Linear Algebra Examples

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}]$

Step 1

Find the eigenvalues.

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Step 1.1

Set up the formula to find the characteristic equation $p (λ)$ .

$p (λ) = determinant (A - λ I_{4})$

Step 1.2

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 1.3

Substitute the known values into $p (λ) = determinant (A - λ I_{4})$ .

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Step 1.3.1

Substitute $[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] - λ I_{4})$

Step 1.3.2

Substitute $[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]$ for $I_{4}$ .

$p (λ) = determinant ([\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 1.4

Simplify.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

Multiply $- λ$ by each element of the matrix.

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

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.5

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.5.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.5.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.6

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.9

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.9.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.9.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.10

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.10.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.10.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.11

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.12

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.12.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.12.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.13

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.13.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.13.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.14

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.14.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.14.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.15

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.15.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.15.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.16

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 + 0 & 0 + 0 & 0 + 0 \\ - 5 + 0 & - 3 - λ & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - λ & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $- 5$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 + 0 & 0 + 0 \\ - 5 + 0 & - 3 - λ & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - λ & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.2

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 + 0 \\ - 5 + 0 & - 3 - λ & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - λ & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.3

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 + 0 & - 3 - λ & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - λ & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.4

Add $- 5$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 & - 3 - λ & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - λ & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.5

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 & - 3 - λ & 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - λ & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.6

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 & - 3 - λ & 0 & 0 \\ 0 + 0 & 0 + 0 & - 3 - λ & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.7

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 & - 3 - λ & 0 & 0 \\ 0 & 0 + 0 & - 3 - λ & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.8

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 & - 3 - λ & 0 & 0 \\ 0 & 0 & - 3 - λ & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.9

Add $- 5$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 & - 3 - λ & 0 & 0 \\ 0 & 0 & - 3 - λ & - 5 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.10

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 & - 3 - λ & 0 & 0 \\ 0 & 0 & - 3 - λ & - 5 \\ 0 & 0 + 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.11

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 & - 3 - λ & 0 & 0 \\ 0 & 0 & - 3 - λ & - 5 \\ 0 & 0 & - 5 + 0 & - 3 - λ \end{array}]$

Step 1.4.3.12

Add $- 5$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 3 - λ & - 5 & 0 & 0 \\ - 5 & - 3 - λ & 0 & 0 \\ 0 & 0 & - 3 - λ & - 5 \\ 0 & 0 & - 5 & - 3 - λ \end{array}]$

Step 1.5

Find the determinant.

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Step 1.5.1

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 1.5.1.1

Consider the corresponding sign chart.

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

Step 1.5.1.2

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

Step 1.5.1.3

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

$| \begin{array}{c} - 3 - λ & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} |$

Step 1.5.1.4

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

$(- 3 - λ) | \begin{array}{c} - 3 - λ & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} |$

Step 1.5.1.5

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

$| \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} |$

Step 1.5.1.6

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

$5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} |$

Step 1.5.1.7

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

$| \begin{array}{c} - 5 & - 3 - λ & 0 \\ 0 & 0 & - 5 \\ 0 & 0 & - 3 - λ \end{array} |$

Step 1.5.1.8

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

$0 | \begin{array}{c} - 5 & - 3 - λ & 0 \\ 0 & 0 & - 5 \\ 0 & 0 & - 3 - λ \end{array} |$

Step 1.5.1.9

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

$| \begin{array}{c} - 5 & - 3 - λ & 0 \\ 0 & 0 & - 3 - λ \\ 0 & 0 & - 5 \end{array} |$

Step 1.5.1.10

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

$0 | \begin{array}{c} - 5 & - 3 - λ & 0 \\ 0 & 0 & - 3 - λ \\ 0 & 0 & - 5 \end{array} |$

Step 1.5.1.11

Add the terms together.

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

Step 1.5.2

Multiply $0$ by $| \begin{array}{c} - 5 & - 3 - λ & 0 \\ 0 & 0 & - 5 \\ 0 & 0 & - 3 - λ \end{array} |$ .

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

Step 1.5.3

Multiply $0$ by $| \begin{array}{c} - 5 & - 3 - λ & 0 \\ 0 & 0 & - 3 - λ \\ 0 & 0 & - 5 \end{array} |$ .

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

Step 1.5.4

Evaluate $| \begin{array}{c} - 3 - λ & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} |$ .

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Step 1.5.4.1

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

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Step 1.5.4.1.1

Consider the corresponding sign chart.

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

Step 1.5.4.1.2

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

Step 1.5.4.1.3

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

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

Step 1.5.4.1.4

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

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

Step 1.5.4.1.5

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

$| \begin{array}{c} 0 & - 5 \\ 0 & - 3 - λ \end{array} |$

Step 1.5.4.1.6

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

$0 | \begin{array}{c} 0 & - 5 \\ 0 & - 3 - λ \end{array} |$

Step 1.5.4.1.7

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

$| \begin{array}{c} 0 & - 3 - λ \\ 0 & - 5 \end{array} |$

Step 1.5.4.1.8

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

$0 | \begin{array}{c} 0 & - 3 - λ \\ 0 & - 5 \end{array} |$

Step 1.5.4.1.9

Add the terms together.

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

Step 1.5.4.2

Multiply $0$ by $| \begin{array}{c} 0 & - 5 \\ 0 & - 3 - λ \end{array} |$ .

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

Step 1.5.4.3

Multiply $0$ by $| \begin{array}{c} 0 & - 3 - λ \\ 0 & - 5 \end{array} |$ .

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

Step 1.5.4.4

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

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Step 1.5.4.4.1

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

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

Step 1.5.4.4.2

Simplify the determinant.

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Step 1.5.4.4.2.1

Simplify each term.

