Linear Algebra Examples

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}]$

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 $[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] - λ I_{3})$

Step 1.3.2

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

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} - 4 - λ & - 8 + 0 & - 6 + 0 \\ - 8 + 0 & 6 - λ & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $- 8$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 4 - λ & - 8 & - 6 + 0 \\ - 8 + 0 & 6 - λ & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - λ \end{array}]$

Step 1.4.3.2

Add $- 6$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 4 - λ & - 8 & - 6 \\ - 8 + 0 & 6 - λ & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - λ \end{array}]$

Step 1.4.3.3

Add $- 8$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 4 - λ & - 8 & - 6 \\ - 8 & 6 - λ & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - λ \end{array}]$

Step 1.4.3.4

Add $- 8$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 4 - λ & - 8 & - 6 \\ - 8 & 6 - λ & - 8 \\ 8 + 0 & - 7 + 0 & 10 - λ \end{array}]$

Step 1.4.3.5

Add $8$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 4 - λ & - 8 & - 6 \\ - 8 & 6 - λ & - 8 \\ 8 & - 7 + 0 & 10 - λ \end{array}]$

Step 1.4.3.6

Add $- 7$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - 4 - λ & - 8 & - 6 \\ - 8 & 6 - λ & - 8 \\ 8 & - 7 & 10 - λ \end{array}]$

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.

$| \begin{array}{c} 6 - λ & - 8 \\ - 7 & 10 - λ \end{array} |$

Step 1.5.1.4

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

$(- 4 - λ) | \begin{array}{c} 6 - λ & - 8 \\ - 7 & 10 - λ \end{array} |$

Step 1.5.1.5

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

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

Step 1.5.1.6

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

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

Step 1.5.1.7

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

$| \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.1.8

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

$- 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.1.9

Add the terms together.

$p (λ) = (- 4 - λ) | \begin{array}{c} 6 - λ & - 8 \\ - 7 & 10 - λ \end{array} | + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2

Evaluate $| \begin{array}{c} 6 - λ & - 8 \\ - 7 & 10 - λ \end{array} |$ .

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Step 1.5.2.1

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

$p (λ) = (- 4 - λ) ((6 - λ) (10 - λ) - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

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

Expand $(6 - λ) (10 - λ)$ using the FOIL Method.

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Step 1.5.2.2.1.1.1

Apply the distributive property.

$p (λ) = (- 4 - λ) (6 (10 - λ) - λ (10 - λ) - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.1.2

Apply the distributive property.

$p (λ) = (- 4 - λ) (6 \cdot 10 + 6 (- λ) - λ (10 - λ) - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.1.3

Apply the distributive property.

$p (λ) = (- 4 - λ) (6 \cdot 10 + 6 (- λ) - λ \cdot 10 - λ (- λ) - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.2

Simplify and combine like terms.

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Step 1.5.2.2.1.2.1

Simplify each term.

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Step 1.5.2.2.1.2.1.1

Multiply $6$ by $10$ .

$p (λ) = (- 4 - λ) (60 + 6 (- λ) - λ \cdot 10 - λ (- λ) - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.2.1.2

Multiply $- 1$ by $6$ .

$p (λ) = (- 4 - λ) (60 - 6 λ - λ \cdot 10 - λ (- λ) - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.2.1.3

Multiply $10$ by $- 1$ .

$p (λ) = (- 4 - λ) (60 - 6 λ - 10 λ - λ (- λ) - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (- 4 - λ) (60 - 6 λ - 10 λ - 1 \cdot - 1 λ \cdot λ - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.2.1.5

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

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Step 1.5.2.2.1.2.1.5.1

Move $λ$ .

$p (λ) = (- 4 - λ) (60 - 6 λ - 10 λ - 1 \cdot - 1 (λ \cdot λ) - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (- 4 - λ) (60 - 6 λ - 10 λ - 1 \cdot - 1 λ^{2} - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = (- 4 - λ) (60 - 6 λ - 10 λ + 1 λ^{2} - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = (- 4 - λ) (60 - 6 λ - 10 λ + λ^{2} - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.2.2

Subtract $10 λ$ from $- 6 λ$ .

$p (λ) = (- 4 - λ) (60 - 16 λ + λ^{2} - (- 7 \cdot - 8)) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.3

Multiply $- (- 7 \cdot - 8)$ .

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Step 1.5.2.2.1.3.1

Multiply $- 7$ by $- 8$ .

$p (λ) = (- 4 - λ) (60 - 16 λ + λ^{2} - 1 \cdot 56) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.1.3.2

Multiply $- 1$ by $56$ .

$p (λ) = (- 4 - λ) (60 - 16 λ + λ^{2} - 56) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.2

Subtract $56$ from $60$ .

$p (λ) = (- 4 - λ) (- 16 λ + λ^{2} + 4) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.2.2.3

Reorder $- 16 λ$ and $λ^{2}$ .

