Linear Algebra Examples

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}]$

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 $[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}]$ for $A$ .

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

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 $I_{4}$ .

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

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.

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

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.5

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.5.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.5.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.6

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.9

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.9.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.9.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.10

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.10.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.10.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.11

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.12

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.12.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.12.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.13

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.13.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.13.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.14

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.14.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.14.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.15

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.15.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.15.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.16

Multiply $- 1$ by $1$ .

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

Step 1.4.2

Add the corresponding elements.

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

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $0$ and $0$ .

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

Step 1.4.3.2

Add $0$ and $0$ .

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

Step 1.4.3.3

Add $0$ and $0$ .

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

Step 1.4.3.4

Add $0$ and $0$ .

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

Step 1.4.3.5

Add $5$ and $0$ .

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

Step 1.4.3.6

Add $- 10$ and $0$ .

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

Step 1.4.3.7

Add $1$ and $0$ .

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

Step 1.4.3.8

Add $0$ and $0$ .

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

Step 1.4.3.9

Add $0$ and $0$ .

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

Step 1.4.3.10

Add $1$ and $0$ .

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

Step 1.4.3.11

Add $0$ and $0$ .

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

Step 1.4.3.12

Add $0$ and $0$ .

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

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.

$| \begin{array}{c} 1 - λ & 5 & - 10 \\ 0 & 2 - λ & 0 \\ 0 & 0 & 3 - λ \end{array} |$

Step 1.5.1.4

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

$(1 - λ) | \begin{array}{c} 1 - λ & 5 & - 10 \\ 0 & 2 - λ & 0 \\ 0 & 0 & 3 - λ \end{array} |$

Step 1.5.1.5

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

$| \begin{array}{c} 0 & 5 & - 10 \\ 1 & 2 - λ & 0 \\ 1 & 0 & 3 - λ \end{array} |$

Step 1.5.1.6

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

$0 | \begin{array}{c} 0 & 5 & - 10 \\ 1 & 2 - λ & 0 \\ 1 & 0 & 3 - λ \end{array} |$

Step 1.5.1.7

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

$| \begin{array}{c} 0 & 1 - λ & - 10 \\ 1 & 0 & 0 \\ 1 & 0 & 3 - λ \end{array} |$

Step 1.5.1.8

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

$0 | \begin{array}{c} 0 & 1 - λ & - 10 \\ 1 & 0 & 0 \\ 1 & 0 & 3 - λ \end{array} |$

Step 1.5.1.9

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

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

Step 1.5.1.10

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

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

Step 1.5.1.11

Add the terms together.

$p (λ) = (1 - λ) | \begin{array}{c} 1 - λ & 5 & - 10 \\ 0 & 2 - λ & 0 \\ 0 & 0 & 3 - λ \end{array} | + 0 | \begin{array}{c} 0 & 5 & - 10 \\ 1 & 2 - λ & 0 \\ 1 & 0 & 3 - λ \end{array} | + 0 | \begin{array}{c} 0 & 1 - λ & - 10 \\ 1 & 0 & 0 \\ 1 & 0 & 3 - λ \end{array} | + 0 | \begin{array}{c} 0 & 1 - λ & 5 \\ 1 & 0 & 2 - λ \\ 1 & 0 & 0 \end{array} |$

Step 1.5.2

Multiply $0$ by $| \begin{array}{c} 0 & 5 & - 10 \\ 1 & 2 - λ & 0 \\ 1 & 0 & 3 - λ \end{array} |$ .

$p (λ) = (1 - λ) | \begin{array}{c} 1 - λ & 5 & - 10 \\ 0 & 2 - λ & 0 \\ 0 & 0 & 3 - λ \end{array} | + 0 + 0 | \begin{array}{c} 0 & 1 - λ & - 10 \\ 1 & 0 & 0 \\ 1 & 0 & 3 - λ \end{array} | + 0 | \begin{array}{c} 0 & 1 - λ & 5 \\ 1 & 0 & 2 - λ \\ 1 & 0 & 0 \end{array} |$

Step 1.5.3

Multiply $0$ by $| \begin{array}{c} 0 & 1 - λ & - 10 \\ 1 & 0 & 0 \\ 1 & 0 & 3 - λ \end{array} |$ .

$p (λ) = (1 - λ) | \begin{array}{c} 1 - λ & 5 & - 10 \\ 0 & 2 - λ & 0 \\ 0 & 0 & 3 - λ \end{array} | + 0 + 0 + 0 | \begin{array}{c} 0 & 1 - λ & 5 \\ 1 & 0 & 2 - λ \\ 1 & 0 & 0 \end{array} |$

Step 1.5.4

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

$p (λ) = (1 - λ) | \begin{array}{c} 1 - λ & 5 & - 10 \\ 0 & 2 - λ & 0 \\ 0 & 0 & 3 - λ \end{array} | + 0 + 0 + 0$

Step 1.5.5

Evaluate $| \begin{array}{c} 1 - λ & 5 & - 10 \\ 0 & 2 - λ & 0 \\ 0 & 0 & 3 - λ \end{array} |$ .

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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 column $1$ 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.

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

Step 1.5.5.1.4

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

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

Step 1.5.5.1.5

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

$| \begin{array}{c} 5 & - 10 \\ 0 & 3 - λ \end{array} |$

Step 1.5.5.1.6

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

$0 | \begin{array}{c} 5 & - 10 \\ 0 & 3 - λ \end{array} |$

Step 1.5.5.1.7

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} 5 & - 10 \\ 2 - λ & 0 \end{array} |$

Step 1.5.5.1.8

Multiply element $a_{31}$ by its cofactor.

$0 | \begin{array}{c} 5 & - 10 \\ 2 - λ & 0 \end{array} |$

Step 1.5.5.1.9

Add the terms together.

