Linear Algebra Examples

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

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

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

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 $I_{4}$ .

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

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.

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.5

Multiply $- λ \cdot 0$ .

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Step 4.1.2.5.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.5.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.6

Multiply $- 1$ by $1$ .

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

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.9

Multiply $- λ \cdot 0$ .

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Step 4.1.2.9.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.9.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.10

Multiply $- λ \cdot 0$ .

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Step 4.1.2.10.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.10.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.11

Multiply $- 1$ by $1$ .

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

Step 4.1.2.12

Multiply $- λ \cdot 0$ .

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Step 4.1.2.12.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.12.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.13

Multiply $- λ \cdot 0$ .

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Step 4.1.2.13.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.13.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.14

Multiply $- λ \cdot 0$ .

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Step 4.1.2.14.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.14.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.15

Multiply $- λ \cdot 0$ .

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Step 4.1.2.15.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.15.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.16

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add $0$ and $0$ .

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

Step 4.3.2

Add $5$ and $0$ .

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

Step 4.3.3

Add $3$ and $0$ .

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

Step 4.3.4

Add $1$ and $0$ .

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

Step 4.3.5

Add $2$ and $0$ .

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

Step 4.3.6

Add $0$ and $0$ .

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

Step 4.3.7

Add $3$ and $0$ .

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

Step 4.3.8

Add $1$ and $0$ .

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

Step 4.3.9

Subtract $λ$ from $0$ .

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

Step 4.3.10

Add $1$ and $0$ .

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

Step 4.3.11

Add $0$ and $0$ .

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

Step 4.3.12

Add $1$ and $0$ .

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

Step 4.3.13

Add $1$ and $0$ .

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

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.

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

Step 5.1.4

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

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

Step 5.1.5

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

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

Step 5.1.6

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

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

Step 5.1.7

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

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

Step 5.1.8

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

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

Step 5.1.9

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

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

Step 5.1.10

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

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

Step 5.1.11

Add the terms together.

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

Step 5.2

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

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

Step 5.3

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

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Step 5.3.1

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 5.3.1.1

Consider the corresponding sign chart.

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

Step 5.3.1.2

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

Step 5.3.1.3

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

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

Step 5.3.1.4

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

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

Step 5.3.1.5

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

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

Step 5.3.1.6

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

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

Step 5.3.1.7

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

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

Step 5.3.1.8

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

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

Step 5.3.1.9

Add the terms together.

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

Step 5.3.2

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

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

Step 5.3.3

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

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Step 5.3.3.1

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

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

Step 5.3.3.2

Simplify the determinant.

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Step 5.3.3.2.1

Simplify each term.

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Step 5.3.3.2.1.1

Apply the distributive property.

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

Step 5.3.3.2.1.2

Multiply $- 1$ by $1$ .

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

Step 5.3.3.2.1.3

Rewrite using the commutative property of multiplication.

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

Step 5.3.3.2.1.4

Simplify each term.

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Step 5.3.3.2.1.4.1

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

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Step 5.3.3.2.1.4.1.1

Move $λ$ .

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

Step 5.3.3.2.1.4.1.2

Multiply $λ$ by $λ$ .

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

Step 5.3.3.2.1.4.2

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.2.1.4.3

Multiply $λ^{2}$ by $1$ .

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

Step 5.3.3.2.1.5

Multiply $- 1$ by $1$ .

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

Step 5.3.3.2.2

Reorder $- λ$ and $λ^{2}$ .

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

Step 5.3.4

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

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Step 5.3.4.1

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

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

Step 5.3.4.2

Simplify the determinant.

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Step 5.3.4.2.1

Simplify each term.

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Step 5.3.4.2.1.1

Multiply $1 - λ$ by $1$ .

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

Step 5.3.4.2.1.2

Multiply $- 1$ by $1$ .

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

Step 5.3.4.2.2

Combine the opposite terms in $1 - λ - 1$ .

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Step 5.3.4.2.2.1

Subtract $1$ from $1$ .

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

Step 5.3.4.2.2.2

Add $- λ$ and $0$ .

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

Step 5.3.5

Simplify the determinant.

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Step 5.3.5.1

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

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

Step 5.3.5.2

Simplify each term.

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Step 5.3.5.2.1

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

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

Step 5.3.5.2.2

Simplify each term.

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Step 5.3.5.2.2.1

Multiply $- 1$ by $3$ .