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Step 1.5.4.4.2.1.1

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

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Step 1.5.4.4.2.1.1.1

Apply the distributive property.

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

Step 1.5.4.4.2.1.1.2

Apply the distributive property.

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

Step 1.5.4.4.2.1.1.3

Apply the distributive property.

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

Step 1.5.4.4.2.1.2

Simplify and combine like terms.

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Step 1.5.4.4.2.1.2.1

Simplify each term.

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Step 1.5.4.4.2.1.2.1.1

Multiply $- 3$ by $- 3$ .

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

Step 1.5.4.4.2.1.2.1.2

Multiply $- 1$ by $- 3$ .

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

Step 1.5.4.4.2.1.2.1.3

Multiply $- 3$ by $- 1$ .

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

Step 1.5.4.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 1.5.4.4.2.1.2.1.5

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

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Step 1.5.4.4.2.1.2.1.5.1

Move $λ$ .

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

Step 1.5.4.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

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

Step 1.5.4.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

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

Step 1.5.4.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

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

Step 1.5.4.4.2.1.2.2

Add $3 λ$ and $3 λ$ .

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

Step 1.5.4.4.2.1.3

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

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Step 1.5.4.4.2.1.3.1

Multiply $- 5$ by $- 5$ .

$p (λ) = (- 3 - λ) ((- 3 - λ) (9 + 6 λ + λ^{2} - 1 \cdot 25) + 0 + 0) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.4.2.1.3.2

Multiply $- 1$ by $25$ .

$p (λ) = (- 3 - λ) ((- 3 - λ) (9 + 6 λ + λ^{2} - 25) + 0 + 0) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.4.2.2

Subtract $25$ from $9$ .

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

Step 1.5.4.4.2.3

Reorder $6 λ$ and $λ^{2}$ .

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

Step 1.5.4.5

Simplify the determinant.

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Step 1.5.4.5.1

Combine the opposite terms in $(- 3 - λ) (λ^{2} + 6 λ - 16) + 0 + 0$ .

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Step 1.5.4.5.1.1

Add $(- 3 - λ) (λ^{2} + 6 λ - 16)$ and $0$ .

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

Step 1.5.4.5.1.2

Add $(- 3 - λ) (λ^{2} + 6 λ - 16)$ and $0$ .

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

Step 1.5.4.5.2

Expand $(- 3 - λ) (λ^{2} + 6 λ - 16)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 1.5.4.5.3

Simplify each term.

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Step 1.5.4.5.3.1

Multiply $6$ by $- 3$ .

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ - 3 \cdot - 16 - λ \cdot λ^{2} - λ (6 λ) - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.2

Multiply $- 3$ by $- 16$ .

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - λ \cdot λ^{2} - λ (6 λ) - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.3

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

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Step 1.5.4.5.3.3.1

Move $λ^{2}$ .

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - (λ^{2} λ) - λ (6 λ) - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.4.5.3.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - (λ^{2} λ^{1}) - λ (6 λ) - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.3.2.2

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

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - λ^{2 + 1} - λ (6 λ) - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.3.3

Add $2$ and $1$ .

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - λ^{3} - λ (6 λ) - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.4

Rewrite using the commutative property of multiplication.

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - λ^{3} - 1 \cdot 6 λ \cdot λ - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.5

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

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Step 1.5.4.5.3.5.1

Move $λ$ .

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - λ^{3} - 1 \cdot 6 (λ \cdot λ) - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - λ^{3} - 1 \cdot 6 λ^{2} - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.6

Multiply $- 1$ by $6$ .

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - λ^{3} - 6 λ^{2} - λ \cdot - 16) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.3.7

Multiply $- 16$ by $- 1$ .

$p (λ) = (- 3 - λ) (- 3 λ^{2} - 18 λ + 48 - λ^{3} - 6 λ^{2} + 16 λ) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.4

Subtract $6 λ^{2}$ from $- 3 λ^{2}$ .

$p (λ) = (- 3 - λ) (- 9 λ^{2} - 18 λ + 48 - λ^{3} + 16 λ) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.5

Add $- 18 λ$ and $16 λ$ .

$p (λ) = (- 3 - λ) (- 9 λ^{2} - 2 λ + 48 - λ^{3}) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.6

Move $48$ .

$p (λ) = (- 3 - λ) (- 9 λ^{2} - 2 λ - λ^{3} + 48) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.7

Move $- 2 λ$ .

$p (λ) = (- 3 - λ) (- 9 λ^{2} - λ^{3} - 2 λ + 48) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.4.5.8

Reorder $- 9 λ^{2}$ and $- λ^{3}$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 | \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} | + 0 + 0$

Step 1.5.5

Evaluate $| \begin{array}{c} - 5 & 0 & 0 \\ 0 & - 3 - λ & - 5 \\ 0 & - 5 & - 3 - λ \end{array} |$ .

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Step 1.5.5.1

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 1.5.5.1.1

Consider the corresponding sign chart.

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

Step 1.5.5.1.2

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

Step 1.5.5.1.3

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

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

Step 1.5.5.1.4

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

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

Step 1.5.5.1.5

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

$| \begin{array}{c} 0 & - 5 \\ 0 & - 3 - λ \end{array} |$

Step 1.5.5.1.6

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

$0 | \begin{array}{c} 0 & - 5 \\ 0 & - 3 - λ \end{array} |$

Step 1.5.5.1.7

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

$| \begin{array}{c} 0 & - 3 - λ \\ 0 & - 5 \end{array} |$

Step 1.5.5.1.8

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

$0 | \begin{array}{c} 0 & - 3 - λ \\ 0 & - 5 \end{array} |$

Step 1.5.5.1.9

Add the terms together.