$p (λ) = (1 - λ) ((1 - λ) | \begin{array}{c} 2 - λ & 0 \\ 0 & 3 - λ \end{array} | + 0 | \begin{array}{c} 5 & - 10 \\ 0 & 3 - λ \end{array} | + 0 | \begin{array}{c} 5 & - 10 \\ 2 - λ & 0 \end{array} |) + 0 + 0 + 0$

Step 1.5.5.2

Multiply $0$ by $| \begin{array}{c} 5 & - 10 \\ 0 & 3 - λ \end{array} |$ .

$p (λ) = (1 - λ) ((1 - λ) | \begin{array}{c} 2 - λ & 0 \\ 0 & 3 - λ \end{array} | + 0 + 0 | \begin{array}{c} 5 & - 10 \\ 2 - λ & 0 \end{array} |) + 0 + 0 + 0$

Step 1.5.5.3

Multiply $0$ by $| \begin{array}{c} 5 & - 10 \\ 2 - λ & 0 \end{array} |$ .

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

Step 1.5.5.4

Evaluate $| \begin{array}{c} 2 - λ & 0 \\ 0 & 3 - λ \end{array} |$ .

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Step 1.5.5.4.1

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

$p (λ) = (1 - λ) ((1 - λ) ((2 - λ) (3 - λ) + 0 \cdot 0) + 0 + 0) + 0 + 0 + 0$

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 $(2 - λ) (3 - λ)$ using the FOIL Method.

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Step 1.5.5.4.2.1.1.1

Apply the distributive property.

$p (λ) = (1 - λ) ((1 - λ) (2 (3 - λ) - λ (3 - λ) + 0 \cdot 0) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.4.2.1.1.2

Apply the distributive property.

$p (λ) = (1 - λ) ((1 - λ) (2 \cdot 3 + 2 (- λ) - λ (3 - λ) + 0 \cdot 0) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.4.2.1.1.3

Apply the distributive property.

$p (λ) = (1 - λ) ((1 - λ) (2 \cdot 3 + 2 (- λ) - λ \cdot 3 - λ (- λ) + 0 \cdot 0) + 0 + 0) + 0 + 0 + 0$

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 $2$ by $3$ .

$p (λ) = (1 - λ) ((1 - λ) (6 + 2 (- λ) - λ \cdot 3 - λ (- λ) + 0 \cdot 0) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.4.2.1.2.1.2

Multiply $- 1$ by $2$ .

$p (λ) = (1 - λ) ((1 - λ) (6 - 2 λ - λ \cdot 3 - λ (- λ) + 0 \cdot 0) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.4.2.1.2.1.3

Multiply $3$ by $- 1$ .

$p (λ) = (1 - λ) ((1 - λ) (6 - 2 λ - 3 λ - λ (- λ) + 0 \cdot 0) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (1 - λ) ((1 - λ) (6 - 2 λ - 3 λ - 1 \cdot - 1 λ \cdot λ + 0 \cdot 0) + 0 + 0) + 0 + 0 + 0$

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 $λ$ .

$p (λ) = (1 - λ) ((1 - λ) (6 - 2 λ - 3 λ - 1 \cdot - 1 (λ \cdot λ) + 0 \cdot 0) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

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

Step 1.5.5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

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

Step 1.5.5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

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

Step 1.5.5.4.2.1.2.2

Subtract $3 λ$ from $- 2 λ$ .

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

Step 1.5.5.4.2.1.3

Multiply $0$ by $0$ .

$p (λ) = (1 - λ) ((1 - λ) (6 - 5 λ + λ^{2} + 0) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.4.2.2

Add $6 - 5 λ + λ^{2}$ and $0$ .

$p (λ) = (1 - λ) ((1 - λ) (6 - 5 λ + λ^{2}) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.4.2.3

Move $6$ .

$p (λ) = (1 - λ) ((1 - λ) (- 5 λ + λ^{2} + 6) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.4.2.4

Reorder $- 5 λ$ and $λ^{2}$ .

$p (λ) = (1 - λ) ((1 - λ) (λ^{2} - 5 λ + 6) + 0 + 0) + 0 + 0 + 0$

Step 1.5.5.5

Simplify the determinant.

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Step 1.5.5.5.1

Combine the opposite terms in $(1 - λ) (λ^{2} - 5 λ + 6) + 0 + 0$ .

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Step 1.5.5.5.1.1

Add $(1 - λ) (λ^{2} - 5 λ + 6)$ and $0$ .

$p (λ) = (1 - λ) ((1 - λ) (λ^{2} - 5 λ + 6) + 0) + 0 + 0 + 0$

Step 1.5.5.5.1.2

Add $(1 - λ) (λ^{2} - 5 λ + 6)$ and $0$ .

$p (λ) = (1 - λ) ((1 - λ) (λ^{2} - 5 λ + 6)) + 0 + 0 + 0$

Step 1.5.5.5.2

Expand $(1 - λ) (λ^{2} - 5 λ + 6)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 1.5.5.5.3

Simplify each term.

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Step 1.5.5.5.3.1

Multiply $λ^{2}$ by $1$ .

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

Step 1.5.5.5.3.2

Multiply $- 5 λ$ by $1$ .

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

Step 1.5.5.5.3.3

Multiply $6$ by $1$ .

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

Step 1.5.5.5.3.4

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

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Step 1.5.5.5.3.4.1

Move $λ^{2}$ .

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

Step 1.5.5.5.3.4.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.5.5.3.4.2.1

Raise $λ$ to the power of $1$ .

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

Step 1.5.5.5.3.4.2.2

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

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

Step 1.5.5.5.3.4.3

Add $2$ and $1$ .

