Linear Algebra Examples

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}]$

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 $[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] - λ I_{4})$

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 $I_{4}$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

Step 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.

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.5.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.9.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.10

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.10.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.10.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.11

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.12

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.12.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.12.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.13

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.13.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.13.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.14

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.14.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.14.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.15

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.15.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.15.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.16

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 & 0 \\ 0 & - λ & 0 & 0 \\ 0 & 0 & - λ & 0 \\ 0 & 0 & 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 0 - λ & 4 + 0 & 10 + 0 & 0 + 0 \\ - 1 + 0 & 4 - λ & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 + 0 & 10 + 0 & 0 + 0 \\ - 1 + 0 & 4 - λ & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.2

Add $4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 + 0 & 0 + 0 \\ - 1 + 0 & 4 - λ & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.3

Add $10$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 + 0 \\ - 1 + 0 & 4 - λ & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.4

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 + 0 & 4 - λ & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.5

Add $- 1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 & 4 - λ & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.6

Add $5$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 & 4 - λ & 5 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.7

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 & 4 - λ & 5 & 0 \\ 7 + 0 & - 14 + 0 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.8

Add $7$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 & 4 - λ & 5 & 0 \\ 7 & - 14 + 0 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.9

Add $- 14$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 & 4 - λ & 5 & 0 \\ 7 & - 14 & 2 - λ & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.10

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 & 4 - λ & 5 & 0 \\ 7 & - 14 & 2 - λ & 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.11

Add $- 12$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 & 4 - λ & 5 & 0 \\ 7 & - 14 & 2 - λ & 0 \\ - 12 & 24 + 0 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.12

Add $24$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 & 4 - λ & 5 & 0 \\ 7 & - 14 & 2 - λ & 0 \\ - 12 & 24 & 6 + 0 & 2 - λ \end{array}]$

Step 1.4.3.13

Add $6$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 4 & 10 & 0 \\ - 1 & 4 - λ & 5 & 0 \\ 7 & - 14 & 2 - λ & 0 \\ - 12 & 24 & 6 & 2 - λ \end{array}]$

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

$| \begin{array}{c} - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \\ - 12 & 24 & 6 \end{array} |$

Step 1.5.1.4

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

$0 | \begin{array}{c} - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \\ - 12 & 24 & 6 \end{array} |$

Step 1.5.1.5

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

$| \begin{array}{c} - λ & 4 & 10 \\ 7 & - 14 & 2 - λ \\ - 12 & 24 & 6 \end{array} |$

Step 1.5.1.6

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

$0 | \begin{array}{c} - λ & 4 & 10 \\ 7 & - 14 & 2 - λ \\ - 12 & 24 & 6 \end{array} |$

Step 1.5.1.7

The minor for $a_{34}$ is the determinant with row $3$ and column $4$ deleted.

$| \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ - 12 & 24 & 6 \end{array} |$

Step 1.5.1.8

Multiply element $a_{34}$ by its cofactor.

$0 | \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ - 12 & 24 & 6 \end{array} |$

Step 1.5.1.9

The minor for $a_{44}$ is the determinant with row $4$ and column $4$ deleted.

$| \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \end{array} |$

Step 1.5.1.10

Multiply element $a_{44}$ by its cofactor.

$(2 - λ) | \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \end{array} |$

Step 1.5.1.11

Add the terms together.

$p (λ) = 0 | \begin{array}{c} - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \\ - 12 & 24 & 6 \end{array} | + 0 | \begin{array}{c} - λ & 4 & 10 \\ 7 & - 14 & 2 - λ \\ - 12 & 24 & 6 \end{array} | + 0 | \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ - 12 & 24 & 6 \end{array} | + (2 - λ) | \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \end{array} |$

Step 1.5.2

Multiply $0$ by $| \begin{array}{c} - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \\ - 12 & 24 & 6 \end{array} |$ .

$p (λ) = 0 + 0 | \begin{array}{c} - λ & 4 & 10 \\ 7 & - 14 & 2 - λ \\ - 12 & 24 & 6 \end{array} | + 0 | \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ - 12 & 24 & 6 \end{array} | + (2 - λ) | \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \end{array} |$

Step 1.5.3

Multiply $0$ by $| \begin{array}{c} - λ & 4 & 10 \\ 7 & - 14 & 2 - λ \\ - 12 & 24 & 6 \end{array} |$ .

