Linear Algebra Examples

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

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

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

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 $I_{4}$ .

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

Step 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.

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.5

Multiply $- λ \cdot 0$ .

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Step 4.1.2.5.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.5.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.6

Multiply $- 1$ by $1$ .

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

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.9

Multiply $- λ \cdot 0$ .

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Step 4.1.2.9.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.9.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.10

Multiply $- λ \cdot 0$ .

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Step 4.1.2.10.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.10.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.11

Multiply $- 1$ by $1$ .

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

Step 4.1.2.12

Multiply $- λ \cdot 0$ .

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Step 4.1.2.12.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.12.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.13

Multiply $- λ \cdot 0$ .

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Step 4.1.2.13.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.13.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.14

Multiply $- λ \cdot 0$ .

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Step 4.1.2.14.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.14.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.15

Multiply $- λ \cdot 0$ .

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Step 4.1.2.15.1

Multiply $0$ by $- 1$ .

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

Step 4.1.2.15.2

Multiply $0$ by $λ$ .

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

Step 4.1.2.16

Multiply $- 1$ by $1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add $0$ and $0$ .

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

Step 4.3.2

Add $0$ and $0$ .

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

Step 4.3.3

Add $1$ and $0$ .

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

Step 4.3.4

Add $0$ and $0$ .

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

Step 4.3.5

Add $1$ and $0$ .

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

Step 4.3.6

Add $0$ and $0$ .

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

Step 4.3.7

Add $0$ and $0$ .

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

Step 4.3.8

Add $1$ and $0$ .

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

Step 4.3.9

Add $0$ and $0$ .

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

Step 4.3.10

Add $1$ and $0$ .

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

Step 4.3.11

Add $0$ and $0$ .

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

Step 4.3.12

Add $0$ and $0$ .

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

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.

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

Step 5.1.4

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

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

Step 5.1.5

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

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

Step 5.1.6

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

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

Step 5.1.7

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

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

Step 5.1.8

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

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

Step 5.1.9

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

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

Step 5.1.10

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

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

Step 5.1.11

Add the terms together.

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

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

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

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

Step 5.4.1.4

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

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

Step 5.4.1.5

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

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

Step 5.4.1.6

Multiply element $a_{32}$ by its cofactor.

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

Step 5.4.1.7

The minor for $a_{33}$ is the determinant with row $3$ and column $3$ deleted.

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

Step 5.4.1.8

Multiply element $a_{33}$ by its cofactor.

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

Step 5.4.1.9

Add the terms together.

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

Step 5.4.2

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

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

Step 5.4.3

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

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

Step 5.4.4

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

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Step 5.4.4.1

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

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

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

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

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Step 5.4.4.2.1.1.1

Apply the distributive property.

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

Step 5.4.4.2.1.1.2

Apply the distributive property.

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

Step 5.4.4.2.1.1.3

Apply the distributive property.

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

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 $1$ by $1$ .

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

Step 5.4.4.2.1.2.1.2

Multiply $- λ$ by $1$ .

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

Step 5.4.4.2.1.2.1.3

Multiply $- 1$ by $1$ .

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

Step 5.4.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

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 $λ$ .

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

Step 5.4.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

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

Step 5.4.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

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

Step 5.4.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

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

Step 5.4.4.2.1.2.2

Subtract $λ$ from $- λ$ .

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

Step 5.4.4.2.1.3

Multiply $- 1$ by $1$ .

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

Step 5.4.4.2.2

Combine the opposite terms in $1 - 2 λ + λ^{2} - 1$ .

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Step 5.4.4.2.2.1

Subtract $1$ from $1$ .

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

Step 5.4.4.2.2.2

Add $- 2 λ + λ^{2}$ and $0$ .

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

Step 5.4.4.2.3

Reorder $- 2 λ$ and $λ^{2}$ .

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

Step 5.4.5

Simplify the determinant.

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Step 5.4.5.1

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

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Step 5.4.5.1.1

Add $0$ and $0$ .

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

Step 5.4.5.1.2

Add $0$ and $(3 - λ) (λ^{2} - 2 λ)$ .

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

Step 5.4.5.2

Expand $(3 - λ) (λ^{2} - 2 λ)$ using the FOIL Method.