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 | \begin{array}{c} - 3 - λ & - 5 \\ - 5 & - 3 - λ \end{array} | + 0 | \begin{array}{c} 0 & - 5 \\ 0 & - 3 - λ \end{array} | + 0 | \begin{array}{c} 0 & - 3 - λ \\ 0 & - 5 \end{array} |) + 0 + 0$

Step 1.5.5.2

Multiply $0$ by $| \begin{array}{c} 0 & - 5 \\ 0 & - 3 - λ \end{array} |$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 | \begin{array}{c} - 3 - λ & - 5 \\ - 5 & - 3 - λ \end{array} | + 0 + 0 | \begin{array}{c} 0 & - 3 - λ \\ 0 & - 5 \end{array} |) + 0 + 0$

Step 1.5.5.3

Multiply $0$ by $| \begin{array}{c} 0 & - 3 - λ \\ 0 & - 5 \end{array} |$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 | \begin{array}{c} - 3 - λ & - 5 \\ - 5 & - 3 - λ \end{array} | + 0 + 0) + 0 + 0$

Step 1.5.5.4

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

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Step 1.5.5.4.1

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

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 ((- 3 - λ) (- 3 - λ) - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2

Simplify the determinant.

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Step 1.5.5.4.2.1

Simplify each term.

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Step 1.5.5.4.2.1.1

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

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Step 1.5.5.4.2.1.1.1

Apply the distributive property.

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (- 3 (- 3 - λ) - λ (- 3 - λ) - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.1.2

Apply the distributive property.

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (- 3 \cdot - 3 - 3 (- λ) - λ (- 3 - λ) - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.1.3

Apply the distributive property.

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (- 3 \cdot - 3 - 3 (- λ) - λ \cdot - 3 - λ (- λ) - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.2

Simplify and combine like terms.

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Step 1.5.5.4.2.1.2.1

Simplify each term.

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Step 1.5.5.4.2.1.2.1.1

Multiply $- 3$ by $- 3$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 - 3 (- λ) - λ \cdot - 3 - λ (- λ) - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.2.1.2

Multiply $- 1$ by $- 3$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 3 λ - λ \cdot - 3 - λ (- λ) - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.2.1.3

Multiply $- 3$ by $- 1$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 3 λ + 3 λ - λ (- λ) - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 3 λ + 3 λ - 1 \cdot - 1 λ \cdot λ - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.2.1.5

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

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Step 1.5.5.4.2.1.2.1.5.1

Move $λ$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 3 λ + 3 λ - 1 \cdot - 1 (λ \cdot λ) - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 3 λ + 3 λ - 1 \cdot - 1 λ^{2} - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 3 λ + 3 λ + 1 λ^{2} - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 3 λ + 3 λ + λ^{2} - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.2.2

Add $3 λ$ and $3 λ$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 6 λ + λ^{2} - (- 5 \cdot - 5)) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.3

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

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Step 1.5.5.4.2.1.3.1

Multiply $- 5$ by $- 5$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 6 λ + λ^{2} - 1 \cdot 25) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.1.3.2

Multiply $- 1$ by $25$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (9 + 6 λ + λ^{2} - 25) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.2

Subtract $25$ from $9$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (6 λ + λ^{2} - 16) + 0 + 0) + 0 + 0$

Step 1.5.5.4.2.3

Reorder $6 λ$ and $λ^{2}$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (λ^{2} + 6 λ - 16) + 0 + 0) + 0 + 0$

Step 1.5.5.5

Simplify the determinant.

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Step 1.5.5.5.1

Combine the opposite terms in $- 5 (λ^{2} + 6 λ - 16) + 0 + 0$ .

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Step 1.5.5.5.1.1

Add $- 5 (λ^{2} + 6 λ - 16)$ and $0$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (λ^{2} + 6 λ - 16) + 0) + 0 + 0$

Step 1.5.5.5.1.2

Add $- 5 (λ^{2} + 6 λ - 16)$ and $0$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 (λ^{2} + 6 λ - 16)) + 0 + 0$

Step 1.5.5.5.2

Apply the distributive property.

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 λ^{2} - 5 (6 λ) - 5 \cdot - 16) + 0 + 0$

Step 1.5.5.5.3

Simplify.

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Step 1.5.5.5.3.1

Multiply $6$ by $- 5$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 λ^{2} - 30 λ - 5 \cdot - 16) + 0 + 0$

Step 1.5.5.5.3.2

Multiply $- 5$ by $- 16$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 λ^{2} - 30 λ + 80) + 0 + 0$

Step 1.5.6

Simplify the determinant.

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Step 1.5.6.1

Combine the opposite terms in $(- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 λ^{2} - 30 λ + 80) + 0 + 0$ .

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Step 1.5.6.1.1

Add $(- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 λ^{2} - 30 λ + 80)$ and $0$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 λ^{2} - 30 λ + 80) + 0$

Step 1.5.6.1.2

Add $(- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 λ^{2} - 30 λ + 80)$ and $0$ .

$p (λ) = (- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48) + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2

Simplify each term.

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Step 1.5.6.2.1

Expand $(- 3 - λ) (- λ^{3} - 9 λ^{2} - 2 λ + 48)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = - 3 (- λ^{3}) - 3 (- 9 λ^{2}) - 3 (- 2 λ) - 3 \cdot 48 - λ (- λ^{3}) - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2

Simplify each term.

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Step 1.5.6.2.2.1

Multiply $- 1$ by $- 3$ .

$p (λ) = 3 λ^{3} - 3 (- 9 λ^{2}) - 3 (- 2 λ) - 3 \cdot 48 - λ (- λ^{3}) - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.2

Multiply $- 9$ by $- 3$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} - 3 (- 2 λ) - 3 \cdot 48 - λ (- λ^{3}) - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.3

Multiply $- 2$ by $- 3$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 3 \cdot 48 - λ (- λ^{3}) - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.4

Multiply $- 3$ by $48$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 - λ (- λ^{3}) - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.5

Rewrite using the commutative property of multiplication.

