Linear Algebra Examples

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

Step 1

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

$p (λ) = determinant (A - λ I_{4})$

Step 2

The identity matrix or unit matrix of size $4$ is the $4 \times 4$ square matrix with ones on the main diagonal and zeros elsewhere.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

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

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.5

Multiply $- λ \cdot 0$ .

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Step 4.1.2.5.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.5.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.6

Multiply $- 1$ by $1$ .

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

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.9

Multiply $- λ \cdot 0$ .

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Step 4.1.2.9.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.9.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.10

Multiply $- λ \cdot 0$ .

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Step 4.1.2.10.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.10.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.11

Multiply $- 1$ by $1$ .

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

Step 4.1.2.12

Multiply $- λ \cdot 0$ .

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Step 4.1.2.12.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.12.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.13

Multiply $- λ \cdot 0$ .

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Step 4.1.2.13.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.13.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.14

Multiply $- λ \cdot 0$ .

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Step 4.1.2.14.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.14.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.15

Multiply $- λ \cdot 0$ .

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Step 4.1.2.15.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.15.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.16

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add $3$ and $0$ .

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

Step 4.3.2

Add $0$ and $0$ .

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

Step 4.3.3

Add $0$ and $0$ .

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

Step 4.3.4

Add $2$ and $0$ .

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

Step 4.3.5

Add $0$ and $0$ .

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

Step 4.3.6

Add $0$ and $0$ .

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

Step 4.3.7

Add $0$ and $0$ .

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

Step 4.3.8

Add $0$ and $0$ .

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

Step 4.3.9

Add $3$ and $0$ .

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

Step 4.3.10

Add $0$ and $0$ .

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

Step 4.3.11

Add $0$ and $0$ .

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

Step 4.3.12

Add $2$ and $0$ .

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

Step 5

Find the determinant.

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Step 5.1

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

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Step 5.1.1

Consider the corresponding sign chart.

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

Step 5.1.2

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

Step 5.1.3

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

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

Step 5.1.4

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

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

Step 5.1.5

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

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

Step 5.1.6

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

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

Step 5.1.7

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

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

Step 5.1.8

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

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

Step 5.1.9

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

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

Step 5.1.10

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

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

Step 5.1.11

Add the terms together.

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

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

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Step 5.4.1

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

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Step 5.4.1.1

Consider the corresponding sign chart.

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

Step 5.4.1.2

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

Step 5.4.1.3

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

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

Step 5.4.1.4

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

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

Step 5.4.1.5

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

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

Step 5.4.1.6

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

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

Step 5.4.1.7

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

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

Step 5.4.1.8

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

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

Step 5.4.1.9

Add the terms together.

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

Step 5.4.2

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

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

Step 5.4.3

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

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

Step 5.4.4

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

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Step 5.4.4.1

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

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

Step 5.4.4.2

Simplify the determinant.

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Step 5.4.4.2.1

Simplify each term.

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Step 5.4.4.2.1.1

Expand $(1 - λ) (2 - λ)$ using the FOIL Method.

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Step 5.4.4.2.1.1.1

Apply the distributive property.

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

Step 5.4.4.2.1.1.2

Apply the distributive property.

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

Step 5.4.4.2.1.1.3

Apply the distributive property.

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

Step 5.4.4.2.1.2

Simplify and combine like terms.

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Step 5.4.4.2.1.2.1

Simplify each term.

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Step 5.4.4.2.1.2.1.1

Multiply $2$ by $1$ .

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

Step 5.4.4.2.1.2.1.2

Multiply $- λ$ by $1$ .

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

Step 5.4.4.2.1.2.1.3

Multiply $2$ by $- 1$ .

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

Step 5.4.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 5.4.4.2.1.2.1.5

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

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Step 5.4.4.2.1.2.1.5.1

Move $λ$ .

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

Step 5.4.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

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

Step 5.4.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

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

Step 5.4.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

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

Step 5.4.4.2.1.2.2

Subtract $2 λ$ from $- λ$ .