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Step 5.4.5.2.1

Apply the distributive property.

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

Step 5.4.5.2.2

Apply the distributive property.

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

Step 5.4.5.2.3

Apply the distributive property.

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

Step 5.4.5.3

Simplify and combine like terms.

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Step 5.4.5.3.1

Simplify each term.

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Step 5.4.5.3.1.1

Multiply $- 2$ by $3$ .

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

Step 5.4.5.3.1.2

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

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Step 5.4.5.3.1.2.1

Move $λ^{2}$ .

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

Step 5.4.5.3.1.2.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.4.5.3.1.2.2.1

Raise $λ$ to the power of $1$ .

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

Step 5.4.5.3.1.2.2.2

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

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

Step 5.4.5.3.1.2.3

Add $2$ and $1$ .

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

Step 5.4.5.3.1.3

Rewrite using the commutative property of multiplication.

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

Step 5.4.5.3.1.4

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

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Step 5.4.5.3.1.4.1

Move $λ$ .

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

Step 5.4.5.3.1.4.2

Multiply $λ$ by $λ$ .

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

Step 5.4.5.3.1.5

Multiply $- 1$ by $- 2$ .

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

Step 5.4.5.3.2

Add $3 λ^{2}$ and $2 λ^{2}$ .

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

Step 5.4.5.4

Move $- 6 λ$ .

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

Step 5.4.5.5

Reorder $5 λ^{2}$ and $- λ^{3}$ .

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

Step 5.5

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

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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.

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

Step 5.5.1.4

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

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

Step 5.5.1.5

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

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

Step 5.5.1.6

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

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

Step 5.5.1.7

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

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

Step 5.5.1.8

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

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

Step 5.5.1.9

Add the terms together.

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

Step 5.5.2

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

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

Step 5.5.3

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

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Step 5.5.3.1

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

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

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 $0$ by $0$ .

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

Step 5.5.3.2.1.2

Apply the distributive property.

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

Step 5.5.3.2.1.3

Multiply $- 1$ by $1$ .

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

Step 5.5.3.2.1.4

Multiply $- - λ$ .

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Step 5.5.3.2.1.4.1

Multiply $- 1$ by $- 1$ .

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

Step 5.5.3.2.1.4.2

Multiply $λ$ by $1$ .

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

Step 5.5.3.2.2

Subtract $1$ from $0$ .

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

Step 5.5.3.2.3

Reorder $- 1$ and $λ$ .

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

Step 5.5.4

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

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Step 5.5.4.1

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

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

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 $0$ by $0$ .

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

Step 5.5.4.2.1.2

Multiply $- 1$ by $1$ .

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

Step 5.5.4.2.2

Subtract $1$ from $0$ .

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

Step 5.5.5

Simplify the determinant.

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Step 5.5.5.1

Subtract $(1 - λ) (λ - 1)$ from $0$ .

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

Step 5.5.5.2

Simplify each term.

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Step 5.5.5.2.1

Apply the distributive property.

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

Step 5.5.5.2.2

Multiply $- 1$ by $1$ .

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

Step 5.5.5.2.3

Multiply $- - λ$ .

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Step 5.5.5.2.3.1

Multiply $- 1$ by $- 1$ .

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

Step 5.5.5.2.3.2

Multiply $λ$ by $1$ .

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

Step 5.5.5.2.4

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

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Step 5.5.5.2.4.1

Apply the distributive property.

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

Step 5.5.5.2.4.2

Apply the distributive property.

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

Step 5.5.5.2.4.3

Apply the distributive property.

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

Step 5.5.5.2.5

Simplify and combine like terms.

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Step 5.5.5.2.5.1

Simplify each term.

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Step 5.5.5.2.5.1.1

Rewrite $- 1 λ$ as $- λ$ .

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

Step 5.5.5.2.5.1.2

Multiply $- 1$ by $- 1$ .

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

Step 5.5.5.2.5.1.3

Multiply $λ$ by $λ$ .

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

Step 5.5.5.2.5.1.4

Move $- 1$ to the left of $λ$ .

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

Step 5.5.5.2.5.1.5

Rewrite $- 1 λ$ as $- λ$ .

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

Step 5.5.5.2.5.2

Subtract $λ$ from $- λ$ .