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 - 1 \cdot - 1 λ \cdot λ^{3} - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.6

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

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Step 1.5.6.2.2.6.1

Move $λ^{3}$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 - 1 \cdot - 1 (λ^{3} λ) - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.6.2

Multiply $λ^{3}$ by $λ$ .

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Step 1.5.6.2.2.6.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 - 1 \cdot - 1 (λ^{3} λ^{1}) - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.6.2.2

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

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 - 1 \cdot - 1 λ^{3 + 1} - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.6.3

Add $3$ and $1$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 - 1 \cdot - 1 λ^{4} - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.7

Multiply $- 1$ by $- 1$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + 1 λ^{4} - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.8

Multiply $λ^{4}$ by $1$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} - λ (- 9 λ^{2}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.9

Rewrite using the commutative property of multiplication.

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} - 1 \cdot - 9 λ \cdot λ^{2} - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.10

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

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Step 1.5.6.2.2.10.1

Move $λ^{2}$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} - 1 \cdot - 9 (λ^{2} λ) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.10.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.6.2.2.10.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} - 1 \cdot - 9 (λ^{2} λ^{1}) - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.10.2.2

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

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} - 1 \cdot - 9 λ^{2 + 1} - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.10.3

Add $2$ and $1$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} - 1 \cdot - 9 λ^{3} - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.11

Multiply $- 1$ by $- 9$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} + 9 λ^{3} - λ (- 2 λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.12

Rewrite using the commutative property of multiplication.

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} + 9 λ^{3} - 1 \cdot - 2 λ \cdot λ - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.13

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

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Step 1.5.6.2.2.13.1

Move $λ$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} + 9 λ^{3} - 1 \cdot - 2 (λ \cdot λ) - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.13.2

Multiply $λ$ by $λ$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} + 9 λ^{3} - 1 \cdot - 2 λ^{2} - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.14

Multiply $- 1$ by $- 2$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} + 9 λ^{3} + 2 λ^{2} - λ \cdot 48 + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.2.15

Multiply $48$ by $- 1$ .

$p (λ) = 3 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} + 9 λ^{3} + 2 λ^{2} - 48 λ + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.3

Add $3 λ^{3}$ and $9 λ^{3}$ .

$p (λ) = 12 λ^{3} + 27 λ^{2} + 6 λ - 144 + λ^{4} + 2 λ^{2} - 48 λ + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.4

Add $27 λ^{2}$ and $2 λ^{2}$ .

$p (λ) = 12 λ^{3} + 29 λ^{2} + 6 λ - 144 + λ^{4} - 48 λ + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.5

Subtract $48 λ$ from $6 λ$ .

$p (λ) = 12 λ^{3} + 29 λ^{2} - 42 λ - 144 + λ^{4} + 5 (- 5 λ^{2} - 30 λ + 80)$

Step 1.5.6.2.6

Apply the distributive property.

$p (λ) = 12 λ^{3} + 29 λ^{2} - 42 λ - 144 + λ^{4} + 5 (- 5 λ^{2}) + 5 (- 30 λ) + 5 \cdot 80$

Step 1.5.6.2.7

Simplify.

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Step 1.5.6.2.7.1

Multiply $- 5$ by $5$ .

$p (λ) = 12 λ^{3} + 29 λ^{2} - 42 λ - 144 + λ^{4} - 25 λ^{2} + 5 (- 30 λ) + 5 \cdot 80$

Step 1.5.6.2.7.2

Multiply $- 30$ by $5$ .

$p (λ) = 12 λ^{3} + 29 λ^{2} - 42 λ - 144 + λ^{4} - 25 λ^{2} - 150 λ + 5 \cdot 80$

Step 1.5.6.2.7.3

Multiply $5$ by $80$ .

$p (λ) = 12 λ^{3} + 29 λ^{2} - 42 λ - 144 + λ^{4} - 25 λ^{2} - 150 λ + 400$

Step 1.5.6.3

Subtract $25 λ^{2}$ from $29 λ^{2}$ .

$p (λ) = 12 λ^{3} + 4 λ^{2} - 42 λ - 144 + λ^{4} - 150 λ + 400$

Step 1.5.6.4

Subtract $150 λ$ from $- 42 λ$ .

$p (λ) = 12 λ^{3} + 4 λ^{2} - 192 λ - 144 + λ^{4} + 400$

Step 1.5.6.5

Add $- 144$ and $400$ .

$p (λ) = 12 λ^{3} + 4 λ^{2} - 192 λ + λ^{4} + 256$

Step 1.5.6.6

Move $- 192 λ$ .

$p (λ) = 12 λ^{3} + 4 λ^{2} + λ^{4} - 192 λ + 256$

Step 1.5.6.7

Move $4 λ^{2}$ .

$p (λ) = 12 λ^{3} + λ^{4} + 4 λ^{2} - 192 λ + 256$

Step 1.5.6.8

Reorder $12 λ^{3}$ and $λ^{4}$ .

$p (λ) = λ^{4} + 12 λ^{3} + 4 λ^{2} - 192 λ + 256$

Step 1.6

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

$λ^{4} + 12 λ^{3} + 4 λ^{2} - 192 λ + 256 = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Factor the left side of the equation.

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Step 1.7.1.1

Factor $λ^{4} + 12 λ^{3} + 4 λ^{2} - 192 λ + 256$ using the rational roots test.