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

Step 5.4.4.2.1.3

Multiply $- 2$ by $3$ .

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

Step 5.4.4.2.2

Subtract $6$ from $2$ .

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

Step 5.4.4.2.3

Reorder $- 3 λ$ and $λ^{2}$ .

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

Step 5.4.5

Simplify the determinant.

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Step 5.4.5.1

Combine the opposite terms in $(2 - λ) (λ^{2} - 3 λ - 4) + 0 + 0$ .

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Step 5.4.5.1.1

Add $(2 - λ) (λ^{2} - 3 λ - 4)$ and $0$ .

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

Step 5.4.5.1.2

Add $(2 - λ) (λ^{2} - 3 λ - 4)$ and $0$ .

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

Step 5.4.5.2

Expand $(2 - λ) (λ^{2} - 3 λ - 4)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 5.4.5.3

Simplify each term.

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Step 5.4.5.3.1

Multiply $- 3$ by $2$ .

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

Step 5.4.5.3.2

Multiply $2$ by $- 4$ .

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

Step 5.4.5.3.3

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

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Step 5.4.5.3.3.1

Move $λ^{2}$ .

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

Step 5.4.5.3.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.4.5.3.3.2.1

Raise $λ$ to the power of $1$ .

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

Step 5.4.5.3.3.2.2

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

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

Step 5.4.5.3.3.3

Add $2$ and $1$ .

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

Step 5.4.5.3.4

Rewrite using the commutative property of multiplication.

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

Step 5.4.5.3.5

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

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Step 5.4.5.3.5.1

Move $λ$ .

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

Step 5.4.5.3.5.2

Multiply $λ$ by $λ$ .

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

Step 5.4.5.3.6

Multiply $- 1$ by $- 3$ .

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

Step 5.4.5.3.7

Multiply $- 4$ by $- 1$ .

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

Step 5.4.5.4

Add $2 λ^{2}$ and $3 λ^{2}$ .

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

Step 5.4.5.5

Add $- 6 λ$ and $4 λ$ .

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

Step 5.4.5.6

Move $- 8$ .

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

Step 5.4.5.7

Move $- 2 λ$ .

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

Step 5.4.5.8

Reorder $5 λ^{2}$ and $- λ^{3}$ .

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

Step 5.5

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

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Step 5.5.1

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

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Step 5.5.1.1

Consider the corresponding sign chart.

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

Step 5.5.1.2

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

Step 5.5.1.3

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

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

Step 5.5.1.4

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

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

Step 5.5.1.5

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

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

Step 5.5.1.6

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

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

Step 5.5.1.7

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

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

Step 5.5.1.8

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

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

Step 5.5.1.9

Add the terms together.

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

Step 5.5.2

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

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

Step 5.5.3

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

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

Step 5.5.4

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

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Step 5.5.4.1

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

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 ((1 - λ) (2 - λ) - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2

Simplify the determinant.

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Step 5.5.4.2.1

Simplify each term.

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Step 5.5.4.2.1.1

Expand $(1 - λ) (2 - λ)$ using the FOIL Method.

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Step 5.5.4.2.1.1.1

Apply the distributive property.

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (1 (2 - λ) - λ (2 - λ) - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.1.2

Apply the distributive property.

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (1 \cdot 2 + 1 (- λ) - λ (2 - λ) - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.1.3

Apply the distributive property.

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (1 \cdot 2 + 1 (- λ) - λ \cdot 2 - λ (- λ) - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.2

Simplify and combine like terms.

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Step 5.5.4.2.1.2.1

Simplify each term.