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

Step 1.5.5.5.3.5

Rewrite using the commutative property of multiplication.

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

Step 1.5.5.5.3.6

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

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Step 1.5.5.5.3.6.1

Move $λ$ .

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

Step 1.5.5.5.3.6.2

Multiply $λ$ by $λ$ .

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

Step 1.5.5.5.3.7

Multiply $- 1$ by $- 5$ .

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

Step 1.5.5.5.3.8

Multiply $6$ by $- 1$ .

$p (λ) = (1 - λ) (λ^{2} - 5 λ + 6 - λ^{3} + 5 λ^{2} - 6 λ) + 0 + 0 + 0$

Step 1.5.5.5.4

Add $λ^{2}$ and $5 λ^{2}$ .

$p (λ) = (1 - λ) (6 λ^{2} - 5 λ + 6 - λ^{3} - 6 λ) + 0 + 0 + 0$

Step 1.5.5.5.5

Subtract $6 λ$ from $- 5 λ$ .

$p (λ) = (1 - λ) (6 λ^{2} - 11 λ + 6 - λ^{3}) + 0 + 0 + 0$

Step 1.5.5.5.6

Move $6$ .

$p (λ) = (1 - λ) (6 λ^{2} - 11 λ - λ^{3} + 6) + 0 + 0 + 0$

Step 1.5.5.5.7

Move $- 11 λ$ .

$p (λ) = (1 - λ) (6 λ^{2} - λ^{3} - 11 λ + 6) + 0 + 0 + 0$

Step 1.5.5.5.8

Reorder $6 λ^{2}$ and $- λ^{3}$ .

$p (λ) = (1 - λ) (- λ^{3} + 6 λ^{2} - 11 λ + 6) + 0 + 0 + 0$

Step 1.5.6

Simplify the determinant.

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Step 1.5.6.1

Combine the opposite terms in $(1 - λ) (- λ^{3} + 6 λ^{2} - 11 λ + 6) + 0 + 0 + 0$ .

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Step 1.5.6.1.1

Add $(1 - λ) (- λ^{3} + 6 λ^{2} - 11 λ + 6)$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 6 λ^{2} - 11 λ + 6) + 0 + 0$

Step 1.5.6.1.2

Add $(1 - λ) (- λ^{3} + 6 λ^{2} - 11 λ + 6)$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 6 λ^{2} - 11 λ + 6) + 0$

Step 1.5.6.1.3

Add $(1 - λ) (- λ^{3} + 6 λ^{2} - 11 λ + 6)$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 6 λ^{2} - 11 λ + 6)$

Step 1.5.6.2

Expand $(1 - λ) (- λ^{3} + 6 λ^{2} - 11 λ + 6)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 1 (- λ^{3}) + 1 (6 λ^{2}) + 1 (- 11 λ) + 1 \cdot 6 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3

Simplify each term.

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Step 1.5.6.3.1

Multiply $- λ^{3}$ by $1$ .

$p (λ) = - λ^{3} + 1 (6 λ^{2}) + 1 (- 11 λ) + 1 \cdot 6 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.2

Multiply $6 λ^{2}$ by $1$ .

$p (λ) = - λ^{3} + 6 λ^{2} + 1 (- 11 λ) + 1 \cdot 6 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.3

Multiply $- 11 λ$ by $1$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 1 \cdot 6 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.4

Multiply $6$ by $1$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.5

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 - 1 \cdot - 1 λ \cdot λ^{3} - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.6

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

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Step 1.5.6.3.6.1

Move $λ^{3}$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 - 1 \cdot - 1 (λ^{3} λ) - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.6.2

Multiply $λ^{3}$ by $λ$ .

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Step 1.5.6.3.6.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 - 1 \cdot - 1 (λ^{3} λ^{1}) - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.6.2.2

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

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 - 1 \cdot - 1 λ^{3 + 1} - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.6.3

Add $3$ and $1$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 - 1 \cdot - 1 λ^{4} - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.7

Multiply $- 1$ by $- 1$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + 1 λ^{4} - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.8

Multiply $λ^{4}$ by $1$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - λ (6 λ^{2}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.9

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 1 \cdot 6 λ \cdot λ^{2} - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.10

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

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Step 1.5.6.3.10.1

Move $λ^{2}$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 1 \cdot 6 (λ^{2} λ) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.10.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.6.3.10.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 1 \cdot 6 (λ^{2} λ^{1}) - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.10.2.2

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

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 1 \cdot 6 λ^{2 + 1} - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.10.3

Add $2$ and $1$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 1 \cdot 6 λ^{3} - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.11

Multiply $- 1$ by $6$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 6 λ^{3} - λ (- 11 λ) - λ \cdot 6$

Step 1.5.6.3.12

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 6 λ^{3} - 1 \cdot - 11 λ \cdot λ - λ \cdot 6$

Step 1.5.6.3.13

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

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Step 1.5.6.3.13.1

Move $λ$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 6 λ^{3} - 1 \cdot - 11 (λ \cdot λ) - λ \cdot 6$

Step 1.5.6.3.13.2

Multiply $λ$ by $λ$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 6 λ^{3} - 1 \cdot - 11 λ^{2} - λ \cdot 6$

Step 1.5.6.3.14

Multiply $- 1$ by $- 11$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 6 λ^{3} + 11 λ^{2} - λ \cdot 6$

Step 1.5.6.3.15

Multiply $6$ by $- 1$ .

$p (λ) = - λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} - 6 λ^{3} + 11 λ^{2} - 6 λ$

Step 1.5.6.4

Subtract $6 λ^{3}$ from $- λ^{3}$ .

$p (λ) = - 7 λ^{3} + 6 λ^{2} - 11 λ + 6 + λ^{4} + 11 λ^{2} - 6 λ$

Step 1.5.6.5

Add $6 λ^{2}$ and $11 λ^{2}$ .