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Step 1.7.1.1.1

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

$p = \pm 1, \pm 256, \pm 2, \pm 128, \pm 4, \pm 64, \pm 8, \pm 32, \pm 16$

$q = \pm 1$

Step 1.7.1.1.2

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

$\pm 1, \pm 256, \pm 2, \pm 128, \pm 4, \pm 64, \pm 8, \pm 32, \pm 16$

Step 1.7.1.1.3

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

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Step 1.7.1.1.3.1

Substitute $2$ into the polynomial.

$2^{4} + 12 \cdot 2^{3} + 4 \cdot 2^{2} - 192 \cdot 2 + 256$

Step 1.7.1.1.3.2

Raise $2$ to the power of $4$ .

$16 + 12 \cdot 2^{3} + 4 \cdot 2^{2} - 192 \cdot 2 + 256$

Step 1.7.1.1.3.3

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

$16 + 12 \cdot 8 + 4 \cdot 2^{2} - 192 \cdot 2 + 256$

Step 1.7.1.1.3.4

Multiply $12$ by $8$ .

$16 + 96 + 4 \cdot 2^{2} - 192 \cdot 2 + 256$

Step 1.7.1.1.3.5

Add $16$ and $96$ .

$112 + 4 \cdot 2^{2} - 192 \cdot 2 + 256$

Step 1.7.1.1.3.6

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

$112 + 4 \cdot 4 - 192 \cdot 2 + 256$

Step 1.7.1.1.3.7

Multiply $4$ by $4$ .

$112 + 16 - 192 \cdot 2 + 256$

Step 1.7.1.1.3.8

Add $112$ and $16$ .

$128 - 192 \cdot 2 + 256$

Step 1.7.1.1.3.9

Multiply $- 192$ by $2$ .

$128 - 384 + 256$

Step 1.7.1.1.3.10

Subtract $384$ from $128$ .

$- 256 + 256$

Step 1.7.1.1.3.11

Add $- 256$ and $256$ .

$0$

Step 1.7.1.1.4

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

$\frac{λ^{4} + 12 λ^{3} + 4 λ^{2} - 192 λ + 256}{λ - 2}$

Step 1.7.1.1.5

Divide $λ^{4} + 12 λ^{3} + 4 λ^{2} - 192 λ + 256$ by $λ - 2$ .

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Step 1.7.1.1.5.1

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

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

Step 1.7.1.1.5.2

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

$λ^{3}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

Step 1.7.1.1.5.3

Multiply the new quotient term by the divisor.

$λ^{3}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

Step 1.7.1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $λ^{4} - 2 λ^{3}$

$λ^{3}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

Step 1.7.1.1.5.5

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

$λ^{3}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

Step 1.7.1.1.5.6

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

$λ^{3}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

Step 1.7.1.1.5.7

Divide the highest order term in the dividend $14 λ^{3}$ by the highest order term in divisor $λ$ .

$λ^{3}$

$14 λ^{2}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

Step 1.7.1.1.5.8

Multiply the new quotient term by the divisor.

$λ^{3}$

$14 λ^{2}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

Step 1.7.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $14 λ^{3} - 28 λ^{2}$

$λ^{3}$

$14 λ^{2}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

Step 1.7.1.1.5.10

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

$λ^{3}$

$14 λ^{2}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

Step 1.7.1.1.5.11

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

$λ^{3}$

$14 λ^{2}$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

Step 1.7.1.1.5.12

Divide the highest order term in the dividend $32 λ^{2}$ by the highest order term in divisor $λ$ .

$λ^{3}$

$14 λ^{2}$

$32 λ$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

Step 1.7.1.1.5.13

Multiply the new quotient term by the divisor.

$λ^{3}$

$14 λ^{2}$

$32 λ$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

$32 λ^{2}$

$64 λ$

Step 1.7.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $32 λ^{2} - 64 λ$

$λ^{3}$

$14 λ^{2}$

$32 λ$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

$32 λ^{2}$

$64 λ$

Step 1.7.1.1.5.15

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

$λ^{3}$

$14 λ^{2}$

$32 λ$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

$32 λ^{2}$

$64 λ$

$128 λ$

Step 1.7.1.1.5.16

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

$λ^{3}$

$14 λ^{2}$

$32 λ$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

$32 λ^{2}$

$64 λ$

$128 λ$

$256$

Step 1.7.1.1.5.17

Divide the highest order term in the dividend $- 128 λ$ by the highest order term in divisor $λ$ .

$λ^{3}$

$14 λ^{2}$

$32 λ$

$128$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

$32 λ^{2}$

$64 λ$

$128 λ$

$256$

Step 1.7.1.1.5.18

Multiply the new quotient term by the divisor.

$λ^{3}$

$14 λ^{2}$

$32 λ$

$128$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

$32 λ^{2}$

$64 λ$

$128 λ$

$256$

$128 λ$

$256$

Step 1.7.1.1.5.19

The expression needs to be subtracted from the dividend, so change all the signs in $- 128 λ + 256$

$λ^{3}$

$14 λ^{2}$

$32 λ$

$128$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

$32 λ^{2}$

$64 λ$

$128 λ$

$256$

$128 λ$

$256$

Step 1.7.1.1.5.20

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

$λ^{3}$

$14 λ^{2}$

$32 λ$

$128$

$λ$

$2$

$λ^{4}$

$12 λ^{3}$

$4 λ^{2}$

$192 λ$

$256$

$λ^{4}$

$2 λ^{3}$

$14 λ^{3}$

$4 λ^{2}$

$14 λ^{3}$

$28 λ^{2}$

$32 λ^{2}$

$192 λ$

$32 λ^{2}$

$64 λ$

$128 λ$

$256$

$128 λ$

$256$

$0$

Step 1.7.1.1.5.21

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

$λ^{3} + 14 λ^{2} + 32 λ - 128$

Step 1.7.1.1.6

Write $λ^{4} + 12 λ^{3} + 4 λ^{2} - 192 λ + 256$ as a set of factors.