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Step 5.5.4.2.1.2.1.1

Multiply $2$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 + 1 (- λ) - λ \cdot 2 - λ (- λ) - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.2.1.2

Multiply $- λ$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 - λ - λ \cdot 2 - λ (- λ) - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.2.1.3

Multiply $2$ by $- 1$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 - λ - 2 λ - λ (- λ) - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 - λ - 2 λ - 1 \cdot - 1 λ \cdot λ - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.2.1.5

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

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Step 5.5.4.2.1.2.1.5.1

Move $λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 - λ - 2 λ - 1 \cdot - 1 (λ \cdot λ) - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 - λ - 2 λ - 1 \cdot - 1 λ^{2} - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 - λ - 2 λ + 1 λ^{2} - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 - λ - 2 λ + λ^{2} - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.2.2

Subtract $2 λ$ from $- λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 - 3 λ + λ^{2} - 2 \cdot 3) + 0 + 0) + 0 + 0$

Step 5.5.4.2.1.3

Multiply $- 2$ by $3$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (2 - 3 λ + λ^{2} - 6) + 0 + 0) + 0 + 0$

Step 5.5.4.2.2

Subtract $6$ from $2$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (- 3 λ + λ^{2} - 4) + 0 + 0) + 0 + 0$

Step 5.5.4.2.3

Reorder $- 3 λ$ and $λ^{2}$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (λ^{2} - 3 λ - 4) + 0 + 0) + 0 + 0$

Step 5.5.5

Simplify the determinant.

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Step 5.5.5.1

Combine the opposite terms in $2 (λ^{2} - 3 λ - 4) + 0 + 0$ .

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Step 5.5.5.1.1

Add $2 (λ^{2} - 3 λ - 4)$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (λ^{2} - 3 λ - 4) + 0) + 0 + 0$

Step 5.5.5.1.2

Add $2 (λ^{2} - 3 λ - 4)$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 (λ^{2} - 3 λ - 4)) + 0 + 0$

Step 5.5.5.2

Apply the distributive property.

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 λ^{2} + 2 (- 3 λ) + 2 \cdot - 4) + 0 + 0$

Step 5.5.5.3

Simplify.

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Step 5.5.5.3.1

Multiply $- 3$ by $2$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 λ^{2} - 6 λ + 2 \cdot - 4) + 0 + 0$

Step 5.5.5.3.2

Multiply $2$ by $- 4$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 λ^{2} - 6 λ - 8) + 0 + 0$

Step 5.6

Simplify the determinant.

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Step 5.6.1

Combine the opposite terms in $(1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 λ^{2} - 6 λ - 8) + 0 + 0$ .

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Step 5.6.1.1

Add $(1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 λ^{2} - 6 λ - 8)$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 λ^{2} - 6 λ - 8) + 0$

Step 5.6.1.2

Add $(1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 λ^{2} - 6 λ - 8)$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8) - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2

Simplify each term.

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Step 5.6.2.1

Expand $(1 - λ) (- λ^{3} + 5 λ^{2} - 2 λ - 8)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 1 (- λ^{3}) + 1 (5 λ^{2}) + 1 (- 2 λ) + 1 \cdot - 8 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2

Simplify each term.

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Step 5.6.2.2.1

Multiply $- λ^{3}$ by $1$ .

$p (λ) = - λ^{3} + 1 (5 λ^{2}) + 1 (- 2 λ) + 1 \cdot - 8 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.2

Multiply $5 λ^{2}$ by $1$ .

$p (λ) = - λ^{3} + 5 λ^{2} + 1 (- 2 λ) + 1 \cdot - 8 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.3

Multiply $- 2 λ$ by $1$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ + 1 \cdot - 8 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.4

Multiply $- 8$ by $1$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 - λ (- λ^{3}) - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.5

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 - 1 \cdot - 1 λ \cdot λ^{3} - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.6

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

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Step 5.6.2.2.6.1

Move $λ^{3}$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 - 1 \cdot - 1 (λ^{3} λ) - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.6.2

Multiply $λ^{3}$ by $λ$ .