$p (λ) = - 7 λ^{3} + 17 λ^{2} - 11 λ + 6 + λ^{4} - 6 λ$

Step 1.5.6.6

Subtract $6 λ$ from $- 11 λ$ .

$p (λ) = - 7 λ^{3} + 17 λ^{2} - 17 λ + 6 + λ^{4}$

Step 1.5.6.7

Move $6$ .

$p (λ) = - 7 λ^{3} + 17 λ^{2} - 17 λ + λ^{4} + 6$

Step 1.5.6.8

Move $- 17 λ$ .

$p (λ) = - 7 λ^{3} + 17 λ^{2} + λ^{4} - 17 λ + 6$

Step 1.5.6.9

Move $17 λ^{2}$ .

$p (λ) = - 7 λ^{3} + λ^{4} + 17 λ^{2} - 17 λ + 6$

Step 1.5.6.10

Reorder $- 7 λ^{3}$ and $λ^{4}$ .

$p (λ) = λ^{4} - 7 λ^{3} + 17 λ^{2} - 17 λ + 6$

Step 1.6

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

$λ^{4} - 7 λ^{3} + 17 λ^{2} - 17 λ + 6 = 0$

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

Regroup terms.

$17 λ^{2} - 17 λ + λ^{4} - 7 λ^{3} + 6 = 0$

Step 1.7.1.2

Factor $17 λ$ out of $17 λ^{2} - 17 λ$ .

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Step 1.7.1.2.1

Factor $17 λ$ out of $17 λ^{2}$ .

$17 λ (λ) - 17 λ + λ^{4} - 7 λ^{3} + 6 = 0$

Step 1.7.1.2.2

Factor $17 λ$ out of $- 17 λ$ .

$17 λ (λ) + 17 λ (- 1) + λ^{4} - 7 λ^{3} + 6 = 0$

Step 1.7.1.2.3

Factor $17 λ$ out of $17 λ (λ) + 17 λ (- 1)$ .

$17 λ (λ - 1) + λ^{4} - 7 λ^{3} + 6 = 0$

Step 1.7.1.3

Factor $λ^{4} - 7 λ^{3} + 6$ using the rational roots test.

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Step 1.7.1.3.1

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.

$p = \pm 1, \pm 6, \pm 2, \pm 3$

$q = \pm 1$

Step 1.7.1.3.2

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

$\pm 1, \pm 6, \pm 2, \pm 3$

Step 1.7.1.3.3

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

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Step 1.7.1.3.3.1

Substitute $1$ into the polynomial.

$1^{4} - 7 \cdot 1^{3} + 6$

Step 1.7.1.3.3.2

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

$1 - 7 \cdot 1^{3} + 6$

Step 1.7.1.3.3.3

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

$1 - 7 \cdot 1 + 6$

Step 1.7.1.3.3.4

Multiply $- 7$ by $1$ .

$1 - 7 + 6$

Step 1.7.1.3.3.5

Subtract $7$ from $1$ .

$- 6 + 6$

Step 1.7.1.3.3.6

Add $- 6$ and $6$ .

$0$

Step 1.7.1.3.4

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

$\frac{λ^{4} - 7 λ^{3} + 6}{λ - 1}$

Step 1.7.1.3.5

Divide $λ^{4} - 7 λ^{3} + 6$ by $λ - 1$ .

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Step 1.7.1.3.5.1

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

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

Step 1.7.1.3.5.2

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

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

Step 1.7.1.3.5.3

Multiply the new quotient term by the divisor.

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

Step 1.7.1.3.5.4

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

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

Step 1.7.1.3.5.5

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

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

Step 1.7.1.3.5.6

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

$λ^{3}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

Step 1.7.1.3.5.7

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

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

Step 1.7.1.3.5.8

Multiply the new quotient term by the divisor.

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

Step 1.7.1.3.5.9

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

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

Step 1.7.1.3.5.10

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

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

Step 1.7.1.3.5.11

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

$λ^{3}$

$6 λ^{2}$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

Step 1.7.1.3.5.12

Divide the highest order term in the dividend $- 6 λ^{2}$ by the highest order term in divisor $λ$ .

$λ^{3}$

$6 λ^{2}$

$6 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

Step 1.7.1.3.5.13

Multiply the new quotient term by the divisor.

$λ^{3}$

$6 λ^{2}$

$6 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

$6 λ^{2}$

$6 λ$

Step 1.7.1.3.5.14

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

$λ^{3}$

$6 λ^{2}$

$6 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

$6 λ^{2}$

$6 λ$

Step 1.7.1.3.5.15

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

$λ^{3}$

$6 λ^{2}$

$6 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

$6 λ^{2}$

$6 λ$

Step 1.7.1.3.5.16

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

$λ^{3}$

$6 λ^{2}$

$6 λ$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

$6 λ^{2}$

$6 λ$

$6$

Step 1.7.1.3.5.17

Divide the highest order term in the dividend $- 6 λ$ by the highest order term in divisor $λ$ .

$λ^{3}$

$6 λ^{2}$

$6 λ$

$6$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

$6 λ^{2}$

$6 λ$

$6$

Step 1.7.1.3.5.18

Multiply the new quotient term by the divisor.