$(λ - 2) (λ^{3} + 14 λ^{2} + 32 λ - 128) = 0$

Step 1.7.1.2

Factor $λ^{3} + 14 λ^{2} + 32 λ - 128$ using the rational roots test.

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Step 1.7.1.2.1

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

$p = \pm 1, \pm 128, \pm 2, \pm 64, \pm 4, \pm 32, \pm 8, \pm 16$

$q = \pm 1$

Step 1.7.1.2.2

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

$\pm 1, \pm 128, \pm 2, \pm 64, \pm 4, \pm 32, \pm 8, \pm 16$

Step 1.7.1.2.3

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

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Step 1.7.1.2.3.1

Substitute $2$ into the polynomial.

$2^{3} + 14 \cdot 2^{2} + 32 \cdot 2 - 128$

Step 1.7.1.2.3.2

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

$8 + 14 \cdot 2^{2} + 32 \cdot 2 - 128$

Step 1.7.1.2.3.3

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

$8 + 14 \cdot 4 + 32 \cdot 2 - 128$

Step 1.7.1.2.3.4

Multiply $14$ by $4$ .

$8 + 56 + 32 \cdot 2 - 128$

Step 1.7.1.2.3.5

Add $8$ and $56$ .

$64 + 32 \cdot 2 - 128$

Step 1.7.1.2.3.6

Multiply $32$ by $2$ .

$64 + 64 - 128$

Step 1.7.1.2.3.7

Add $64$ and $64$ .

$128 - 128$

Step 1.7.1.2.3.8

Subtract $128$ from $128$ .

$0$

Step 1.7.1.2.4

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

$\frac{λ^{3} + 14 λ^{2} + 32 λ - 128}{λ - 2}$

Step 1.7.1.2.5

Divide $λ^{3} + 14 λ^{2} + 32 λ - 128$ by $λ - 2$ .

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Step 1.7.1.2.5.1

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

$λ$

$2$

$λ^{3}$

$14 λ^{2}$

$32 λ$

$128$

Step 1.7.1.2.5.2

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

					$λ^{2}$
	$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$

Step 1.7.1.2.5.3

Multiply the new quotient term by the divisor.

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			+	$λ^{3}$	-	$2 λ^{2}$

Step 1.7.1.2.5.4

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

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$

Step 1.7.1.2.5.5

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

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$

Step 1.7.1.2.5.6

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

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$

Step 1.7.1.2.5.7

Divide the highest order term in the dividend $16 λ^{2}$ by the highest order term in divisor $λ$ .

				$λ^{2}$	+	$16 λ$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$

Step 1.7.1.2.5.8

Multiply the new quotient term by the divisor.

				$λ^{2}$	+	$16 λ$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$
					+	$16 λ^{2}$	-	$32 λ$

Step 1.7.1.2.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $16 λ^{2} - 32 λ$

				$λ^{2}$	+	$16 λ$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$
					-	$16 λ^{2}$	+	$32 λ$

Step 1.7.1.2.5.10

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

				$λ^{2}$	+	$16 λ$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$
					-	$16 λ^{2}$	+	$32 λ$
							+	$64 λ$

Step 1.7.1.2.5.11

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

				$λ^{2}$	+	$16 λ$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$
					-	$16 λ^{2}$	+	$32 λ$
							+	$64 λ$	-	$128$

Step 1.7.1.2.5.12

Divide the highest order term in the dividend $64 λ$ by the highest order term in divisor $λ$ .

				$λ^{2}$	+	$16 λ$	+	$64$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$
					-	$16 λ^{2}$	+	$32 λ$
							+	$64 λ$	-	$128$

Step 1.7.1.2.5.13

Multiply the new quotient term by the divisor.

				$λ^{2}$	+	$16 λ$	+	$64$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$
					-	$16 λ^{2}$	+	$32 λ$
							+	$64 λ$	-	$128$
							+	$64 λ$	-	$128$

Step 1.7.1.2.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $64 λ - 128$

				$λ^{2}$	+	$16 λ$	+	$64$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$
					-	$16 λ^{2}$	+	$32 λ$
							+	$64 λ$	-	$128$
							-	$64 λ$	+	$128$

Step 1.7.1.2.5.15

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

				$λ^{2}$	+	$16 λ$	+	$64$
$λ$	-	$2$		$λ^{3}$	+	$14 λ^{2}$	+	$32 λ$	-	$128$
			-	$λ^{3}$	+	$2 λ^{2}$
					+	$16 λ^{2}$	+	$32 λ$
					-	$16 λ^{2}$	+	$32 λ$
							+	$64 λ$	-	$128$
							-	$64 λ$	+	$128$
										$0$

Step 1.7.1.2.5.16

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

$λ^{2} + 16 λ + 64$

Step 1.7.1.2.6

Write $λ^{3} + 14 λ^{2} + 32 λ - 128$ as a set of factors.

$(λ - 2) ((λ - 2) (λ^{2} + 16 λ + 64)) = 0$

Step 1.7.1.3

Factor using the perfect square rule.

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Step 1.7.1.3.1

Factor using the perfect square rule.

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Step 1.7.1.3.1.1

Rewrite $64$ as $8^{2}$ .

$(λ - 2) ((λ - 2) (λ^{2} + 16 λ + 8^{2})) = 0$

Step 1.7.1.3.1.2

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

$16 λ = 2 \cdot λ \cdot 8$

Step 1.7.1.3.1.3

Rewrite the polynomial.

$(λ - 2) ((λ - 2) (λ^{2} + 2 \cdot λ \cdot 8 + 8^{2})) = 0$

Step 1.7.1.3.1.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = λ$ and $b = 8$ .