$p (λ) = 0 + 0 + 0 | \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ - 12 & 24 & 6 \end{array} | + (2 - λ) | \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \end{array} |$

Step 1.5.4

Multiply $0$ by $| \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ - 12 & 24 & 6 \end{array} |$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) | \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \end{array} |$

Step 1.5.5

Evaluate $| \begin{array}{c} - λ & 4 & 10 \\ - 1 & 4 - λ & 5 \\ 7 & - 14 & 2 - λ \end{array} |$ .

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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.

$| \begin{array}{c} 4 - λ & 5 \\ - 14 & 2 - λ \end{array} |$

Step 1.5.5.1.4

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

$- λ | \begin{array}{c} 4 - λ & 5 \\ - 14 & 2 - λ \end{array} |$

Step 1.5.5.1.5

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

$| \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} |$

Step 1.5.5.1.6

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

$- 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} |$

Step 1.5.5.1.7

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

$| \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |$

Step 1.5.5.1.8

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

$10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |$

Step 1.5.5.1.9

Add the terms together.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ | \begin{array}{c} 4 - λ & 5 \\ - 14 & 2 - λ \end{array} | - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2

Evaluate $| \begin{array}{c} 4 - λ & 5 \\ - 14 & 2 - λ \end{array} |$ .

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Step 1.5.5.2.1

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

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ ((4 - λ) (2 - λ) - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2

Simplify the determinant.

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Step 1.5.5.2.2.1

Simplify each term.

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Step 1.5.5.2.2.1.1

Expand $(4 - λ) (2 - λ)$ using the FOIL Method.

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Step 1.5.5.2.2.1.1.1

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (4 (2 - λ) - λ (2 - λ) - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.1.2

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (4 \cdot 2 + 4 (- λ) - λ (2 - λ) - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.1.3

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (4 \cdot 2 + 4 (- λ) - λ \cdot 2 - λ (- λ) - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.2

Simplify and combine like terms.

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Step 1.5.5.2.2.1.2.1

Simplify each term.

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Step 1.5.5.2.2.1.2.1.1

Multiply $4$ by $2$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 + 4 (- λ) - λ \cdot 2 - λ (- λ) - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.2.1.2

Multiply $- 1$ by $4$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 4 λ - λ \cdot 2 - λ (- λ) - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.2.1.3

Multiply $2$ by $- 1$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 4 λ - 2 λ - λ (- λ) - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 4 λ - 2 λ - 1 \cdot - 1 λ \cdot λ - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.2.1.5

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

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Step 1.5.5.2.2.1.2.1.5.1

Move $λ$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 4 λ - 2 λ - 1 \cdot - 1 (λ \cdot λ) - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 4 λ - 2 λ - 1 \cdot - 1 λ^{2} - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 4 λ - 2 λ + 1 λ^{2} - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 4 λ - 2 λ + λ^{2} - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.2.2

Subtract $2 λ$ from $- 4 λ$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 6 λ + λ^{2} - (- 14 \cdot 5)) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.3

Multiply $- (- 14 \cdot 5)$ .

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Step 1.5.5.2.2.1.3.1

Multiply $- 14$ by $5$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 6 λ + λ^{2} - - 70) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.1.3.2

Multiply $- 1$ by $- 70$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (8 - 6 λ + λ^{2} + 70) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.2

Add $8$ and $70$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (- 6 λ + λ^{2} + 78) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.2.2.3

Reorder $- 6 λ$ and $λ^{2}$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 | \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} | + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.3

Evaluate $| \begin{array}{c} - 1 & 5 \\ 7 & 2 - λ \end{array} |$ .

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Step 1.5.5.3.1

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

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (- (2 - λ) - 7 \cdot 5) + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.3.2

Simplify the determinant.

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Step 1.5.5.3.2.1

Simplify each term.

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Step 1.5.5.3.2.1.1

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (- 1 \cdot 2 - - λ - 7 \cdot 5) + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.3.2.1.2

Multiply $- 1$ by $2$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (- 2 - - λ - 7 \cdot 5) + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.3.2.1.3

Multiply $- - λ$ .

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Step 1.5.5.3.2.1.3.1

Multiply $- 1$ by $- 1$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (- 2 + 1 λ - 7 \cdot 5) + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.3.2.1.3.2

Multiply $λ$ by $1$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (- 2 + λ - 7 \cdot 5) + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.3.2.1.4

Multiply $- 7$ by $5$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (- 2 + λ - 35) + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.3.2.2

Subtract $35$ from $- 2$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (λ - 37) + 10 | \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |)$

Step 1.5.5.4

Evaluate $| \begin{array}{c} - 1 & 4 - λ \\ 7 & - 14 \end{array} |$ .