$λ^{3}$

$6 λ^{2}$

$6 λ$

$6$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

$6 λ^{2}$

$6 λ$

$6$

$6 λ$

$6$

Step 1.7.1.3.5.19

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

$λ^{3}$

$6 λ^{2}$

$6 λ$

$6$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

$6 λ^{2}$

$6 λ$

$6$

$6 λ$

$6$

Step 1.7.1.3.5.20

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

$λ^{3}$

$6 λ^{2}$

$6 λ$

$6$

$λ$

$1$

$λ^{4}$

$7 λ^{3}$

$0 λ^{2}$

$0 λ$

$6$

$λ^{4}$

$λ^{3}$

$6 λ^{3}$

$0 λ^{2}$

$6 λ^{3}$

$6 λ^{2}$

$0 λ$

$6 λ^{2}$

$6 λ$

$6$

$6 λ$

$6$

$0$

Step 1.7.1.3.5.21

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

$λ^{3} - 6 λ^{2} - 6 λ - 6$

Step 1.7.1.3.6

Write $λ^{4} - 7 λ^{3} + 6$ as a set of factors.

$17 λ (λ - 1) + (λ - 1) (λ^{3} - 6 λ^{2} - 6 λ - 6) = 0$

Step 1.7.1.4

Factor $λ - 1$ out of $17 λ (λ - 1) + (λ - 1) (λ^{3} - 6 λ^{2} - 6 λ - 6)$ .

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Step 1.7.1.4.1

Factor $λ - 1$ out of $17 λ (λ - 1)$ .

$(λ - 1) (17 λ) + (λ - 1) (λ^{3} - 6 λ^{2} - 6 λ - 6) = 0$

Step 1.7.1.4.2

Factor $λ - 1$ out of $(λ - 1) (17 λ) + (λ - 1) (λ^{3} - 6 λ^{2} - 6 λ - 6)$ .

$(λ - 1) (17 λ + λ^{3} - 6 λ^{2} - 6 λ - 6) = 0$

Step 1.7.1.5

Subtract $6 λ$ from $17 λ$ .

$(λ - 1) (λ^{3} - 6 λ^{2} + 11 λ - 6) = 0$

Step 1.7.1.6

Factor.

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Step 1.7.1.6.1

Rewrite $λ^{3} - 6 λ^{2} + 11 λ - 6$ in a factored form.

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Step 1.7.1.6.1.1

Factor $λ^{3} - 6 λ^{2} + 11 λ - 6$ using the rational roots test.

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Step 1.7.1.6.1.1.1

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.

$p = \pm 1, \pm 6, \pm 2, \pm 3$

$q = \pm 1$

Step 1.7.1.6.1.1.2

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

$\pm 1, \pm 6, \pm 2, \pm 3$

Step 1.7.1.6.1.1.3

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

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Step 1.7.1.6.1.1.3.1

Substitute $1$ into the polynomial.

$1^{3} - 6 \cdot 1^{2} + 11 \cdot 1 - 6$

Step 1.7.1.6.1.1.3.2

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

$1 - 6 \cdot 1^{2} + 11 \cdot 1 - 6$

Step 1.7.1.6.1.1.3.3

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

$1 - 6 \cdot 1 + 11 \cdot 1 - 6$

Step 1.7.1.6.1.1.3.4

Multiply $- 6$ by $1$ .

$1 - 6 + 11 \cdot 1 - 6$

Step 1.7.1.6.1.1.3.5

Subtract $6$ from $1$ .

$- 5 + 11 \cdot 1 - 6$

Step 1.7.1.6.1.1.3.6

Multiply $11$ by $1$ .

$- 5 + 11 - 6$

Step 1.7.1.6.1.1.3.7

Add $- 5$ and $11$ .

$6 - 6$

Step 1.7.1.6.1.1.3.8

Subtract $6$ from $6$ .

$0$

Step 1.7.1.6.1.1.4

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

$\frac{λ^{3} - 6 λ^{2} + 11 λ - 6}{λ - 1}$

Step 1.7.1.6.1.1.5

Divide $λ^{3} - 6 λ^{2} + 11 λ - 6$ by $λ - 1$ .

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Step 1.7.1.6.1.1.5.1

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

$λ$

$1$

$λ^{3}$

$6 λ^{2}$

$11 λ$

$6$

Step 1.7.1.6.1.1.5.2

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

					$λ^{2}$
	$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$

Step 1.7.1.6.1.1.5.3

Multiply the new quotient term by the divisor.

				$λ^{2}$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			+	$λ^{3}$	-	$λ^{2}$

Step 1.7.1.6.1.1.5.4

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

				$λ^{2}$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$

Step 1.7.1.6.1.1.5.5

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

				$λ^{2}$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$

Step 1.7.1.6.1.1.5.6

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

				$λ^{2}$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$

Step 1.7.1.6.1.1.5.7

Divide the highest order term in the dividend $- 5 λ^{2}$ by the highest order term in divisor $λ$ .

				$λ^{2}$	-	$5 λ$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$

Step 1.7.1.6.1.1.5.8

Multiply the new quotient term by the divisor.

				$λ^{2}$	-	$5 λ$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$
					-	$5 λ^{2}$	+	$5 λ$

Step 1.7.1.6.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 λ^{2} + 5 λ$

				$λ^{2}$	-	$5 λ$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$
					+	$5 λ^{2}$	-	$5 λ$

Step 1.7.1.6.1.1.5.10

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

				$λ^{2}$	-	$5 λ$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$
					+	$5 λ^{2}$	-	$5 λ$
							+	$6 λ$

Step 1.7.1.6.1.1.5.11

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

				$λ^{2}$	-	$5 λ$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$
					+	$5 λ^{2}$	-	$5 λ$
							+	$6 λ$	-	$6$

Step 1.7.1.6.1.1.5.12

Divide the highest order term in the dividend $6 λ$ by the highest order term in divisor $λ$ .

				$λ^{2}$	-	$5 λ$	+	$6$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$
					+	$5 λ^{2}$	-	$5 λ$
							+	$6 λ$	-	$6$

Step 1.7.1.6.1.1.5.13

Multiply the new quotient term by the divisor.