$(λ - 2) ((λ - 2) {(λ + 8)}^{2}) = 0$

Step 1.7.1.3.2

Remove unnecessary parentheses.

$(λ - 2) (λ - 2) {(λ + 8)}^{2} = 0$

Step 1.7.1.4

Combine like factors.

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Step 1.7.1.4.1

Raise $λ - 2$ to the power of $1$ .

$(λ - 2) (λ - 2) {(λ + 8)}^{2} = 0$

Step 1.7.1.4.2

Raise $λ - 2$ to the power of $1$ .

$(λ - 2) (λ - 2) {(λ + 8)}^{2} = 0$

Step 1.7.1.4.3

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

${(λ - 2)}^{1 + 1} {(λ + 8)}^{2} = 0$

Step 1.7.1.4.4

Add $1$ and $1$ .

${(λ - 2)}^{2} {(λ + 8)}^{2} = 0$

Step 1.7.2

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

${(λ - 2)}^{2} = 0$

${(λ + 8)}^{2} = 0$

Step 1.7.3

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

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Step 1.7.3.1

Set ${(λ - 2)}^{2}$ equal to $0$ .

${(λ - 2)}^{2} = 0$

Step 1.7.3.2

Solve ${(λ - 2)}^{2} = 0$ for $λ$ .

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Step 1.7.3.2.1

Set the $λ - 2$ equal to $0$ .

$λ - 2 = 0$

Step 1.7.3.2.2

Add $2$ to both sides of the equation.

$λ = 2$

Step 1.7.4

Set ${(λ + 8)}^{2}$ equal to $0$ and solve for $λ$ .

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Step 1.7.4.1

Set ${(λ + 8)}^{2}$ equal to $0$ .

${(λ + 8)}^{2} = 0$

Step 1.7.4.2

Solve ${(λ + 8)}^{2} = 0$ for $λ$ .

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Step 1.7.4.2.1

Set the $λ + 8$ equal to $0$ .

$λ + 8 = 0$

Step 1.7.4.2.2

Subtract $8$ from both sides of the equation.

$λ = - 8$

Step 1.7.5

The final solution is all the values that make ${(λ - 2)}^{2} {(λ + 8)}^{2} = 0$ true.

$λ = 2, - 8$

Step 2

The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where $N$ is the null space and $I$ is the identity matrix.

$ε_{A} = N (A - λ I_{4})$

Step 3

Find the eigenvector using the eigenvalue $λ = 2$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] - 2 [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply $- 2$ by each element of the matrix.

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $- 2$ by $1$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.5

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.6

Multiply $- 2$ by $1$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.7

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.8

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.9

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.10

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.11

Multiply $- 2$ by $1$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.12

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.13

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.14

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.15

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.16

Multiply $- 2$ by $1$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & - 2 \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} - 3 - 2 & - 5 + 0 & 0 + 0 & 0 + 0 \\ - 5 + 0 & - 3 - 2 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Subtract $2$ from $- 3$ .

$[\begin{array}{c} - 5 & - 5 + 0 & 0 + 0 & 0 + 0 \\ - 5 + 0 & - 3 - 2 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.2

Add $- 5$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 + 0 & 0 + 0 \\ - 5 + 0 & - 3 - 2 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.3

Add $0$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 + 0 \\ - 5 + 0 & - 3 - 2 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.4

Add $0$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 + 0 & - 3 - 2 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.5

Add $- 5$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 3 - 2 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.6

Subtract $2$ from $- 3$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.7

Add $0$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.8

Add $0$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 & 0 \\ 0 + 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.9

Add $0$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 & 0 \\ 0 & 0 + 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.10

Add $0$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 & 0 \\ 0 & 0 & - 3 - 2 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.11

Subtract $2$ from $- 3$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 & 0 \\ 0 & 0 & - 5 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.12

Add $- 5$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 & 0 \\ 0 & 0 & - 5 & - 5 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.13

Add $0$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 & 0 \\ 0 & 0 & - 5 & - 5 \\ 0 & 0 + 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.14

Add $0$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 & 0 \\ 0 & 0 & - 5 & - 5 \\ 0 & 0 & - 5 + 0 & - 3 - 2 \end{array}]$

Step 3.2.3.15

Add $- 5$ and $0$ .

$[\begin{array}{c} - 5 & - 5 & 0 & 0 \\ - 5 & - 5 & 0 & 0 \\ 0 & 0 & - 5 & - 5 \\ 0 & 0 & - 5 & - 3 - 2 \end{array}]$

Step 3.2.3.16

Subtract $2$ from $- 3$ .

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

Step 3.3

Find the null space when $λ = 2$ .

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Step 3.3.1

Write as an augmented matrix for $A x = 0$ .

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

Step 3.3.2

Find the reduced row echelon form.

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Step 3.3.2.1

Multiply each element of $R_{1}$ by $- \frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Step 3.3.2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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

Step 3.3.2.1.2

Simplify $R_{1}$ .

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

Step 3.3.2.2

Perform the row operation $R_{2} = R_{2} + 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 3.3.2.2.2

Simplify $R_{2}$ .

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

Step 3.3.2.3

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 3$ .

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

Step 3.3.2.4

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 3$ a $1$ .

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Step 3.3.2.4.1

Multiply each element of $R_{2}$ by $- \frac{1}{5}$ to make the entry at $2, 3$ a $1$ .

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

Step 3.3.2.4.2

Simplify $R_{2}$ .

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

Step 3.3.2.5

Perform the row operation $R_{4} = R_{4} + 5 R_{2}$ to make the entry at $4, 3$ a $0$ .

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Step 3.3.2.5.1

Perform the row operation $R_{4} = R_{4} + 5 R_{2}$ to make the entry at $4, 3$ a $0$ .