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Step 1.5.5.4.1

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

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (λ - 37) + 10 (- - 14 - 7 (4 - λ)))$

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

Multiply $- 1$ by $- 14$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (λ - 37) + 10 (14 - 7 (4 - λ)))$

Step 1.5.5.4.2.1.2

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (λ - 37) + 10 (14 - 7 \cdot 4 - 7 (- λ)))$

Step 1.5.5.4.2.1.3

Multiply $- 7$ by $4$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (λ - 37) + 10 (14 - 28 - 7 (- λ)))$

Step 1.5.5.4.2.1.4

Multiply $- 1$ by $- 7$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (λ - 37) + 10 (14 - 28 + 7 λ))$

Step 1.5.5.4.2.2

Subtract $28$ from $14$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (λ - 37) + 10 (- 14 + 7 λ))$

Step 1.5.5.4.2.3

Reorder $- 14$ and $7 λ$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ (λ^{2} - 6 λ + 78) - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5

Simplify the determinant.

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Step 1.5.5.5.1

Simplify each term.

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Step 1.5.5.5.1.1

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ \cdot λ^{2} - λ (- 6 λ) - λ \cdot 78 - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.2

Simplify.

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Step 1.5.5.5.1.2.1

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

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Step 1.5.5.5.1.2.1.1

Move $λ^{2}$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- (λ^{2} λ) - λ (- 6 λ) - λ \cdot 78 - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.2.1.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.5.5.1.2.1.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- (λ^{2} λ^{1}) - λ (- 6 λ) - λ \cdot 78 - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.2.1.2.2

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

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{2 + 1} - λ (- 6 λ) - λ \cdot 78 - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.2.1.3

Add $2$ and $1$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} - λ (- 6 λ) - λ \cdot 78 - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.2.2

Rewrite using the commutative property of multiplication.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} - 1 \cdot - 6 λ \cdot λ - λ \cdot 78 - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.2.3

Multiply $78$ by $- 1$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} - 1 \cdot - 6 λ \cdot λ - 78 λ - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.3

Simplify each term.

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Step 1.5.5.5.1.3.1

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

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Step 1.5.5.5.1.3.1.1

Move $λ$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} - 1 \cdot - 6 (λ \cdot λ) - 78 λ - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.3.1.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} - 1 \cdot - 6 λ^{2} - 78 λ - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.3.2

Multiply $- 1$ by $- 6$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} + 6 λ^{2} - 78 λ - 4 (λ - 37) + 10 (7 λ - 14))$

Step 1.5.5.5.1.4

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} + 6 λ^{2} - 78 λ - 4 λ - 4 \cdot - 37 + 10 (7 λ - 14))$

Step 1.5.5.5.1.5

Multiply $- 4$ by $- 37$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} + 6 λ^{2} - 78 λ - 4 λ + 148 + 10 (7 λ - 14))$

Step 1.5.5.5.1.6

Apply the distributive property.

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} + 6 λ^{2} - 78 λ - 4 λ + 148 + 10 (7 λ) + 10 \cdot - 14)$

Step 1.5.5.5.1.7

Multiply $7$ by $10$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} + 6 λ^{2} - 78 λ - 4 λ + 148 + 70 λ + 10 \cdot - 14)$

Step 1.5.5.5.1.8

Multiply $10$ by $- 14$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} + 6 λ^{2} - 78 λ - 4 λ + 148 + 70 λ - 140)$

Step 1.5.5.5.2

Subtract $4 λ$ from $- 78 λ$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} + 6 λ^{2} - 82 λ + 148 + 70 λ - 140)$

Step 1.5.5.5.3

Add $- 82 λ$ and $70 λ$ .

$p (λ) = 0 + 0 + 0 + (2 - λ) (- λ^{3} + 6 λ^{2} - 12 λ + 148 - 140)$

Step 1.5.5.5.4

Subtract $140$ from $148$ .

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

Step 1.5.6

Simplify the determinant.

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Step 1.5.6.1

Combine the opposite terms in $0 + 0 + 0 + (2 - λ) (- λ^{3} + 6 λ^{2} - 12 λ + 8)$ .