				$λ^{2}$	-	$5 λ$	+	$6$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$
					+	$5 λ^{2}$	-	$5 λ$
							+	$6 λ$	-	$6$
							+	$6 λ$	-	$6$

Step 1.7.1.6.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $6 λ - 6$

				$λ^{2}$	-	$5 λ$	+	$6$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$
					+	$5 λ^{2}$	-	$5 λ$
							+	$6 λ$	-	$6$
							-	$6 λ$	+	$6$

Step 1.7.1.6.1.1.5.15

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

				$λ^{2}$	-	$5 λ$	+	$6$
$λ$	-	$1$		$λ^{3}$	-	$6 λ^{2}$	+	$11 λ$	-	$6$
			-	$λ^{3}$	+	$λ^{2}$
					-	$5 λ^{2}$	+	$11 λ$
					+	$5 λ^{2}$	-	$5 λ$
							+	$6 λ$	-	$6$
							-	$6 λ$	+	$6$
										$0$

Step 1.7.1.6.1.1.5.16

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

$λ^{2} - 5 λ + 6$

Step 1.7.1.6.1.1.6

Write $λ^{3} - 6 λ^{2} + 11 λ - 6$ as a set of factors.

$(λ - 1) ((λ - 1) (λ^{2} - 5 λ + 6)) = 0$

Step 1.7.1.6.1.2

Factor $λ^{2} - 5 λ + 6$ using the AC method.

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Step 1.7.1.6.1.2.1

Factor $λ^{2} - 5 λ + 6$ using the AC method.

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Step 1.7.1.6.1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 5$ .

$- 3, - 2$

Step 1.7.1.6.1.2.1.2

Write the factored form using these integers.

$(λ - 1) ((λ - 1) ((λ - 3) (λ - 2))) = 0$

Step 1.7.1.6.1.2.2

Remove unnecessary parentheses.

$(λ - 1) ((λ - 1) (λ - 3) (λ - 2)) = 0$

Step 1.7.1.6.2

Remove unnecessary parentheses.

$(λ - 1) (λ - 1) (λ - 3) (λ - 2) = 0$

Step 1.7.1.7

Combine exponents.

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Step 1.7.1.7.1

Raise $λ - 1$ to the power of $1$ .

$(λ - 1) (λ - 1) (λ - 3) (λ - 2) = 0$

Step 1.7.1.7.2

Raise $λ - 1$ to the power of $1$ .

$(λ - 1) (λ - 1) (λ - 3) (λ - 2) = 0$

Step 1.7.1.7.3

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

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

Step 1.7.1.7.4

Add $1$ and $1$ .

${(λ - 1)}^{2} (λ - 3) (λ - 2) = 0$

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 $0$ .

${(λ - 1)}^{2} = 0$

$λ - 3 = 0$

$λ - 2 = 0$

Step 1.7.3

Set ${(λ - 1)}^{2}$ equal to $0$ and solve for $λ$ .

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Step 1.7.3.1

Set ${(λ - 1)}^{2}$ equal to $0$ .

${(λ - 1)}^{2} = 0$

Step 1.7.3.2

Solve ${(λ - 1)}^{2} = 0$ for $λ$ .

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Step 1.7.3.2.1

Set the $λ - 1$ equal to $0$ .

$λ - 1 = 0$

Step 1.7.3.2.2

Add $1$ to both sides of the equation.

$λ = 1$

Step 1.7.4

Set $λ - 3$ equal to $0$ and solve for $λ$ .

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Step 1.7.4.1

Set $λ - 3$ equal to $0$ .

$λ - 3 = 0$

Step 1.7.4.2

Add $3$ to both sides of the equation.

$λ = 3$

Step 1.7.5

Set $λ - 2$ equal to $0$ and solve for $λ$ .

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Step 1.7.5.1

Set $λ - 2$ equal to $0$ .

$λ - 2 = 0$

Step 1.7.5.2

Add $2$ to both sides of the equation.

$λ = 2$

Step 1.7.6

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

$λ = 1, 3, 2$

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 $λ = 1$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] - [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Subtract the corresponding elements.

$[\begin{array}{c} 1 - 1 & 0 - 0 & 0 - 0 & 0 - 0 \\ 0 - 0 & 1 - 1 & 5 - 0 & - 10 - 0 \\ 1 - 0 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2

Simplify each element.

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Step 3.2.2.1

Subtract $1$ from $1$ .

$[\begin{array}{c} 0 & 0 - 0 & 0 - 0 & 0 - 0 \\ 0 - 0 & 1 - 1 & 5 - 0 & - 10 - 0 \\ 1 - 0 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.2

Subtract $0$ from $0$ .

$[\begin{array}{c} 0 & 0 & 0 - 0 & 0 - 0 \\ 0 - 0 & 1 - 1 & 5 - 0 & - 10 - 0 \\ 1 - 0 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.3

Subtract $0$ from $0$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 - 0 \\ 0 - 0 & 1 - 1 & 5 - 0 & - 10 - 0 \\ 1 - 0 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.4

Subtract $0$ from $0$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 - 0 & 1 - 1 & 5 - 0 & - 10 - 0 \\ 1 - 0 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.5

Subtract $0$ from $0$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 - 1 & 5 - 0 & - 10 - 0 \\ 1 - 0 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.6

Subtract $1$ from $1$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 - 0 & - 10 - 0 \\ 1 - 0 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.7

Subtract $0$ from $5$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 - 0 \\ 1 - 0 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.8

Subtract $0$ from $- 10$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 \\ 1 - 0 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.9

Subtract $0$ from $1$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 \\ 1 & 0 - 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.10

Subtract $0$ from $0$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 \\ 1 & 0 & 2 - 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.11

Subtract $1$ from $2$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 \\ 1 & 0 & 1 & 0 - 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.12

Subtract $0$ from $0$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 \\ 1 & 0 & 1 & 0 \\ 1 - 0 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.13

Subtract $0$ from $1$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 \\ 1 & 0 & 1 & 0 \\ 1 & 0 - 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.14

Subtract $0$ from $0$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 - 0 & 3 - 1 \end{array}]$

Step 3.2.2.15

Subtract $0$ from $0$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 3 - 1 \end{array}]$

Step 3.2.2.16

Subtract $1$ from $3$ .

