Linear Algebra Examples

$[\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$

Step 1

Find the eigenvalues.

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Step 1.1

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

$p (λ) = determinant (A - λ I_{3})$

Step 1.2

The identity matrix or unit matrix of size $3$ is the $3 \times 3$ square matrix with ones on the main diagonal and zeros elsewhere.

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

Step 1.3

Substitute the known values into $p (λ) = determinant (A - λ I_{3})$ .

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Step 1.3.1

Substitute $[\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] - λ I_{3})$

Step 1.3.2

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

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 1.4

Simplify.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

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

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 2 - λ & 4 + 0 & 6 + 0 \\ 8 + 0 & 10 - λ & 12 + 0 \\ 1 + 0 & 2 + 0 & 0 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 4 & 6 + 0 \\ 8 + 0 & 10 - λ & 12 + 0 \\ 1 + 0 & 2 + 0 & 0 - λ \end{array}]$

Step 1.4.3.2

Add $6$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 4 & 6 \\ 8 + 0 & 10 - λ & 12 + 0 \\ 1 + 0 & 2 + 0 & 0 - λ \end{array}]$

Step 1.4.3.3

Add $8$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 4 & 6 \\ 8 & 10 - λ & 12 + 0 \\ 1 + 0 & 2 + 0 & 0 - λ \end{array}]$

Step 1.4.3.4

Add $12$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 4 & 6 \\ 8 & 10 - λ & 12 \\ 1 + 0 & 2 + 0 & 0 - λ \end{array}]$

Step 1.4.3.5

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 4 & 6 \\ 8 & 10 - λ & 12 \\ 1 & 2 + 0 & 0 - λ \end{array}]$

Step 1.4.3.6

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 4 & 6 \\ 8 & 10 - λ & 12 \\ 1 & 2 & 0 - λ \end{array}]$

Step 1.4.3.7

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} 2 - λ & 4 & 6 \\ 8 & 10 - λ & 12 \\ 1 & 2 & - λ \end{array}]$

Step 1.5

Find the determinant.

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Step 1.5.1

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

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Step 1.5.1.1

Consider the corresponding sign chart.

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

Step 1.5.1.2

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

Step 1.5.1.3

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

$| \begin{array}{c} 10 - λ & 12 \\ 2 & - λ \end{array} |$

Step 1.5.1.4

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

$(2 - λ) | \begin{array}{c} 10 - λ & 12 \\ 2 & - λ \end{array} |$

Step 1.5.1.5

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

$| \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} |$

Step 1.5.1.6

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

$- 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} |$

Step 1.5.1.7

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

$| \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.1.8

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

$6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.1.9

Add the terms together.

$p (λ) = (2 - λ) | \begin{array}{c} 10 - λ & 12 \\ 2 & - λ \end{array} | - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2

Evaluate $| \begin{array}{c} 10 - λ & 12 \\ 2 & - λ \end{array} |$ .

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Step 1.5.2.1

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

$p (λ) = (2 - λ) ((10 - λ) (- λ) - 2 \cdot 12) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2.2

Simplify the determinant.

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Step 1.5.2.2.1

Simplify each term.

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Step 1.5.2.2.1.1

Apply the distributive property.

$p (λ) = (2 - λ) (10 (- λ) - λ (- λ) - 2 \cdot 12) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2.2.1.2

Multiply $- 1$ by $10$ .

$p (λ) = (2 - λ) (- 10 λ - λ (- λ) - 2 \cdot 12) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2.2.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (2 - λ) (- 10 λ - 1 \cdot - 1 λ \cdot λ - 2 \cdot 12) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2.2.1.4

Simplify each term.

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Step 1.5.2.2.1.4.1

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

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Step 1.5.2.2.1.4.1.1

Move $λ$ .

$p (λ) = (2 - λ) (- 10 λ - 1 \cdot - 1 (λ \cdot λ) - 2 \cdot 12) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2.2.1.4.1.2

Multiply $λ$ by $λ$ .

$p (λ) = (2 - λ) (- 10 λ - 1 \cdot - 1 λ^{2} - 2 \cdot 12) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2.2.1.4.2

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (- 10 λ + 1 λ^{2} - 2 \cdot 12) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2.2.1.4.3

Multiply $λ^{2}$ by $1$ .

