Finite Math Examples

$[\begin{array}{c} 1 & 2 & - 1 & - 2 \\ 0 & 0 & 0 & 0 \\ 1 & - 1 & 2 & 5 \\ - 1 & 1 & - 1 & 2 \end{array}]$

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 $[\begin{array}{c} 1 & 2 & - 1 & - 2 \\ 0 & 0 & 0 & 0 \\ 1 & - 1 & 2 & 5 \\ - 1 & 1 & - 1 & 2 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 & - 1 & - 2 \\ 0 & 0 & 0 & 0 \\ 1 & - 1 & 2 & 5 \\ - 1 & 1 & - 1 & 2 \end{array}] - λ I_{4})$

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 $I_{4}$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 & - 1 & - 2 \\ 0 & 0 & 0 & 0 \\ 1 & - 1 & 2 & 5 \\ - 1 & 1 & - 1 & 2 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

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.

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.5

Multiply $- λ \cdot 0$ .

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Step 4.1.2.5.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.5.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.6

Multiply $- 1$ by $1$ .

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

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.9

Multiply $- λ \cdot 0$ .

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Step 4.1.2.9.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.9.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.10

Multiply $- λ \cdot 0$ .

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Step 4.1.2.10.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.10.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.11

Multiply $- 1$ by $1$ .

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

Step 4.1.2.12

Multiply $- λ \cdot 0$ .

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Step 4.1.2.12.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.12.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.13

Multiply $- λ \cdot 0$ .

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Step 4.1.2.13.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.13.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.14

Multiply $- λ \cdot 0$ .

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Step 4.1.2.14.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.14.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.15

Multiply $- λ \cdot 0$ .

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Step 4.1.2.15.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.15.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.16

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 1 & 2 & - 1 & - 2 \\ 0 & 0 & 0 & 0 \\ 1 & - 1 & 2 & 5 \\ - 1 & 1 & - 1 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 + 0 & - 1 + 0 & - 2 + 0 \\ 0 + 0 & 0 - λ & 0 + 0 & 0 + 0 \\ 1 + 0 & - 1 + 0 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 + 0 & - 2 + 0 \\ 0 + 0 & 0 - λ & 0 + 0 & 0 + 0 \\ 1 + 0 & - 1 + 0 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.2

Add $- 1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 + 0 \\ 0 + 0 & 0 - λ & 0 + 0 & 0 + 0 \\ 1 + 0 & - 1 + 0 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.3

Add $- 2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 + 0 & 0 - λ & 0 + 0 & 0 + 0 \\ 1 + 0 & - 1 + 0 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.4

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & 0 - λ & 0 + 0 & 0 + 0 \\ 1 + 0 & - 1 + 0 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.5

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & - λ & 0 + 0 & 0 + 0 \\ 1 + 0 & - 1 + 0 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.6

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & - λ & 0 & 0 + 0 \\ 1 + 0 & - 1 + 0 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.7

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & - λ & 0 & 0 \\ 1 + 0 & - 1 + 0 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.8

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & - λ & 0 & 0 \\ 1 & - 1 + 0 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.9

Add $- 1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & - λ & 0 & 0 \\ 1 & - 1 & 2 - λ & 5 + 0 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.10

Add $5$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & - λ & 0 & 0 \\ 1 & - 1 & 2 - λ & 5 \\ - 1 + 0 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.11

Add $- 1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & - λ & 0 & 0 \\ 1 & - 1 & 2 - λ & 5 \\ - 1 & 1 + 0 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.12

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & - λ & 0 & 0 \\ 1 & - 1 & 2 - λ & 5 \\ - 1 & 1 & - 1 + 0 & 2 - λ \end{array}]$

Step 4.3.13

Add $- 1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 1 - λ & 2 & - 1 & - 2 \\ 0 & - λ & 0 & 0 \\ 1 & - 1 & 2 - λ & 5 \\ - 1 & 1 & - 1 & 2 - λ \end{array}]$

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

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

Step 5.1.4

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

$0 | \begin{array}{c} 2 & - 1 & - 2 \\ - 1 & 2 - λ & 5 \\ 1 & - 1 & 2 - λ \end{array} |$

Step 5.1.5

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

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

Step 5.1.6

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

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

Step 5.1.7

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

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

Step 5.1.8

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

$0 | \begin{array}{c} 1 - λ & 2 & - 2 \\ 1 & - 1 & 5 \\ - 1 & 1 & 2 - λ \end{array} |$

Step 5.1.9

The minor for $a_{24}$ is the determinant with row $2$ and column $4$ deleted.