$[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 2 \end{array}]$

Step 3.3

Find the null space when $λ = 1$ .

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Step 3.3.1

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

$[\begin{array}{c} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 5 & - 10 & 0 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 2 & 0 \end{array}]$

Step 3.3.2

Find the reduced row echelon form.

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Step 3.3.2.1

Swap $R_{3}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 5 & - 10 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 2 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation $R_{4} = R_{4} - R_{1}$ to make the entry at $4, 1$ a $0$ .

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Step 3.3.2.2.1

Perform the row operation $R_{4} = R_{4} - R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 5 & - 10 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 - 1 & 0 - 0 & 0 - 1 & 2 - 0 & 0 - 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 5 & - 10 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 & 2 & 0 \end{array}]$

Step 3.3.2.3

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

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Step 3.3.2.3.1

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

$[\begin{array}{c} 1 & 0 & 1 & 0 & 0 \\ \frac{0}{5} & \frac{0}{5} & \frac{5}{5} & \frac{- 10}{5} & \frac{0}{5} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 & 2 & 0 \end{array}]$

Step 3.3.2.3.2

Simplify $R_{2}$ .

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

Step 3.3.2.4

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

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Step 3.3.2.4.1

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

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

Step 3.3.2.4.2

Simplify $R_{4}$ .

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

Step 3.3.2.5

Perform the row operation $R_{1} = R_{1} - R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 3.3.2.5.1

Perform the row operation $R_{1} = R_{1} - R_{2}$ to make the entry at $1, 3$ a $0$ .

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

Step 3.3.2.5.2

Simplify $R_{1}$ .

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

Step 3.3.3

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

$x_{1} + 2 x_{4} = 0$

$x_{3} - 2 x_{4} = 0$

$0 = 0$

Step 3.3.4

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

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

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

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

Step 3.3.7

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

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

Step 4

Find the eigenvector using the eigenvalue $λ = 3$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] - 3 [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

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 $- 3$ by each element of the matrix.

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $- 3$ by $1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.5

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.6

Multiply $- 3$ by $1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.7

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.8

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.9

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.10

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.11

Multiply $- 3$ by $1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.12

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.13

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ 0 & - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.14

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ 0 & 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.15

Multiply $- 3$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ 0 & 0 & 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.16

Multiply $- 3$ by $1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & - 3 & 0 \\ 0 & 0 & 0 & - 3 \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} 1 - 3 & 0 + 0 & 0 + 0 & 0 + 0 \\ 0 + 0 & 1 - 3 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Subtract $3$ from $1$ .

$[\begin{array}{c} - 2 & 0 + 0 & 0 + 0 & 0 + 0 \\ 0 + 0 & 1 - 3 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.2

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 + 0 & 0 + 0 \\ 0 + 0 & 1 - 3 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.3

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 + 0 \\ 0 + 0 & 1 - 3 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.4

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 + 0 & 1 - 3 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.5

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & 1 - 3 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.6

Subtract $3$ from $1$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.7

Add $5$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.8

Add $- 10$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 \\ 1 + 0 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.9

Add $1$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 \\ 1 & 0 + 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.10

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 \\ 1 & 0 & 2 - 3 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.11

Subtract $3$ from $2$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 \\ 1 & 0 & - 1 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.12

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 \\ 1 & 0 & - 1 & 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.13

Add $1$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 \\ 1 & 0 & - 1 & 0 \\ 1 & 0 + 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.14

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 \\ 1 & 0 & - 1 & 0 \\ 1 & 0 & 0 + 0 & 3 - 3 \end{array}]$

Step 4.2.3.15

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 \\ 1 & 0 & - 1 & 0 \\ 1 & 0 & 0 & 3 - 3 \end{array}]$

Step 4.2.3.16

Subtract $3$ from $3$ .

$[\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 \\ 1 & 0 & - 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}]$

Step 4.3

Find the null space when $λ = 3$ .

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Step 4.3.1

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

$[\begin{array}{c} - 2 & 0 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 & 0 \\ 1 & 0 & - 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}]$

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 $- \frac{1}{2}$ 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 $- \frac{1}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 \\ 0 & - 2 & 5 & - 10 & 0 \\ 1 & 0 & - 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 & 0 \\ 1 & 0 & - 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.3.2.2.1

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 & 0 \\ 1 - 1 & 0 - 0 & - 1 - 0 & 0 - 0 & 0 - 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 & 0 \\ 0 & 0 & - 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.3

Perform the row operation $R_{4} = R_{4} - R_{1}$ to make the entry at $4, 1$ a $0$ .

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Step 4.3.2.3.1

Perform the row operation $R_{4} = R_{4} - R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 & 0 \\ 0 & 0 & - 1 & 0 & 0 \\ 1 - 1 & 0 - 0 & 0 - 0 & 0 - 0 & 0 - 0 \end{array}]$

Step 4.3.2.3.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 2 & 5 & - 10 & 0 \\ 0 & 0 & - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.4

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

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Step 4.3.2.4.1

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

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 5 & - \frac{1}{2} \cdot - 10 & - \frac{1}{2} \cdot 0 \\ 0 & 0 & - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.4.2

Simplify $R_{2}$ .