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

Step 3.3.2.5.2

Simplify $R_{4}$ .

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

Step 3.3.3

Use the result matrix to declare the final solution to the system of equations.

$x_{1} + x_{2} = 0$

$x_{3} + x_{4} = 0$

$0 = 0$

Step 3.3.4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} - x_{2} \\ x_{2} \\ - x_{4} \\ x_{4} \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

${x_{2} [\begin{array}{c} - 1 \\ 1 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 0 \\ 0 \\ - 1 \\ 1 \end{array}] | x_{2}, x_{4} \in R}$

Step 3.3.7

The solution is the set of vectors created from the free variables of the system.

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

Step 4

Find the eigenvector using the eigenvalue $λ = - 8$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + 8 [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 4.2

Simplify.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $8$ by each element of the matrix.

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 \cdot 1 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $8$ by $1$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.5

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 \cdot 1 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.6

Multiply $8$ by $1$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 8 \cdot 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.7

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.8

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.9

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 0 & 8 \cdot 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.10

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 0 & 0 & 8 \cdot 1 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.11

Multiply $8$ by $1$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 0 & 0 & 8 & 8 \cdot 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.12

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 0 & 0 & 8 & 0 \\ 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.13

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 8 \cdot 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.14

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & 8 \cdot 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.15

Multiply $8$ by $0$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & 0 & 8 \cdot 1 \end{array}]$

Step 4.2.1.2.16

Multiply $8$ by $1$ .

$[\begin{array}{c} - 3 & - 5 & 0 & 0 \\ - 5 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 5 \\ 0 & 0 & - 5 & - 3 \end{array}] + [\begin{array}{c} 8 & 0 & 0 & 0 \\ 0 & 8 & 0 & 0 \\ 0 & 0 & 8 & 0 \\ 0 & 0 & 0 & 8 \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} - 3 + 8 & - 5 + 0 & 0 + 0 & 0 + 0 \\ - 5 + 0 & - 3 + 8 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Add $- 3$ and $8$ .

$[\begin{array}{c} 5 & - 5 + 0 & 0 + 0 & 0 + 0 \\ - 5 + 0 & - 3 + 8 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.2

Add $- 5$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 + 0 & 0 + 0 \\ - 5 + 0 & - 3 + 8 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.3

Add $0$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 + 0 \\ - 5 + 0 & - 3 + 8 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.4

Add $0$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 + 0 & - 3 + 8 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.5

Add $- 5$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & - 3 + 8 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.6

Add $- 3$ and $8$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 + 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.7

Add $0$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 & 0 + 0 \\ 0 + 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.8

Add $0$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 & 0 \\ 0 + 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.9

Add $0$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 & 0 \\ 0 & 0 + 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.10

Add $0$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 & 0 \\ 0 & 0 & - 3 + 8 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.11

Add $- 3$ and $8$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 & 0 \\ 0 & 0 & 5 & - 5 + 0 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.12

Add $- 5$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 & 0 \\ 0 & 0 & 5 & - 5 \\ 0 + 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.13

Add $0$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 & 0 \\ 0 & 0 & 5 & - 5 \\ 0 & 0 + 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.14

Add $0$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 & 0 \\ 0 & 0 & 5 & - 5 \\ 0 & 0 & - 5 + 0 & - 3 + 8 \end{array}]$

Step 4.2.3.15

Add $- 5$ and $0$ .

$[\begin{array}{c} 5 & - 5 & 0 & 0 \\ - 5 & 5 & 0 & 0 \\ 0 & 0 & 5 & - 5 \\ 0 & 0 & - 5 & - 3 + 8 \end{array}]$

Step 4.2.3.16

Add $- 3$ and $8$ .

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

Step 4.3

Find the null space when $λ = - 8$ .

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Step 4.3.1

Write as an augmented matrix for $A x = 0$ .

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

Step 4.3.2

Find the reduced row echelon form.

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Step 4.3.2.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Step 4.3.2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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

Step 4.3.2.1.2

Simplify $R_{1}$ .

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

Step 4.3.2.2

Perform the row operation $R_{2} = R_{2} + 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 5 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 4.3.2.2.2

Simplify $R_{2}$ .

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

Step 4.3.2.3

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 3$ .

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

Step 4.3.2.4

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 3$ a $1$ .

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Step 4.3.2.4.1

Multiply each element of $R_{2}$ by $\frac{1}{5}$ to make the entry at $2, 3$ a $1$ .

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

Step 4.3.2.4.2

Simplify $R_{2}$ .

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

Step 4.3.2.5

Perform the row operation $R_{4} = R_{4} + 5 R_{2}$ to make the entry at $4, 3$ a $0$ .

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Step 4.3.2.5.1

Perform the row operation $R_{4} = R_{4} + 5 R_{2}$ to make the entry at $4, 3$ a $0$ .

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

Step 4.3.2.5.2

Simplify $R_{4}$ .

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

Step 4.3.3

Use the result matrix to declare the final solution to the system of equations.

$x_{1} - x_{2} = 0$

$x_{3} - x_{4} = 0$

$0 = 0$

Step 4.3.4

Write a solution vector by solving in terms of the free variables in each row.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} x_{2} \\ x_{2} \\ x_{4} \\ x_{4} \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = x_{2} [\begin{array}{c} 1 \\ 1 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 0 \\ 0 \\ 1 \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${x_{2} [\begin{array}{c} 1 \\ 1 \\ 0 \\ 0 \end{array}] + x_{4} [\begin{array}{c} 0 \\ 0 \\ 1 \\ 1 \end{array}] | x_{2}, x_{4} \in R}$

Step 4.3.7

The solution is the set of vectors created from the free variables of the system.

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

Step 5

The eigenspace of $A$ is the list of the vector space for each eigenvalue.

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