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

Step 5.1.10

Multiply element $a_{24}$ by its cofactor.

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

Step 5.1.11

Add the terms together.

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

Step 5.2

Multiply $0$ by $| \begin{array}{c} 2 & - 1 & - 2 \\ - 1 & 2 - λ & 5 \\ 1 & - 1 & 2 - λ \end{array} |$ .

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

Step 5.3

Multiply $0$ by $| \begin{array}{c} 1 - λ & 2 & - 2 \\ 1 & - 1 & 5 \\ - 1 & 1 & 2 - λ \end{array} |$ .

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

Step 5.4

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

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

Step 5.5

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

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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.

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

Step 5.5.1.4

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

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

Step 5.5.1.5

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

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

Step 5.5.1.6

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

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

Step 5.5.1.7

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

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

Step 5.5.1.8

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

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

Step 5.5.1.9

Add the terms together.

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

Step 5.5.2

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

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Step 5.5.2.1

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

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

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

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

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Step 5.5.2.2.1.1.1

Apply the distributive property.

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

Step 5.5.2.2.1.1.2

Apply the distributive property.

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

Step 5.5.2.2.1.1.3

Apply the distributive property.

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

Step 5.5.2.2.1.2

Simplify and combine like terms.

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Step 5.5.2.2.1.2.1

Simplify each term.

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Step 5.5.2.2.1.2.1.1

Multiply $2$ by $2$ .

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

Step 5.5.2.2.1.2.1.2

Multiply $- 1$ by $2$ .

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

Step 5.5.2.2.1.2.1.3

Multiply $2$ by $- 1$ .

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

Step 5.5.2.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 5.5.2.2.1.2.1.5

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

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Step 5.5.2.2.1.2.1.5.1

Move $λ$ .

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

Step 5.5.2.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

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

Step 5.5.2.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

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

Step 5.5.2.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

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

Step 5.5.2.2.1.2.2

Subtract $2 λ$ from $- 2 λ$ .

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

Step 5.5.2.2.1.3

Multiply $- (- 1 \cdot 5)$ .

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Step 5.5.2.2.1.3.1

Multiply $- 1$ by $5$ .

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

Step 5.5.2.2.1.3.2

Multiply $- 1$ by $- 5$ .

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

Step 5.5.2.2.2

Add $4$ and $5$ .

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

Step 5.5.2.2.3

Reorder $- 4 λ$ and $λ^{2}$ .

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

Step 5.5.3

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

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Step 5.5.3.1

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

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

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 $2 - λ$ by $1$ .

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

Step 5.5.3.2.1.2

Multiply $- (- 1 \cdot 5)$ .

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Step 5.5.3.2.1.2.1

Multiply $- 1$ by $5$ .

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

Step 5.5.3.2.1.2.2

Multiply $- 1$ by $- 5$ .

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

Step 5.5.3.2.2

Add $2$ and $5$ .

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 | \begin{array}{c} 1 & 2 - λ \\ - 1 & - 1 \end{array} |) + 0 + 0$

Step 5.5.4

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

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Step 5.5.4.1

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

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (1 \cdot - 1 - - (2 - λ))) + 0 + 0$

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 $- 1$ by $1$ .

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (- 1 - - (2 - λ))) + 0 + 0$

Step 5.5.4.2.1.2

Apply the distributive property.

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (- 1 - (- 1 \cdot 2 - - λ))) + 0 + 0$

Step 5.5.4.2.1.3

Multiply $- 1$ by $2$ .

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (- 1 - (- 2 - - λ))) + 0 + 0$

Step 5.5.4.2.1.4

Multiply $- - λ$ .

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Step 5.5.4.2.1.4.1

Multiply $- 1$ by $- 1$ .

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (- 1 - (- 2 + 1 λ))) + 0 + 0$

Step 5.5.4.2.1.4.2

Multiply $λ$ by $1$ .

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (- 1 - (- 2 + λ))) + 0 + 0$

Step 5.5.4.2.1.5

Apply the distributive property.

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (- 1 - - 2 - λ)) + 0 + 0$

Step 5.5.4.2.1.6

Multiply $- 1$ by $- 2$ .

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (- 1 + 2 - λ)) + 0 + 0$

Step 5.5.4.2.2

Add $- 1$ and $2$ .

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (1 - λ)) + 0 + 0$

Step 5.5.4.2.3

Reorder $1$ and $- λ$ .