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

Step 5.3.5.2.2.2

Multiply $3$ by $- 1$ .

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

Step 5.3.5.2.2.3

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

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Step 5.3.5.2.2.3.1

Move $λ^{2}$ .

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

Step 5.3.5.2.2.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.3.5.2.2.3.2.1

Raise $λ$ to the power of $1$ .

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

Step 5.3.5.2.2.3.2.2

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

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

Step 5.3.5.2.2.3.3

Add $2$ and $1$ .

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

Step 5.3.5.2.2.4

Rewrite using the commutative property of multiplication.

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

Step 5.3.5.2.2.5

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

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Step 5.3.5.2.2.5.1

Move $λ$ .

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

Step 5.3.5.2.2.5.2

Multiply $λ$ by $λ$ .

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

Step 5.3.5.2.2.6

Multiply $- 1$ by $- 1$ .

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

Step 5.3.5.2.2.7

Multiply $λ^{2}$ by $1$ .

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

Step 5.3.5.2.2.8

Multiply $- λ \cdot - 1$ .

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Step 5.3.5.2.2.8.1

Multiply $- 1$ by $- 1$ .

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

Step 5.3.5.2.2.8.2

Multiply $λ$ by $1$ .

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

Step 5.3.5.2.3

Add $3 λ^{2}$ and $λ^{2}$ .

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

Step 5.3.5.2.4

Add $- 3 λ$ and $λ$ .

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

Step 5.3.5.2.5

Multiply $- 1$ by $- 2$ .

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

Step 5.3.5.3

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

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Step 5.3.5.3.1

Add $- 2 λ$ and $2 λ$ .

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

Step 5.3.5.3.2

Add $4 λ^{2} - 3 - λ^{3}$ and $0$ .

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

Step 5.3.5.4

Move $- 3$ .

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

Step 5.3.5.5

Reorder $4 λ^{2}$ and $- λ^{3}$ .

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

Step 5.4

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

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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.

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

Step 5.4.1.4

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

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

Step 5.4.1.5

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

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

Step 5.4.1.6

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

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

Step 5.4.1.7

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

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

Step 5.4.1.8

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

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

Step 5.4.1.9

Add the terms together.

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

Step 5.4.2

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

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

Step 5.4.3

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

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Step 5.4.3.1

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

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

Step 5.4.3.2

Simplify the determinant.

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Step 5.4.3.2.1

Simplify each term.

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Step 5.4.3.2.1.1

Multiply $1 - λ$ by $1$ .

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

Step 5.4.3.2.1.2

Multiply $- 1$ by $1$ .

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

Step 5.4.3.2.2

Combine the opposite terms in $1 - λ - 1$ .

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Step 5.4.3.2.2.1

Subtract $1$ from $1$ .

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

Step 5.4.3.2.2.2

Add $- λ$ and $0$ .

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

Step 5.4.4

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

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Step 5.4.4.1

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

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

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

Apply the distributive property.

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

Step 5.4.4.2.1.2

Multiply $3$ by $1$ .

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

Step 5.4.4.2.1.3

Multiply $- 1$ by $3$ .

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

Step 5.4.4.2.1.4

Multiply $0$ by $1$ .

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

Step 5.4.4.2.2

Add $3 - 3 λ$ and $0$ .

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

Step 5.4.4.2.3

Reorder $3$ and $- 3 λ$ .

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

Step 5.4.5

Simplify the determinant.

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Step 5.4.5.1

Add $1 (- λ) - (3 - λ) (- 3 λ + 3)$ and $0$ .

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

Step 5.4.5.2

Simplify each term.

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Step 5.4.5.2.1

Multiply $- λ$ by $1$ .

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

Step 5.4.5.2.2

Apply the distributive property.

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

Step 5.4.5.2.3

Multiply $- 1$ by $3$ .

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

Step 5.4.5.2.4

Multiply $- - λ$ .

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Step 5.4.5.2.4.1

Multiply $- 1$ by $- 1$ .

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

Step 5.4.5.2.4.2

Multiply $λ$ by $1$ .

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

Step 5.4.5.2.5

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

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Step 5.4.5.2.5.1

Apply the distributive property.

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

Step 5.4.5.2.5.2

Apply the distributive property.

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

Step 5.4.5.2.5.3

Apply the distributive property.

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

Step 5.4.5.2.6

Simplify and combine like terms.

