdot - 14 λ \cdot λ - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.13

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

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Step 5.6.2.2.13.1

Move $λ$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 5 λ^{3} - 1 \cdot - 14 (λ \cdot λ) - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.13.2

Multiply $λ$ by $λ$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 5 λ^{3} - 1 \cdot - 14 λ^{2} - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.14

Multiply $- 1$ by $- 14$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 5 λ^{3} + 14 λ^{2} - λ \cdot 5 - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.2.15

Multiply $5$ by $- 1$ .

$p (λ) = - 2 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} - 5 λ^{3} + 14 λ^{2} - 5 λ - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.3

Subtract $5 λ^{3}$ from $- 2 λ^{3}$ .

$p (λ) = - 7 λ^{3} + 10 λ^{2} - 28 λ + 10 + λ^{4} + 14 λ^{2} - 5 λ - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.4

Add $10 λ^{2}$ and $14 λ^{2}$ .

$p (λ) = - 7 λ^{3} + 24 λ^{2} - 28 λ + 10 + λ^{4} - 5 λ - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.5

Subtract $5 λ$ from $- 28 λ$ .

$p (λ) = - 7 λ^{3} + 24 λ^{2} - 33 λ + 10 + λ^{4} - 3 (λ^{2} - 14 λ + 25) - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.6

Apply the distributive property.

$p (λ) = - 7 λ^{3} + 24 λ^{2} - 33 λ + 10 + λ^{4} - 3 λ^{2} - 3 (- 14 λ) - 3 \cdot 25 - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.7

Simplify.

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Step 5.6.2.7.1

Multiply $- 14$ by $- 3$ .

$p (λ) = - 7 λ^{3} + 24 λ^{2} - 33 λ + 10 + λ^{4} - 3 λ^{2} + 42 λ - 3 \cdot 25 - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.7.2

Multiply $- 3$ by $25$ .

$p (λ) = - 7 λ^{3} + 24 λ^{2} - 33 λ + 10 + λ^{4} - 3 λ^{2} + 42 λ - 75 - 1 (2 λ^{2} - 3 λ + 15)$

Step 5.6.2.8

Apply the distributive property.

$p (λ) = - 7 λ^{3} + 24 λ^{2} - 33 λ + 10 + λ^{4} - 3 λ^{2} + 42 λ - 75 - 1 (2 λ^{2}) - 1 (- 3 λ) - 1 \cdot 15$

Step 5.6.2.9

Simplify.

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Step 5.6.2.9.1

Multiply $2$ by $- 1$ .

$p (λ) = - 7 λ^{3} + 24 λ^{2} - 33 λ + 10 + λ^{4} - 3 λ^{2} + 42 λ - 75 - 2 λ^{2} - 1 (- 3 λ) - 1 \cdot 15$

Step 5.6.2.9.2

Multiply $- 3$ by $- 1$ .

$p (λ) = - 7 λ^{3} + 24 λ^{2} - 33 λ + 10 + λ^{4} - 3 λ^{2} + 42 λ - 75 - 2 λ^{2} + 3 λ - 1 \cdot 15$

Step 5.6.2.9.3

Multiply $- 1$ by $15$ .

$p (λ) = - 7 λ^{3} + 24 λ^{2} - 33 λ + 10 + λ^{4} - 3 λ^{2} + 42 λ - 75 - 2 λ^{2} + 3 λ - 15$

Step 5.6.3

Subtract $3 λ^{2}$ from $24 λ^{2}$ .

$p (λ) = - 7 λ^{3} + 21 λ^{2} - 33 λ + 10 + λ^{4} + 42 λ - 75 - 2 λ^{2} + 3 λ - 15$

Step 5.6.4

Subtract $2 λ^{2}$ from $21 λ^{2}$ .

$p (λ) = - 7 λ^{3} + 19 λ^{2} - 33 λ + 10 + λ^{4} + 42 λ - 75 + 3 λ - 15$

Step 5.6.5

Add $- 33 λ$ and $42 λ$ .

$p (λ) = - 7 λ^{3} + 19 λ^{2} + 9 λ + 10 + λ^{4} - 75 + 3 λ - 15$

Step 5.6.6

Add $9 λ$ and $3 λ$ .

$p (λ) = - 7 λ^{3} + 19 λ^{2} + 12 λ + 10 + λ^{4} - 75 - 15$

Step 5.6.7

Subtract $75$ from $10$ .

$p (λ) = - 7 λ^{3} + 19 λ^{2} + 12 λ + λ^{4} - 65 - 15$

Step 5.6.8

Subtract $15$ from $- 65$ .

$p (λ) = - 7 λ^{3} + 19 λ^{2} + 12 λ + λ^{4} - 80$

Step 5.6.9

Move $12 λ$ .

$p (λ) = - 7 λ^{3} + 19 λ^{2} + λ^{4} + 12 λ - 80$

Step 5.6.10

Move $19 λ^{2}$ .

$p (λ) = - 7 λ^{3} + λ^{4} + 19 λ^{2} + 12 λ - 80$

Step 5.6.11

Reorder $- 7 λ^{3}$ and $λ^{4}$ .

$p (λ) = λ^{4} - 7 λ^{3} + 19 λ^{2} + 12 λ - 80$