$p (λ) = 0 - λ ((1 - λ) (λ^{2} - 4 λ + 9) + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

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

Expand $(1 - λ) (λ^{2} - 4 λ + 9)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 0 - λ (1 λ^{2} + 1 (- 4 λ) + 1 \cdot 9 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2

Simplify each term.

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Step 5.5.5.1.2.1

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0 - λ (λ^{2} + 1 (- 4 λ) + 1 \cdot 9 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.2

Multiply $- 4 λ$ by $1$ .

$p (λ) = 0 - λ (λ^{2} - 4 λ + 1 \cdot 9 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.3

Multiply $9$ by $1$ .

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.4

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

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Step 5.5.5.1.2.4.1

Move $λ^{2}$ .

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - (λ^{2} λ) - λ (- 4 λ) - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.4.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.5.5.1.2.4.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - (λ^{2} λ^{1}) - λ (- 4 λ) - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.4.2.2

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

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - λ^{2 + 1} - λ (- 4 λ) - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.4.3

Add $2$ and $1$ .

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - λ^{3} - λ (- 4 λ) - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.5

Rewrite using the commutative property of multiplication.

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - λ^{3} - 1 \cdot - 4 λ \cdot λ - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.6

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

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Step 5.5.5.1.2.6.1

Move $λ$ .

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - λ^{3} - 1 \cdot - 4 (λ \cdot λ) - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.6.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - λ^{3} - 1 \cdot - 4 λ^{2} - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.7

Multiply $- 1$ by $- 4$ .

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - λ^{3} + 4 λ^{2} - λ \cdot 9 + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.2.8

Multiply $9$ by $- 1$ .

$p (λ) = 0 - λ (λ^{2} - 4 λ + 9 - λ^{3} + 4 λ^{2} - 9 λ + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.3

Add $λ^{2}$ and $4 λ^{2}$ .

$p (λ) = 0 - λ (5 λ^{2} - 4 λ + 9 - λ^{3} - 9 λ + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.4

Subtract $9 λ$ from $- 4 λ$ .

$p (λ) = 0 - λ (5 λ^{2} - 13 λ + 9 - λ^{3} + 1 (- λ + 7) - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.5

Multiply $- λ + 7$ by $1$ .

$p (λ) = 0 - λ (5 λ^{2} - 13 λ + 9 - λ^{3} - λ + 7 - 2 (- λ + 1)) + 0 + 0$

Step 5.5.5.1.6

Apply the distributive property.

$p (λ) = 0 - λ (5 λ^{2} - 13 λ + 9 - λ^{3} - λ + 7 - 2 (- λ) - 2 \cdot 1) + 0 + 0$

Step 5.5.5.1.7

Multiply $- 1$ by $- 2$ .

$p (λ) = 0 - λ (5 λ^{2} - 13 λ + 9 - λ^{3} - λ + 7 + 2 λ - 2 \cdot 1) + 0 + 0$

Step 5.5.5.1.8

Multiply $- 2$ by $1$ .

$p (λ) = 0 - λ (5 λ^{2} - 13 λ + 9 - λ^{3} - λ + 7 + 2 λ - 2) + 0 + 0$

Step 5.5.5.2

Subtract $λ$ from $- 13 λ$ .

$p (λ) = 0 - λ (5 λ^{2} - 14 λ + 9 - λ^{3} + 7 + 2 λ - 2) + 0 + 0$

Step 5.5.5.3

Add $- 14 λ$ and $2 λ$ .

$p (λ) = 0 - λ (5 λ^{2} - 12 λ + 9 - λ^{3} + 7 - 2) + 0 + 0$

Step 5.5.5.4

Add $9$ and $7$ .

$p (λ) = 0 - λ (5 λ^{2} - 12 λ - λ^{3} + 16 - 2) + 0 + 0$

Step 5.5.5.5

Subtract $2$ from $16$ .

$p (λ) = 0 - λ (5 λ^{2} - 12 λ - λ^{3} + 14) + 0 + 0$

Step 5.5.5.6

Move $- 12 λ$ .

$p (λ) = 0 - λ (5 λ^{2} - λ^{3} - 12 λ + 14) + 0 + 0$

Step 5.5.5.7

Reorder $5 λ^{2}$ and $- λ^{3}$ .

$p (λ) = 0 - λ (- λ^{3} + 5 λ^{2} - 12 λ + 14) + 0 + 0$

Step 5.6

Simplify the determinant.

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Step 5.6.1

Combine the opposite terms in $0 - λ (- λ^{3} + 5 λ^{2} - 12 λ + 14) + 0 + 0$ .