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

Step 4.3.2.5

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

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Step 4.3.2.5.1

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

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

Step 4.3.2.5.2

Simplify $R_{3}$ .

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

Step 4.3.2.6

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

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Step 4.3.2.6.1

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

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

Step 4.3.2.6.2

Simplify $R_{2}$ .

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

Step 4.3.3

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

$x_{1} = 0$

$x_{2} + 5 x_{4} = 0$

$x_{3} = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} 0 \\ - 5 x_{4} \\ 0 \\ x_{4} \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = x_{4} [\begin{array}{c} 0 \\ - 5 \\ 0 \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${x_{4} [\begin{array}{c} 0 \\ - 5 \\ 0 \\ 1 \end{array}] | x_{4} \in R}$

Step 4.3.7

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

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

Step 5

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

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Step 5.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] - 2 [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 5.2

Simplify.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

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

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2

Simplify each element in the matrix.

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Step 5.2.1.2.1

Multiply $- 2$ by $1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.2

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.3

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.4

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.5

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.6

Multiply $- 2$ by $1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.7

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.8

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.9

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.10

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.11

Multiply $- 2$ by $1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.12

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.13

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.14

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.15

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & - 2 \cdot 1 \end{array}]$

Step 5.2.1.2.16

Multiply $- 2$ by $1$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 5 & - 10 \\ 1 & 0 & 2 & 0 \\ 1 & 0 & 0 & 3 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & - 2 \end{array}]$

Step 5.2.2

Add the corresponding elements.

$[\begin{array}{c} 1 - 2 & 0 + 0 & 0 + 0 & 0 + 0 \\ 0 + 0 & 1 - 2 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3

Simplify each element.

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Step 5.2.3.1

Subtract $2$ from $1$ .

$[\begin{array}{c} - 1 & 0 + 0 & 0 + 0 & 0 + 0 \\ 0 + 0 & 1 - 2 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.2

Add $0$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 + 0 & 0 + 0 \\ 0 + 0 & 1 - 2 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.3

Add $0$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 + 0 \\ 0 + 0 & 1 - 2 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.4

Add $0$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 + 0 & 1 - 2 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.5

Add $0$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & 1 - 2 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.6

Subtract $2$ from $1$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 + 0 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.7

Add $5$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 + 0 \\ 1 + 0 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.8

Add $- 10$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 \\ 1 + 0 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.9

Add $1$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 \\ 1 & 0 + 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.10

Add $0$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 \\ 1 & 0 & 2 - 2 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.11

Subtract $2$ from $2$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 \\ 1 & 0 & 0 & 0 + 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.12

Add $0$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 \\ 1 & 0 & 0 & 0 \\ 1 + 0 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.13

Add $1$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 \\ 1 & 0 & 0 & 0 \\ 1 & 0 + 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.14

Add $0$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 + 0 & 3 - 2 \end{array}]$

Step 5.2.3.15

Add $0$ and $0$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 3 - 2 \end{array}]$

Step 5.2.3.16

Subtract $2$ from $3$ .

$[\begin{array}{c} - 1 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{array}]$

Step 5.3

Find the null space when $λ = 2$ .

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Step 5.3.1

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

$[\begin{array}{c} - 1 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \end{array}]$

Step 5.3.2

Find the reduced row echelon form.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

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

$[\begin{array}{c} - - 1 & - 0 & - 0 & - 0 & - 0 \\ 0 & - 1 & 5 & - 10 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \end{array}]$

Step 5.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \end{array}]$

Step 5.3.2.2

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 5.3.2.2.1

Perform the row operation $R_{3} = R_{3} - R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 & 0 \\ 1 - 1 & 0 - 0 & 0 - 0 & 0 - 0 & 0 - 0 \\ 1 & 0 & 0 & 1 & 0 \end{array}]$

Step 5.3.2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \end{array}]$

Step 5.3.2.3

Perform the row operation $R_{4} = R_{4} - R_{1}$ to make the entry at $4, 1$ a $0$ .

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Step 5.3.2.3.1

Perform the row operation $R_{4} = R_{4} - R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 - 1 & 0 - 0 & 0 - 0 & 1 - 0 & 0 - 0 \end{array}]$

Step 5.3.2.3.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 5 & - 10 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}]$

Step 5.3.2.4

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

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Step 5.3.2.4.1

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

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ - 0 & - - 1 & - 1 \cdot 5 & - - 10 & - 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}]$

Step 5.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & - 5 & 10 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}]$

Step 5.3.2.5

Swap $R_{4}$ with $R_{3}$ to put a nonzero entry at $3, 4$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & - 5 & 10 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.2.6

Perform the row operation $R_{2} = R_{2} - 10 R_{3}$ to make the entry at $2, 4$ a $0$ .

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Step 5.3.2.6.1

Perform the row operation $R_{2} = R_{2} - 10 R_{3}$ to make the entry at $2, 4$ a $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 - 10 \cdot 0 & 1 - 10 \cdot 0 & - 5 - 10 \cdot 0 & 10 - 10 \cdot 1 & 0 - 10 \cdot 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.2.6.2

Simplify $R_{2}$ .

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

Step 5.3.3

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

$x_{1} = 0$

$x_{2} - 5 x_{3} = 0$

$x_{4} = 0$

$0 = 0$

Step 5.3.4

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

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = [\begin{array}{c} 0 \\ 5 x_{3} \\ x_{3} \\ 0 \end{array}]$

Step 5.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] = x_{3} [\begin{array}{c} 0 \\ 5 \\ 1 \\ 0 \end{array}]$

Step 5.3.6

Write as a solution set.

${x_{3} [\begin{array}{c} 0 \\ 5 \\ 1 \\ 0 \end{array}] | x_{3} \in R}$

Step 5.3.7

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

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

Step 6

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

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