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Step 1.5.6.1.1

Add $0$ and $0$ .

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

Step 1.5.6.1.2

Add $0$ and $0$ .

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

Step 1.5.6.1.3

Add $0$ and $(2 - λ) (- λ^{3} + 6 λ^{2} - 12 λ + 8)$ .

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

Step 1.5.6.2

Expand $(2 - λ) (- λ^{3} + 6 λ^{2} - 12 λ + 8)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 2 (- λ^{3}) + 2 (6 λ^{2}) + 2 (- 12 λ) + 2 \cdot 8 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3

Simplify each term.

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Step 1.5.6.3.1

Multiply $- 1$ by $2$ .

$p (λ) = - 2 λ^{3} + 2 (6 λ^{2}) + 2 (- 12 λ) + 2 \cdot 8 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.2

Multiply $6$ by $2$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} + 2 (- 12 λ) + 2 \cdot 8 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.3

Multiply $- 12$ by $2$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 2 \cdot 8 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.4

Multiply $2$ by $8$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 - λ (- λ^{3}) - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.5

Rewrite using the commutative property of multiplication.

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 - 1 \cdot - 1 λ \cdot λ^{3} - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

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 $λ^{3}$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 - 1 \cdot - 1 (λ^{3} λ) - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

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 $1$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 - 1 \cdot - 1 (λ^{3} λ^{1}) - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.6.2.2

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

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 - 1 \cdot - 1 λ^{3 + 1} - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.6.3

Add $3$ and $1$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 - 1 \cdot - 1 λ^{4} - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.7

Multiply $- 1$ by $- 1$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + 1 λ^{4} - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.8

Multiply $λ^{4}$ by $1$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - λ (6 λ^{2}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.9

Rewrite using the commutative property of multiplication.

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 1 \cdot 6 λ \cdot λ^{2} - λ (- 12 λ) - λ \cdot 8$

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 $λ^{2}$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 1 \cdot 6 (λ^{2} λ) - λ (- 12 λ) - λ \cdot 8$

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 $1$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 1 \cdot 6 (λ^{2} λ^{1}) - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.10.2.2

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

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 1 \cdot 6 λ^{2 + 1} - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.10.3

Add $2$ and $1$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 1 \cdot 6 λ^{3} - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.11

Multiply $- 1$ by $6$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 6 λ^{3} - λ (- 12 λ) - λ \cdot 8$

Step 1.5.6.3.12

Rewrite using the commutative property of multiplication.

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 6 λ^{3} - 1 \cdot - 12 λ \cdot λ - λ \cdot 8$

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 $λ$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 6 λ^{3} - 1 \cdot - 12 (λ \cdot λ) - λ \cdot 8$

Step 1.5.6.3.13.2

Multiply $λ$ by $λ$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 6 λ^{3} - 1 \cdot - 12 λ^{2} - λ \cdot 8$

Step 1.5.6.3.14

Multiply $- 1$ by $- 12$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 6 λ^{3} + 12 λ^{2} - λ \cdot 8$

Step 1.5.6.3.15

Multiply $8$ by $- 1$ .

$p (λ) = - 2 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} - 6 λ^{3} + 12 λ^{2} - 8 λ$

Step 1.5.6.4

Subtract $6 λ^{3}$ from $- 2 λ^{3}$ .

$p (λ) = - 8 λ^{3} + 12 λ^{2} - 24 λ + 16 + λ^{4} + 12 λ^{2} - 8 λ$

Step 1.5.6.5

Add $12 λ^{2}$ and $12 λ^{2}$ .

$p (λ) = - 8 λ^{3} + 24 λ^{2} - 24 λ + 16 + λ^{4} - 8 λ$

Step 1.5.6.6

Subtract $8 λ$ from $- 24 λ$ .

$p (λ) = - 8 λ^{3} + 24 λ^{2} - 32 λ + 16 + λ^{4}$

Step 1.5.6.7

Move $16$ .

$p (λ) = - 8 λ^{3} + 24 λ^{2} - 32 λ + λ^{4} + 16$

Step 1.5.6.8

Move $- 32 λ$ .

$p (λ) = - 8 λ^{3} + 24 λ^{2} + λ^{4} - 32 λ + 16$

Step 1.5.6.9

Move $24 λ^{2}$ .

$p (λ) = - 8 λ^{3} + λ^{4} + 24 λ^{2} - 32 λ + 16$

Step 1.5.6.10

Reorder $- 8 λ^{3}$ and $λ^{4}$ .