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Step 5.6.1.1

Subtract $λ (- λ^{3} + 5 λ^{2} - 12 λ + 14)$ from $0$ .

$p (λ) = - λ (- λ^{3} + 5 λ^{2} - 12 λ + 14) + 0 + 0$

Step 5.6.1.2

Add $- λ (- λ^{3} + 5 λ^{2} - 12 λ + 14)$ and $0$ .

$p (λ) = - λ (- λ^{3} + 5 λ^{2} - 12 λ + 14) + 0$

Step 5.6.1.3

Add $- λ (- λ^{3} + 5 λ^{2} - 12 λ + 14)$ and $0$ .

$p (λ) = - λ (- λ^{3} + 5 λ^{2} - 12 λ + 14)$

Step 5.6.2

Apply the distributive property.

$p (λ) = - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 12 λ) - λ \cdot 14$

Step 5.6.3

Simplify.

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Step 5.6.3.1

Rewrite using the commutative property of multiplication.

$p (λ) = - 1 \cdot - 1 λ \cdot λ^{3} - λ (5 λ^{2}) - λ (- 12 λ) - λ \cdot 14$

Step 5.6.3.2

Rewrite using the commutative property of multiplication.

$p (λ) = - 1 \cdot - 1 λ \cdot λ^{3} - 1 \cdot 5 λ \cdot λ^{2} - λ (- 12 λ) - λ \cdot 14$

Step 5.6.3.3

Rewrite using the commutative property of multiplication.

$p (λ) = - 1 \cdot - 1 λ \cdot λ^{3} - 1 \cdot 5 λ \cdot λ^{2} - 1 \cdot - 12 λ \cdot λ - λ \cdot 14$

Step 5.6.3.4

Multiply $14$ by $- 1$ .

$p (λ) = - 1 \cdot - 1 λ \cdot λ^{3} - 1 \cdot 5 λ \cdot λ^{2} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4

Simplify each term.

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Step 5.6.4.1

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

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Step 5.6.4.1.1

Move $λ^{3}$ .

$p (λ) = - 1 \cdot - 1 (λ^{3} λ) - 1 \cdot 5 λ \cdot λ^{2} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.1.2

Multiply $λ^{3}$ by $λ$ .

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Step 5.6.4.1.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - 1 \cdot - 1 (λ^{3} λ^{1}) - 1 \cdot 5 λ \cdot λ^{2} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.1.2.2

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

$p (λ) = - 1 \cdot - 1 λ^{3 + 1} - 1 \cdot 5 λ \cdot λ^{2} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.1.3

Add $3$ and $1$ .

$p (λ) = - 1 \cdot - 1 λ^{4} - 1 \cdot 5 λ \cdot λ^{2} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.2

Multiply $- 1$ by $- 1$ .

$p (λ) = 1 λ^{4} - 1 \cdot 5 λ \cdot λ^{2} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.3

Multiply $λ^{4}$ by $1$ .

$p (λ) = λ^{4} - 1 \cdot 5 λ \cdot λ^{2} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.4

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

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Step 5.6.4.4.1

Move $λ^{2}$ .

$p (λ) = λ^{4} - 1 \cdot 5 (λ^{2} λ) - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.4.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.6.4.4.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = λ^{4} - 1 \cdot 5 (λ^{2} λ^{1}) - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.4.2.2

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

$p (λ) = λ^{4} - 1 \cdot 5 λ^{2 + 1} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.4.3

Add $2$ and $1$ .

$p (λ) = λ^{4} - 1 \cdot 5 λ^{3} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.5

Multiply $- 1$ by $5$ .

$p (λ) = λ^{4} - 5 λ^{3} - 1 \cdot - 12 λ \cdot λ - 14 λ$

Step 5.6.4.6

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

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Step 5.6.4.6.1

Move $λ$ .

$p (λ) = λ^{4} - 5 λ^{3} - 1 \cdot - 12 (λ \cdot λ) - 14 λ$

Step 5.6.4.6.2

Multiply $λ$ by $λ$ .

$p (λ) = λ^{4} - 5 λ^{3} - 1 \cdot - 12 λ^{2} - 14 λ$

Step 5.6.4.7

Multiply $- 1$ by $- 12$ .

$p (λ) = λ^{4} - 5 λ^{3} + 12 λ^{2} - 14 λ$

Step 6

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

$λ^{4} - 5 λ^{3} + 12 λ^{2} - 14 λ = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Graph each side of the equation. The solution is the x-value of the point of intersection.

$λ \approx 0, 2.43287625$