$p (λ) = (2 - λ) (- 10 λ + λ^{2} - 2 \cdot 12) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2.2.1.5

Multiply $- 2$ by $12$ .

$p (λ) = (2 - λ) (- 10 λ + λ^{2} - 24) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.2.2.2

Reorder $- 10 λ$ and $λ^{2}$ .

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 | \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} | + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.3

Evaluate $| \begin{array}{c} 8 & 12 \\ 1 & - λ \end{array} |$ .

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Step 1.5.3.1

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

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (8 (- λ) - 1 \cdot 12) + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.3.2

Simplify each term.

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Step 1.5.3.2.1

Multiply $- 1$ by $8$ .

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 1 \cdot 12) + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.3.2.2

Multiply $- 1$ by $12$ .

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 12) + 6 | \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$

Step 1.5.4

Evaluate $| \begin{array}{c} 8 & 10 - λ \\ 1 & 2 \end{array} |$ .

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Step 1.5.4.1

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

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 12) + 6 (8 \cdot 2 - (10 - λ))$

Step 1.5.4.2

Simplify the determinant.

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Step 1.5.4.2.1

Simplify each term.

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Step 1.5.4.2.1.1

Multiply $8$ by $2$ .

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 12) + 6 (16 - (10 - λ))$

Step 1.5.4.2.1.2

Apply the distributive property.

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 12) + 6 (16 - 1 \cdot 10 - - λ)$

Step 1.5.4.2.1.3

Multiply $- 1$ by $10$ .

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 12) + 6 (16 - 10 - - λ)$

Step 1.5.4.2.1.4

Multiply $- - λ$ .

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Step 1.5.4.2.1.4.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 12) + 6 (16 - 10 + 1 λ)$

Step 1.5.4.2.1.4.2

Multiply $λ$ by $1$ .

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 12) + 6 (16 - 10 + λ)$

Step 1.5.4.2.2

Subtract $10$ from $16$ .

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 12) + 6 (6 + λ)$

Step 1.5.4.2.3

Reorder $6$ and $λ$ .

$p (λ) = (2 - λ) (λ^{2} - 10 λ - 24) - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5

Simplify the determinant.

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Step 1.5.5.1

Simplify each term.

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Step 1.5.5.1.1

Expand $(2 - λ) (λ^{2} - 10 λ - 24)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 2 λ^{2} + 2 (- 10 λ) + 2 \cdot - 24 - λ \cdot λ^{2} - λ (- 10 λ) - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2

Simplify each term.

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Step 1.5.5.1.2.1

Multiply $- 10$ by $2$ .

$p (λ) = 2 λ^{2} - 20 λ + 2 \cdot - 24 - λ \cdot λ^{2} - λ (- 10 λ) - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.2

Multiply $2$ by $- 24$ .

$p (λ) = 2 λ^{2} - 20 λ - 48 - λ \cdot λ^{2} - λ (- 10 λ) - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.3

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

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Step 1.5.5.1.2.3.1

Move $λ^{2}$ .

$p (λ) = 2 λ^{2} - 20 λ - 48 - (λ^{2} λ) - λ (- 10 λ) - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.5.1.2.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 2 λ^{2} - 20 λ - 48 - (λ^{2} λ^{1}) - λ (- 10 λ) - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.3.2.2

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

$p (λ) = 2 λ^{2} - 20 λ - 48 - λ^{2 + 1} - λ (- 10 λ) - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.3.3

Add $2$ and $1$ .

$p (λ) = 2 λ^{2} - 20 λ - 48 - λ^{3} - λ (- 10 λ) - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = 2 λ^{2} - 20 λ - 48 - λ^{3} - 1 \cdot - 10 λ \cdot λ - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.5

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

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Step 1.5.5.1.2.5.1

Move $λ$ .

$p (λ) = 2 λ^{2} - 20 λ - 48 - λ^{3} - 1 \cdot - 10 (λ \cdot λ) - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 2 λ^{2} - 20 λ - 48 - λ^{3} - 1 \cdot - 10 λ^{2} - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.6

Multiply $- 1$ by $- 10$ .