$p (λ) = λ^{4} - 8 λ^{3} + 24 λ^{2} - 32 λ + 16$

Step 1.6

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

$λ^{4} - 8 λ^{3} + 24 λ^{2} - 32 λ + 16 = 0$

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

Make each term match the terms from the binomial theorem formula.

$λ^{4} - 4 \cdot (2 λ^{3}) + 6 \cdot (2^{2} λ^{2}) - 4 \cdot (2^{3} λ) + 2^{4} = 0$

Step 1.7.1.2

Factor $λ^{4} - 4 \cdot 2 λ^{3} + 6 \cdot 2^{2} λ^{2} - 4 \cdot 2^{3} λ + 2^{4}$ using the binomial theorem.

${(2 - λ)}^{4} = 0$

Step 1.7.2

Set the $2 - λ$ equal to $0$ .

$2 - λ = 0$

Step 1.7.3

Solve for $λ$ .

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Step 1.7.3.1

Subtract $2$ from both sides of the equation.

$- λ = - 2$

Step 1.7.3.2

Divide each term in $- λ = - 2$ by $- 1$ and simplify.

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Step 1.7.3.2.1

Divide each term in $- λ = - 2$ by $- 1$ .

$\frac{- λ}{- 1} = \frac{- 2}{- 1}$

Step 1.7.3.2.2

Simplify the left side.

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Step 1.7.3.2.2.1

Dividing two negative values results in a positive value.

$\frac{λ}{1} = \frac{- 2}{- 1}$

Step 1.7.3.2.2.2

Divide $λ$ by $1$ .

$λ = \frac{- 2}{- 1}$

Step 1.7.3.2.3

Simplify the right side.

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Step 1.7.3.2.3.1

Divide $- 2$ by $- 1$ .

$λ = 2$

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 $λ = 2$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] - 2 [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}])$

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.

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $- 2$ by $1$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.5

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.6

Multiply $- 2$ by $1$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.7

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.8

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.9

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.10

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.11

Multiply $- 2$ by $1$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.12

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.13

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.14

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.15

Multiply $- 2$ by $0$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.16

Multiply $- 2$ by $1$ .

$[\begin{array}{c} 0 & 4 & 10 & 0 \\ - 1 & 4 & 5 & 0 \\ 7 & - 14 & 2 & 0 \\ - 12 & 24 & 6 & 2 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 & 0 \\ 0 & - 2 & 0 & 0 \\ 0 & 0 & - 2 & 0 \\ 0 & 0 & 0 & - 2 \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} 0 - 2 & 4 + 0 & 10 + 0 & 0 + 0 \\ - 1 + 0 & 4 - 2 & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Subtract $2$ from $0$ .

$[\begin{array}{c} - 2 & 4 + 0 & 10 + 0 & 0 + 0 \\ - 1 + 0 & 4 - 2 & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.2

Add $4$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 + 0 & 0 + 0 \\ - 1 + 0 & 4 - 2 & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.3

Add $10$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 + 0 \\ - 1 + 0 & 4 - 2 & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.4

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 + 0 & 4 - 2 & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.5

Add $- 1$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 4 - 2 & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.6

Subtract $2$ from $4$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 + 0 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.7

Add $5$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 + 0 \\ 7 + 0 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.8

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 \\ 7 + 0 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.9

Add $7$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 \\ 7 & - 14 + 0 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.10

Add $- 14$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 \\ 7 & - 14 & 2 - 2 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.11

Subtract $2$ from $2$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 \\ 7 & - 14 & 0 & 0 + 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.12

Add $0$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 \\ 7 & - 14 & 0 & 0 \\ - 12 + 0 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.13

Add $- 12$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 \\ 7 & - 14 & 0 & 0 \\ - 12 & 24 + 0 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.14

Add $24$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 \\ 7 & - 14 & 0 & 0 \\ - 12 & 24 & 6 + 0 & 2 - 2 \end{array}]$

Step 3.2.3.15

Add $6$ and $0$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 \\ 7 & - 14 & 0 & 0 \\ - 12 & 24 & 6 & 2 - 2 \end{array}]$

Step 3.2.3.16

Subtract $2$ from $2$ .

$[\begin{array}{c} - 2 & 4 & 10 & 0 \\ - 1 & 2 & 5 & 0 \\ 7 & - 14 & 0 & 0 \\ - 12 & 24 & 6 & 0 \end{array}]$

Step 3.3

Find the null space when $λ = 2$ .