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Step 5.4.5.2.6.1

Simplify each term.

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Step 5.4.5.2.6.1.1

Multiply $- 3$ by $- 3$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- λ + 9 λ - 3 \cdot 3 + λ (- 3 λ) + λ \cdot 3) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.4.5.2.6.1.2

Multiply $- 3$ by $3$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- λ + 9 λ - 9 + λ (- 3 λ) + λ \cdot 3) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.4.5.2.6.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- λ + 9 λ - 9 - 3 λ \cdot λ + λ \cdot 3) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.4.5.2.6.1.4

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

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Step 5.4.5.2.6.1.4.1

Move $λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- λ + 9 λ - 9 - 3 (λ \cdot λ) + λ \cdot 3) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.4.5.2.6.1.4.2

Multiply $λ$ by $λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- λ + 9 λ - 9 - 3 λ^{2} + λ \cdot 3) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.4.5.2.6.1.5

Move $3$ to the left of $λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- λ + 9 λ - 9 - 3 λ^{2} + 3 λ) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.4.5.2.6.2

Add $9 λ$ and $3 λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- λ + 12 λ - 9 - 3 λ^{2}) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.4.5.3

Add $- λ$ and $12 λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (11 λ - 9 - 3 λ^{2}) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.4.5.4

Move $- 9$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (11 λ - 3 λ^{2} - 9) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.4.5.5

Reorder $11 λ$ and $- 3 λ^{2}$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 | \begin{array}{c} 1 & 3 - λ & 2 \\ 3 & 1 & - λ \\ 0 & 1 & 1 \end{array} |$

Step 5.5

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

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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.

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

Step 5.5.1.4

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

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

Step 5.5.1.5

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

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

Step 5.5.1.6

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

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

Step 5.5.1.7

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

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

Step 5.5.1.8

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

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

Step 5.5.1.9

Add the terms together.

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 | \begin{array}{c} 1 & - λ \\ 1 & 1 \end{array} | - (3 - λ) | \begin{array}{c} 3 & - λ \\ 0 & 1 \end{array} | + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.2

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

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Step 5.5.2.1

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

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (1 \cdot 1 - - λ) - (3 - λ) | \begin{array}{c} 3 & - λ \\ 0 & 1 \end{array} | + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.2.2

Simplify the determinant.

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Step 5.5.2.2.1

Simplify each term.

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Step 5.5.2.2.1.1

Multiply $1$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (1 - - λ) - (3 - λ) | \begin{array}{c} 3 & - λ \\ 0 & 1 \end{array} | + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.2.2.1.2

Multiply $- - λ$ .

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Step 5.5.2.2.1.2.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (1 + 1 λ) - (3 - λ) | \begin{array}{c} 3 & - λ \\ 0 & 1 \end{array} | + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.2.2.1.2.2

Multiply $λ$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (1 + λ) - (3 - λ) | \begin{array}{c} 3 & - λ \\ 0 & 1 \end{array} | + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.2.2.2

Reorder $1$ and $λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) | \begin{array}{c} 3 & - λ \\ 0 & 1 \end{array} | + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.3

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

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Step 5.5.3.1

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

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) (3 \cdot 1 + 0 (- λ)) + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.3.2

Simplify the determinant.

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Step 5.5.3.2.1

Simplify each term.

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Step 5.5.3.2.1.1

Multiply $3$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) (3 + 0 (- λ)) + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.3.2.1.2

Multiply $0 (- λ)$ .

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Step 5.5.3.2.1.2.1

Multiply $- 1$ by $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) (3 + 0 λ) + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.3.2.1.2.2

Multiply $0$ by $λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) (3 + 0) + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.3.2.2

Add $3$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) \cdot 3 + 2 | \begin{array}{c} 3 & 1 \\ 0 & 1 \end{array} |)$

Step 5.5.4

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

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Step 5.5.4.1

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

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) \cdot 3 + 2 (3 \cdot 1 + 0 \cdot 1))$

Step 5.5.4.2

Simplify the determinant.

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Step 5.5.4.2.1

Simplify each term.

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Step 5.5.4.2.1.1

Multiply $3$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) \cdot 3 + 2 (3 + 0 \cdot 1))$

Step 5.5.4.2.1.2

Multiply $0$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) \cdot 3 + 2 (3 + 0))$

Step 5.5.4.2.2

Add $3$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (1 (λ + 1) - (3 - λ) \cdot 3 + 2 \cdot 3)$

Step 5.5.5

Simplify the determinant.