$p (λ) = 2 λ^{2} - 20 λ - 48 - λ^{3} + 10 λ^{2} - λ \cdot - 24 - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.2.7

Multiply $- 24$ by $- 1$ .

$p (λ) = 2 λ^{2} - 20 λ - 48 - λ^{3} + 10 λ^{2} + 24 λ - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.3

Add $2 λ^{2}$ and $10 λ^{2}$ .

$p (λ) = 12 λ^{2} - 20 λ - 48 - λ^{3} + 24 λ - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.4

Add $- 20 λ$ and $24 λ$ .

$p (λ) = 12 λ^{2} + 4 λ - 48 - λ^{3} - 4 (- 8 λ - 12) + 6 (λ + 6)$

Step 1.5.5.1.5

Apply the distributive property.

$p (λ) = 12 λ^{2} + 4 λ - 48 - λ^{3} - 4 (- 8 λ) - 4 \cdot - 12 + 6 (λ + 6)$

Step 1.5.5.1.6

Multiply $- 8$ by $- 4$ .

$p (λ) = 12 λ^{2} + 4 λ - 48 - λ^{3} + 32 λ - 4 \cdot - 12 + 6 (λ + 6)$

Step 1.5.5.1.7

Multiply $- 4$ by $- 12$ .

$p (λ) = 12 λ^{2} + 4 λ - 48 - λ^{3} + 32 λ + 48 + 6 (λ + 6)$

Step 1.5.5.1.8

Apply the distributive property.

$p (λ) = 12 λ^{2} + 4 λ - 48 - λ^{3} + 32 λ + 48 + 6 λ + 6 \cdot 6$

Step 1.5.5.1.9

Multiply $6$ by $6$ .

$p (λ) = 12 λ^{2} + 4 λ - 48 - λ^{3} + 32 λ + 48 + 6 λ + 36$

Step 1.5.5.2

Combine the opposite terms in $12 λ^{2} + 4 λ - 48 - λ^{3} + 32 λ + 48 + 6 λ + 36$ .

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Step 1.5.5.2.1

Add $- 48$ and $48$ .

$p (λ) = 12 λ^{2} + 4 λ - λ^{3} + 32 λ + 0 + 6 λ + 36$

Step 1.5.5.2.2

Add $12 λ^{2} + 4 λ - λ^{3} + 32 λ$ and $0$ .

$p (λ) = 12 λ^{2} + 4 λ - λ^{3} + 32 λ + 6 λ + 36$

Step 1.5.5.3

Add $4 λ$ and $32 λ$ .

$p (λ) = 12 λ^{2} - λ^{3} + 36 λ + 6 λ + 36$

Step 1.5.5.4

Add $36 λ$ and $6 λ$ .

$p (λ) = 12 λ^{2} - λ^{3} + 42 λ + 36$

Step 1.5.5.5

Reorder $12 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 12 λ^{2} + 42 λ + 36$

Step 1.6

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

$- λ^{3} + 12 λ^{2} + 42 λ + 36 = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

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

$λ \approx 14.96690066$

Step 2

The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where $N$ is the null space and $I$ is the identity matrix.

$ε_{A} = N (A - λ I_{3})$

Step 3

Find the eigenvector using the eigenvalue $λ = 14.96690066$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] - 14.96690066 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply $- 14.96690066$ by each element of the matrix.

$[\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - 14.96690066 \cdot 1 & - 14.96690066 \cdot 0 & - 14.96690066 \cdot 0 \\ - 14.96690066 \cdot 0 & - 14.96690066 \cdot 1 & - 14.96690066 \cdot 0 \\ - 14.96690066 \cdot 0 & - 14.96690066 \cdot 0 & - 14.96690066 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $- 14.96690066$ by $1$ .

$[\begin{array}{c} 2 & 4 & 6 \\ 8 & 10 & 12 \\ 1 & 2 & 0 \end{array}] + [\begin{array}{c} - 14.96690066 & - 14.96690066 \cdot 0 & - 14.96690066 \cdot 0 \\ - 14.96690066 \cdot 0 & - 14.96690066 \cdot 1 & - 14.96690066 \cdot 0 \\ - 14.96690066 \cdot 0 & - 14.96690066 \cdot 0 & - 14.96690066 \cdot 1 \end{array}]$

Step 3.2.1.2.2