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

Step 5.5.5.2.6

Multiply $- 1$ by $1$ .

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

Step 5.5.5.3

Combine the opposite terms in $- 2 λ + 1 + λ^{2} - 1$ .

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Step 5.5.5.3.1

Subtract $1$ from $1$ .

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

Step 5.5.5.3.2

Add $- 2 λ + λ^{2}$ and $0$ .

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

Step 5.5.5.4

Reorder $- 2 λ$ and $λ^{2}$ .

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

Step 5.6

Simplify the determinant.

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Step 5.6.1

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

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Step 5.6.1.1

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

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

Step 5.6.1.2

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

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

Step 5.6.2

Simplify each term.

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Step 5.6.2.1

Expand $(1 - λ) (- λ^{3} + 5 λ^{2} - 6 λ)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 5.6.2.2

Simplify each term.

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Step 5.6.2.2.1

Multiply $- λ^{3}$ by $1$ .

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

Step 5.6.2.2.2

Multiply $5 λ^{2}$ by $1$ .

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

Step 5.6.2.2.3

Multiply $- 6 λ$ by $1$ .

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

Step 5.6.2.2.4

Rewrite using the commutative property of multiplication.

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

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 $λ^{3}$ .

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

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 $1$ .

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

Step 5.6.2.2.5.2.2

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

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

Step 5.6.2.2.5.3

Add $3$ and $1$ .

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

Step 5.6.2.2.6

Multiply $- 1$ by $- 1$ .

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

Step 5.6.2.2.7

Multiply $λ^{4}$ by $1$ .

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

Step 5.6.2.2.8

Rewrite using the commutative property of multiplication.

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

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 $λ^{2}$ .

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

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 $1$ .

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

Step 5.6.2.2.9.2.2

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

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

Step 5.6.2.2.9.3

Add $2$ and $1$ .

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

Step 5.6.2.2.10

Multiply $- 1$ by $5$ .

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

Step 5.6.2.2.11

Rewrite using the commutative property of multiplication.

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

Step 5.6.2.2.12

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

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Step 5.6.2.2.12.1

Move $λ$ .

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

Step 5.6.2.2.12.2

Multiply $λ$ by $λ$ .

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

Step 5.6.2.2.13

Multiply $- 1$ by $- 6$ .

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

Step 5.6.2.3

Subtract $5 λ^{3}$ from $- λ^{3}$ .

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

Step 5.6.2.4

Add $5 λ^{2}$ and $6 λ^{2}$ .

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

Step 5.6.2.5

Apply the distributive property.

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

Step 5.6.2.6

Rewrite $- 1 λ^{2}$ as $- λ^{2}$ .

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

Step 5.6.2.7

Multiply $- 2$ by $- 1$ .

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

Step 5.6.3

Subtract $λ^{2}$ from $11 λ^{2}$ .

$p (λ) = - 6 λ^{3} + 10 λ^{2} - 6 λ + λ^{4} + 2 λ$

Step 5.6.4

Add $- 6 λ$ and $2 λ$ .

$p (λ) = - 6 λ^{3} + 10 λ^{2} + λ^{4} - 4 λ$

Step 5.6.5

Move $10 λ^{2}$ .

$p (λ) = - 6 λ^{3} + λ^{4} + 10 λ^{2} - 4 λ$

Step 5.6.6

Reorder $- 6 λ^{3}$ and $λ^{4}$ .

$p (λ) = λ^{4} - 6 λ^{3} + 10 λ^{2} - 4 λ$

Step 6

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

$λ^{4} - 6 λ^{3} + 10 λ^{2} - 4 λ = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Factor the left side of the equation.

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Step 7.1.1

Factor $λ$ out of $λ^{4} - 6 λ^{3} + 10 λ^{2} - 4 λ$ .

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Step 7.1.1.1

Factor $λ$ out of $λ^{4}$ .

$λ \cdot λ^{3} - 6 λ^{3} + 10 λ^{2} - 4 λ = 0$

Step 7.1.1.2

Factor $λ$ out of $- 6 λ^{3}$ .

$λ \cdot λ^{3} + λ (- 6 λ^{2}) + 10 λ^{2} - 4 λ = 0$

Step 7.1.1.3

Factor $λ$ out of $10 λ^{2}$ .