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Step 5.5.5.1

Simplify each term.

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Step 5.5.5.1.1

Multiply $λ + 1$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (λ + 1 - (3 - λ) \cdot 3 + 2 \cdot 3)$

Step 5.5.5.1.2

Apply the distributive property.

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (λ + 1 + (- 1 \cdot 3 - - λ) \cdot 3 + 2 \cdot 3)$

Step 5.5.5.1.3

Multiply $- 1$ by $3$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (λ + 1 + (- 3 - - λ) \cdot 3 + 2 \cdot 3)$

Step 5.5.5.1.4

Multiply $- - λ$ .

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Step 5.5.5.1.4.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (λ + 1 + (- 3 + 1 λ) \cdot 3 + 2 \cdot 3)$

Step 5.5.5.1.4.2

Multiply $λ$ by $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (λ + 1 + (- 3 + λ) \cdot 3 + 2 \cdot 3)$

Step 5.5.5.1.5

Apply the distributive property.

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (λ + 1 - 3 \cdot 3 + λ \cdot 3 + 2 \cdot 3)$

Step 5.5.5.1.6

Multiply $- 3$ by $3$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (λ + 1 - 9 + λ \cdot 3 + 2 \cdot 3)$

Step 5.5.5.1.7

Move $3$ to the left of $λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (λ + 1 - 9 + 3 \cdot λ + 2 \cdot 3)$

Step 5.5.5.1.8

Multiply $2$ by $3$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (λ + 1 - 9 + 3 λ + 6)$

Step 5.5.5.2

Add $λ$ and $3 λ$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ + 1 - 9 + 6)$

Step 5.5.5.3

Subtract $9$ from $1$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 8 + 6)$

Step 5.5.5.4

Add $- 8$ and $6$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 0 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6

Simplify the determinant.

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Step 5.6.1

Add $(1 - λ) (- λ^{3} + 4 λ^{2} - 3)$ and $0$ .

$p (λ) = (1 - λ) (- λ^{3} + 4 λ^{2} - 3) + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2

Simplify each term.

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Step 5.6.2.1

Expand $(1 - λ) (- λ^{3} + 4 λ^{2} - 3)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 1 (- λ^{3}) + 1 (4 λ^{2}) + 1 \cdot - 3 - λ (- λ^{3}) - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2

Simplify each term.

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Step 5.6.2.2.1

Multiply $- λ^{3}$ by $1$ .

$p (λ) = - λ^{3} + 1 (4 λ^{2}) + 1 \cdot - 3 - λ (- λ^{3}) - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.2

Multiply $4 λ^{2}$ by $1$ .

$p (λ) = - λ^{3} + 4 λ^{2} + 1 \cdot - 3 - λ (- λ^{3}) - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.3

Multiply $- 3$ by $1$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 - λ (- λ^{3}) - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{3} + 4 λ^{2} - 3 - 1 \cdot - 1 λ \cdot λ^{3} - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.5

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

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Step 5.6.2.2.5.1

Move $λ^{3}$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 - 1 \cdot - 1 (λ^{3} λ) - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.5.2

Multiply $λ^{3}$ by $λ$ .

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Step 5.6.2.2.5.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 - 1 \cdot - 1 (λ^{3} λ^{1}) - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.5.2.2

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

$p (λ) = - λ^{3} + 4 λ^{2} - 3 - 1 \cdot - 1 λ^{3 + 1} - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.5.3

Add $3$ and $1$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 - 1 \cdot - 1 λ^{4} - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.6

Multiply $- 1$ by $- 1$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 + 1 λ^{4} - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.7

Multiply $λ^{4}$ by $1$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 + λ^{4} - λ (4 λ^{2}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.8

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{3} + 4 λ^{2} - 3 + λ^{4} - 1 \cdot 4 λ \cdot λ^{2} - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.9

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

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Step 5.6.2.2.9.1

Move $λ^{2}$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 + λ^{4} - 1 \cdot 4 (λ^{2} λ) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.9.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.6.2.2.9.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 + λ^{4} - 1 \cdot 4 (λ^{2} λ^{1}) - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.9.2.2

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

$p (λ) = - λ^{3} + 4 λ^{2} - 3 + λ^{4} - 1 \cdot 4 λ^{2 + 1} - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.9.3

Add $2$ and $1$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 + λ^{4} - 1 \cdot 4 λ^{3} - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.10

Multiply $- 1$ by $4$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 + λ^{4} - 4 λ^{3} - λ \cdot - 3 + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.2.11

Multiply $- 3$ by $- 1$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 + λ^{4} - 4 λ^{3} + 3 λ + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.3

Subtract $4 λ^{3}$ from $- λ^{3}$ .