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Step 5.6.2.2.6.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 - 1 \cdot - 1 (λ^{3} λ^{1}) - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.6.2.2

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

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 - 1 \cdot - 1 λ^{3 + 1} - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.6.3

Add $3$ and $1$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 - 1 \cdot - 1 λ^{4} - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.7

Multiply $- 1$ by $- 1$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + 1 λ^{4} - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.8

Multiply $λ^{4}$ by $1$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - λ (5 λ^{2}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.9

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 1 \cdot 5 λ \cdot λ^{2} - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.10

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

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Step 5.6.2.2.10.1

Move $λ^{2}$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 1 \cdot 5 (λ^{2} λ) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.10.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.6.2.2.10.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 1 \cdot 5 (λ^{2} λ^{1}) - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.10.2.2

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

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 1 \cdot 5 λ^{2 + 1} - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.10.3

Add $2$ and $1$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 1 \cdot 5 λ^{3} - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.11

Multiply $- 1$ by $5$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 5 λ^{3} - λ (- 2 λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.12

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 5 λ^{3} - 1 \cdot - 2 λ \cdot λ - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.13

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

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Step 5.6.2.2.13.1

Move $λ$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 5 λ^{3} - 1 \cdot - 2 (λ \cdot λ) - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.13.2

Multiply $λ$ by $λ$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 5 λ^{3} - 1 \cdot - 2 λ^{2} - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.14

Multiply $- 1$ by $- 2$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 5 λ^{3} + 2 λ^{2} - λ \cdot - 8 - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.2.15

Multiply $- 8$ by $- 1$ .

$p (λ) = - λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} - 5 λ^{3} + 2 λ^{2} + 8 λ - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.3

Subtract $5 λ^{3}$ from $- λ^{3}$ .

$p (λ) = - 6 λ^{3} + 5 λ^{2} - 2 λ - 8 + λ^{4} + 2 λ^{2} + 8 λ - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.4

Add $5 λ^{2}$ and $2 λ^{2}$ .

$p (λ) = - 6 λ^{3} + 7 λ^{2} - 2 λ - 8 + λ^{4} + 8 λ - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.5

Add $- 2 λ$ and $8 λ$ .

$p (λ) = - 6 λ^{3} + 7 λ^{2} + 6 λ - 8 + λ^{4} - 3 (2 λ^{2} - 6 λ - 8)$

Step 5.6.2.6

Apply the distributive property.

$p (λ) = - 6 λ^{3} + 7 λ^{2} + 6 λ - 8 + λ^{4} - 3 (2 λ^{2}) - 3 (- 6 λ) - 3 \cdot - 8$

Step 5.6.2.7

Simplify.

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Step 5.6.2.7.1

Multiply $2$ by $- 3$ .

$p (λ) = - 6 λ^{3} + 7 λ^{2} + 6 λ - 8 + λ^{4} - 6 λ^{2} - 3 (- 6 λ) - 3 \cdot - 8$

Step 5.6.2.7.2

Multiply $- 6$ by $- 3$ .

$p (λ) = - 6 λ^{3} + 7 λ^{2} + 6 λ - 8 + λ^{4} - 6 λ^{2} + 18 λ - 3 \cdot - 8$

Step 5.6.2.7.3

Multiply $- 3$ by $- 8$ .

$p (λ) = - 6 λ^{3} + 7 λ^{2} + 6 λ - 8 + λ^{4} - 6 λ^{2} + 18 λ + 24$

Step 5.6.3

Subtract $6 λ^{2}$ from $7 λ^{2}$ .

$p (λ) = - 6 λ^{3} + λ^{2} + 6 λ - 8 + λ^{4} + 18 λ + 24$

Step 5.6.4

Add $6 λ$ and $18 λ$ .

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

Step 5.6.5

Add $- 8$ and $24$ .

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

Step 5.6.6

Move $24 λ$ .

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

Step 5.6.7

Move $λ^{2}$ .

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

Step 5.6.8

Reorder $- 6 λ^{3}$ and $λ^{4}$ .

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