$λ \cdot λ^{3} + λ (- 6 λ^{2}) + λ (10 λ) - 4 λ = 0$

Step 7.1.1.4

Factor $λ$ out of $- 4 λ$ .

$λ \cdot λ^{3} + λ (- 6 λ^{2}) + λ (10 λ) + λ \cdot - 4 = 0$

Step 7.1.1.5

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

$λ (λ^{3} - 6 λ^{2}) + λ (10 λ) + λ \cdot - 4 = 0$

Step 7.1.1.6

Factor $λ$ out of $λ (λ^{3} - 6 λ^{2}) + λ (10 λ)$ .

$λ (λ^{3} - 6 λ^{2} + 10 λ) + λ \cdot - 4 = 0$

Step 7.1.1.7

Factor $λ$ out of $λ (λ^{3} - 6 λ^{2} + 10 λ) + λ \cdot - 4$ .

$λ (λ^{3} - 6 λ^{2} + 10 λ - 4) = 0$

Step 7.1.2

Factor.

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Step 7.1.2.1

Factor $λ^{3} - 6 λ^{2} + 10 λ - 4$ using the rational roots test.

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Step 7.1.2.1.1

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

$p = \pm 1, \pm 4, \pm 2$

$q = \pm 1$

Step 7.1.2.1.2

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

$\pm 1, \pm 4, \pm 2$

Step 7.1.2.1.3

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

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Step 7.1.2.1.3.1

Substitute $2$ into the polynomial.

$2^{3} - 6 \cdot 2^{2} + 10 \cdot 2 - 4$

Step 7.1.2.1.3.2

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

$8 - 6 \cdot 2^{2} + 10 \cdot 2 - 4$

Step 7.1.2.1.3.3

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

$8 - 6 \cdot 4 + 10 \cdot 2 - 4$

Step 7.1.2.1.3.4

Multiply $- 6$ by $4$ .

$8 - 24 + 10 \cdot 2 - 4$

Step 7.1.2.1.3.5

Subtract $24$ from $8$ .

$- 16 + 10 \cdot 2 - 4$

Step 7.1.2.1.3.6

Multiply $10$ by $2$ .

$- 16 + 20 - 4$

Step 7.1.2.1.3.7

Add $- 16$ and $20$ .

$4 - 4$

Step 7.1.2.1.3.8

Subtract $4$ from $4$ .

$0$

Step 7.1.2.1.4

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

$\frac{λ^{3} - 6 λ^{2} + 10 λ - 4}{λ - 2}$

Step 7.1.2.1.5

Divide $λ^{3} - 6 λ^{2} + 10 λ - 4$ by $λ - 2$ .

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Step 7.1.2.1.5.1

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

$λ$

$2$

$λ^{3}$

$6 λ^{2}$

$10 λ$

$4$

Step 7.1.2.1.5.2

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

					$λ^{2}$
	$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$

Step 7.1.2.1.5.3

Multiply the new quotient term by the divisor.

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			+	$λ^{3}$	-	$2 λ^{2}$

Step 7.1.2.1.5.4

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

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$

Step 7.1.2.1.5.5

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

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$

Step 7.1.2.1.5.6

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

				$λ^{2}$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$

Step 7.1.2.1.5.7

Divide the highest order term in the dividend $- 4 λ^{2}$ by the highest order term in divisor $λ$ .

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$

Step 7.1.2.1.5.8

Multiply the new quotient term by the divisor.

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$
					-	$4 λ^{2}$	+	$8 λ$

Step 7.1.2.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 λ^{2} + 8 λ$

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$
					+	$4 λ^{2}$	-	$8 λ$

Step 7.1.2.1.5.10

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

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$2 λ$

Step 7.1.2.1.5.11

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

				$λ^{2}$	-	$4 λ$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$2 λ$	-	$4$

Step 7.1.2.1.5.12

Divide the highest order term in the dividend $2 λ$ by the highest order term in divisor $λ$ .

				$λ^{2}$	-	$4 λ$	+	$2$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$2 λ$	-	$4$

Step 7.1.2.1.5.13

Multiply the new quotient term by the divisor.