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Step 3.3.1

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

$[\begin{array}{c} - 2 & 4 & 10 & 0 & 0 \\ - 1 & 2 & 5 & 0 & 0 \\ 7 & - 14 & 0 & 0 & 0 \\ - 12 & 24 & 6 & 0 & 0 \end{array}]$

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

$[\begin{array}{c} - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 4 & - \frac{1}{2} \cdot 10 & - \frac{1}{2} \cdot 0 & - \frac{1}{2} \cdot 0 \\ - 1 & 2 & 5 & 0 & 0 \\ 7 & - 14 & 0 & 0 & 0 \\ - 12 & 24 & 6 & 0 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ - 1 & 2 & 5 & 0 & 0 \\ 7 & - 14 & 0 & 0 & 0 \\ - 12 & 24 & 6 & 0 & 0 \end{array}]$

Step 3.3.2.2

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

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Step 3.3.2.2.1

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

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ - 1 + 1 \cdot 1 & 2 - 2 & 5 - 5 & 0 + 0 & 0 + 0 \\ 7 & - 14 & 0 & 0 & 0 \\ - 12 & 24 & 6 & 0 & 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 7 & - 14 & 0 & 0 & 0 \\ - 12 & 24 & 6 & 0 & 0 \end{array}]$

Step 3.3.2.3

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.2.3.1

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 7 - 7 \cdot 1 & - 14 - 7 \cdot - 2 & 0 - 7 \cdot - 5 & 0 - 7 \cdot 0 & 0 - 7 \cdot 0 \\ - 12 & 24 & 6 & 0 & 0 \end{array}]$

Step 3.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 35 & 0 & 0 \\ - 12 & 24 & 6 & 0 & 0 \end{array}]$

Step 3.3.2.4

Perform the row operation $R_{4} = R_{4} + 12 R_{1}$ to make the entry at $4, 1$ a $0$ .

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Step 3.3.2.4.1

Perform the row operation $R_{4} = R_{4} + 12 R_{1}$ to make the entry at $4, 1$ a $0$ .

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 35 & 0 & 0 \\ - 12 + 12 \cdot 1 & 24 + 12 \cdot - 2 & 6 + 12 \cdot - 5 & 0 + 12 \cdot 0 & 0 + 12 \cdot 0 \end{array}]$

Step 3.3.2.4.2

Simplify $R_{4}$ .

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 35 & 0 & 0 \\ 0 & 0 & - 54 & 0 & 0 \end{array}]$

Step 3.3.2.5

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

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ 0 & 0 & 35 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & - 54 & 0 & 0 \end{array}]$

Step 3.3.2.6

Multiply each element of $R_{2}$ by $\frac{1}{35}$ to make the entry at $2, 3$ a $1$ .

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Step 3.3.2.6.1

Multiply each element of $R_{2}$ by $\frac{1}{35}$ to make the entry at $2, 3$ a $1$ .

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ \frac{0}{35} & \frac{0}{35} & \frac{35}{35} & \frac{0}{35} & \frac{0}{35} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & - 54 & 0 & 0 \end{array}]$

Step 3.3.2.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & - 54 & 0 & 0 \end{array}]$

Step 3.3.2.7

Perform the row operation $R_{4} = R_{4} + 54 R_{2}$ to make the entry at $4, 3$ a $0$ .

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Step 3.3.2.7.1

Perform the row operation $R_{4} = R_{4} + 54 R_{2}$ to make the entry at $4, 3$ a $0$ .

$[\begin{array}{c} 1 & - 2 & - 5 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 + 54 \cdot 0 & 0 + 54 \cdot 0 & - 54 + 54 \cdot 1 & 0 + 54 \cdot 0 & 0 + 54 \cdot 0 \end{array}]$

Step 3.3.2.7.2

Simplify $R_{4}$ .

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

Step 3.3.2.8

Perform the row operation $R_{1} = R_{1} + 5 R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 3.3.2.8.1

Perform the row operation $R_{1} = R_{1} + 5 R_{2}$ to make the entry at $1, 3$ a $0$ .

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

Step 3.3.2.8.2

Simplify $R_{1}$ .

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

Step 3.3.3

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

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

$x_{3} = 0$

$0 = 0$

Step 3.3.4

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

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

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

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

Step 3.3.7

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

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

Step 4

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

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