$p (λ) = - 5 λ^{3} + 4 λ^{2} - 3 + λ^{4} + 3 λ + 5 (- 3 λ^{2} + 11 λ - 9) - 3 (4 λ - 2)$

Step 5.6.2.4

Apply the distributive property.

$p (λ) = - 5 λ^{3} + 4 λ^{2} - 3 + λ^{4} + 3 λ + 5 (- 3 λ^{2}) + 5 (11 λ) + 5 \cdot - 9 - 3 (4 λ - 2)$

Step 5.6.2.5

Simplify.

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Step 5.6.2.5.1

Multiply $- 3$ by $5$ .

$p (λ) = - 5 λ^{3} + 4 λ^{2} - 3 + λ^{4} + 3 λ - 15 λ^{2} + 5 (11 λ) + 5 \cdot - 9 - 3 (4 λ - 2)$

Step 5.6.2.5.2

Multiply $11$ by $5$ .

$p (λ) = - 5 λ^{3} + 4 λ^{2} - 3 + λ^{4} + 3 λ - 15 λ^{2} + 55 λ + 5 \cdot - 9 - 3 (4 λ - 2)$

Step 5.6.2.5.3

Multiply $5$ by $- 9$ .

$p (λ) = - 5 λ^{3} + 4 λ^{2} - 3 + λ^{4} + 3 λ - 15 λ^{2} + 55 λ - 45 - 3 (4 λ - 2)$

Step 5.6.2.6

Apply the distributive property.

$p (λ) = - 5 λ^{3} + 4 λ^{2} - 3 + λ^{4} + 3 λ - 15 λ^{2} + 55 λ - 45 - 3 (4 λ) - 3 \cdot - 2$

Step 5.6.2.7

Multiply $4$ by $- 3$ .

$p (λ) = - 5 λ^{3} + 4 λ^{2} - 3 + λ^{4} + 3 λ - 15 λ^{2} + 55 λ - 45 - 12 λ - 3 \cdot - 2$

Step 5.6.2.8

Multiply $- 3$ by $- 2$ .

$p (λ) = - 5 λ^{3} + 4 λ^{2} - 3 + λ^{4} + 3 λ - 15 λ^{2} + 55 λ - 45 - 12 λ + 6$

Step 5.6.3

Subtract $15 λ^{2}$ from $4 λ^{2}$ .

$p (λ) = - 5 λ^{3} - 11 λ^{2} - 3 + λ^{4} + 3 λ + 55 λ - 45 - 12 λ + 6$

Step 5.6.4

Subtract $45$ from $- 3$ .

$p (λ) = - 5 λ^{3} - 11 λ^{2} + λ^{4} + 3 λ + 55 λ - 48 - 12 λ + 6$

Step 5.6.5

Add $3 λ$ and $55 λ$ .

$p (λ) = - 5 λ^{3} - 11 λ^{2} + λ^{4} + 58 λ - 48 - 12 λ + 6$

Step 5.6.6

Subtract $12 λ$ from $58 λ$ .

$p (λ) = - 5 λ^{3} - 11 λ^{2} + λ^{4} + 46 λ - 48 + 6$

Step 5.6.7

Add $- 48$ and $6$ .

$p (λ) = - 5 λ^{3} - 11 λ^{2} + λ^{4} + 46 λ - 42$

Step 5.6.8

Move $- 11 λ^{2}$ .

$p (λ) = - 5 λ^{3} + λ^{4} - 11 λ^{2} + 46 λ - 42$

Step 5.6.9

Reorder $- 5 λ^{3}$ and $λ^{4}$ .

$p (λ) = λ^{4} - 5 λ^{3} - 11 λ^{2} + 46 λ - 42$

Step 6

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

$λ^{4} - 5 λ^{3} - 11 λ^{2} + 46 λ - 42 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

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

$λ \approx - 3.37785181, 5.74238923$