				$λ^{2}$	-	$4 λ$	+	$2$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$2 λ$	-	$4$
							+	$2 λ$	-	$4$

Step 7.1.2.1.5.14

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

				$λ^{2}$	-	$4 λ$	+	$2$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$2 λ$	-	$4$
							-	$2 λ$	+	$4$

Step 7.1.2.1.5.15

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

				$λ^{2}$	-	$4 λ$	+	$2$
$λ$	-	$2$		$λ^{3}$	-	$6 λ^{2}$	+	$10 λ$	-	$4$
			-	$λ^{3}$	+	$2 λ^{2}$
					-	$4 λ^{2}$	+	$10 λ$
					+	$4 λ^{2}$	-	$8 λ$
							+	$2 λ$	-	$4$
							-	$2 λ$	+	$4$
										$0$

Step 7.1.2.1.5.16

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

$λ^{2} - 4 λ + 2$

Step 7.1.2.1.6

Write $λ^{3} - 6 λ^{2} + 10 λ - 4$ as a set of factors.

$λ ((λ - 2) (λ^{2} - 4 λ + 2)) = 0$

Step 7.1.2.2

Remove unnecessary parentheses.

$λ (λ - 2) (λ^{2} - 4 λ + 2) = 0$

Step 7.2

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

$λ = 0$

$λ - 2 = 0$

$λ^{2} - 4 λ + 2 = 0$

Step 7.3

Set $λ$ equal to $0$ .

$λ = 0$

Step 7.4

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

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Step 7.4.1

Set $λ - 2$ equal to $0$ .

$λ - 2 = 0$

Step 7.4.2

Add $2$ to both sides of the equation.

$λ = 2$

Step 7.5

Set $λ^{2} - 4 λ + 2$ equal to $0$ and solve for $λ$ .

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Step 7.5.1

Set $λ^{2} - 4 λ + 2$ equal to $0$ .

$λ^{2} - 4 λ + 2 = 0$

Step 7.5.2

Solve $λ^{2} - 4 λ + 2 = 0$ for $λ$ .

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Step 7.5.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 7.5.2.2

Substitute the values $a = 1$ , $b = - 4$ , and $c = 2$ into the quadratic formula and solve for $λ$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot 2)}}{2 \cdot 1}$

Step 7.5.2.3

Simplify.

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Step 7.5.2.3.1

Simplify the numerator.

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Step 7.5.2.3.1.1

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

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}$

Step 7.5.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 2$ .

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Step 7.5.2.3.1.2.1

Multiply $- 4$ by $1$ .

$λ = \frac{4 \pm \sqrt{16 - 4 \cdot 2}}{2 \cdot 1}$

Step 7.5.2.3.1.2.2

Multiply $- 4$ by $2$ .

$λ = \frac{4 \pm \sqrt{16 - 8}}{2 \cdot 1}$

Step 7.5.2.3.1.3

Subtract $8$ from $16$ .

$λ = \frac{4 \pm \sqrt{8}}{2 \cdot 1}$

Step 7.5.2.3.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 7.5.2.3.1.4.1

Factor $4$ out of $8$ .

$λ = \frac{4 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Step 7.5.2.3.1.4.2

Rewrite $4$ as $2^{2}$ .

$λ = \frac{4 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Step 7.5.2.3.1.5

Pull terms out from under the radical.

$λ = \frac{4 \pm 2 \sqrt{2}}{2 \cdot 1}$

Step 7.5.2.3.2

Multiply $2$ by $1$ .

$λ = \frac{4 \pm 2 \sqrt{2}}{2}$

Step 7.5.2.3.3

Simplify $\frac{4 \pm 2 \sqrt{2}}{2}$ .

$λ = 2 \pm \sqrt{2}$

Step 7.5.2.4

The final answer is the combination of both solutions.

$λ = 2 + \sqrt{2}, 2 - \sqrt{2}$

Step 7.6

The final solution is all the values that make $λ (λ - 2) (λ^{2} - 4 λ + 2) = 0$ true.

$λ = 0, 2, 2 + \sqrt{2}, 2 - \sqrt{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$λ = 0, 2, 2 + \sqrt{2}, 2 - \sqrt{2}$

Decimal Form:

$λ = 0, 2, 3.41421356 \dots, 0.58578643 \dots$