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 | \begin{array}{c} - 8 & - 8 \\ 8 & 10 - λ \end{array} | - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.3

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

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Step 1.5.3.1

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

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (- 8 (10 - λ) - 8 \cdot - 8) - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.3.2

Simplify the determinant.

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Step 1.5.3.2.1

Simplify each term.

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Step 1.5.3.2.1.1

Apply the distributive property.

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (- 8 \cdot 10 - 8 (- λ) - 8 \cdot - 8) - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.3.2.1.2

Multiply $- 8$ by $10$ .

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (- 80 - 8 (- λ) - 8 \cdot - 8) - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.3.2.1.3

Multiply $- 1$ by $- 8$ .

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (- 80 + 8 λ - 8 \cdot - 8) - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.3.2.1.4

Multiply $- 8$ by $- 8$ .

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (- 80 + 8 λ + 64) - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.3.2.2

Add $- 80$ and $64$ .

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (8 λ - 16) - 6 | \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$

Step 1.5.4

Evaluate $| \begin{array}{c} - 8 & 6 - λ \\ 8 & - 7 \end{array} |$ .

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Step 1.5.4.1

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

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (8 λ - 16) - 6 (- 8 \cdot - 7 - 8 (6 - λ))$

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 $- 7$ .

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (8 λ - 16) - 6 (56 - 8 (6 - λ))$

Step 1.5.4.2.1.2

Apply the distributive property.

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (8 λ - 16) - 6 (56 - 8 \cdot 6 - 8 (- λ))$

Step 1.5.4.2.1.3

Multiply $- 8$ by $6$ .

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (8 λ - 16) - 6 (56 - 48 - 8 (- λ))$

Step 1.5.4.2.1.4

Multiply $- 1$ by $- 8$ .

$p (λ) = (- 4 - λ) (λ^{2} - 16 λ + 4) + 8 (8 λ - 16) - 6 (56 - 48 + 8 λ)$

Step 1.5.4.2.2

Subtract $48$ from $56$ .

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

Step 1.5.4.2.3

Reorder $8$ and $8 λ$ .

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

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 $(- 4 - λ) (λ^{2} - 16 λ + 4)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 1.5.5.1.2

Simplify each term.

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Step 1.5.5.1.2.1

Multiply $- 16$ by $- 4$ .

$p (λ) = - 4 λ^{2} + 64 λ - 4 \cdot 4 - λ \cdot λ^{2} - λ (- 16 λ) - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.2.2

Multiply $- 4$ by $4$ .

$p (λ) = - 4 λ^{2} + 64 λ - 16 - λ \cdot λ^{2} - λ (- 16 λ) - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

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 $λ^{2}$ .

$p (λ) = - 4 λ^{2} + 64 λ - 16 - (λ^{2} λ) - λ (- 16 λ) - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

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 $1$ .

$p (λ) = - 4 λ^{2} + 64 λ - 16 - (λ^{2} λ^{1}) - λ (- 16 λ) - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.2.3.2.2

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

$p (λ) = - 4 λ^{2} + 64 λ - 16 - λ^{2 + 1} - λ (- 16 λ) - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.2.3.3

Add $2$ and $1$ .

$p (λ) = - 4 λ^{2} + 64 λ - 16 - λ^{3} - λ (- 16 λ) - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = - 4 λ^{2} + 64 λ - 16 - λ^{3} - 1 \cdot - 16 λ \cdot λ - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

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 $λ$ .

$p (λ) = - 4 λ^{2} + 64 λ - 16 - λ^{3} - 1 \cdot - 16 (λ \cdot λ) - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.2.5.2

Multiply $λ$ by $λ$ .

$p (λ) = - 4 λ^{2} + 64 λ - 16 - λ^{3} - 1 \cdot - 16 λ^{2} - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.2.6

Multiply $- 1$ by $- 16$ .

$p (λ) = - 4 λ^{2} + 64 λ - 16 - λ^{3} + 16 λ^{2} - λ \cdot 4 + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.2.7

Multiply $4$ by $- 1$ .

$p (λ) = - 4 λ^{2} + 64 λ - 16 - λ^{3} + 16 λ^{2} - 4 λ + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.3

Add $- 4 λ^{2}$ and $16 λ^{2}$ .

$p (λ) = 12 λ^{2} + 64 λ - 16 - λ^{3} - 4 λ + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.4

Subtract $4 λ$ from $64 λ$ .

$p (λ) = 12 λ^{2} + 60 λ - 16 - λ^{3} + 8 (8 λ - 16) - 6 (8 λ + 8)$

Step 1.5.5.1.5

Apply the distributive property.

$p (λ) = 12 λ^{2} + 60 λ - 16 - λ^{3} + 8 (8 λ) + 8 \cdot - 16 - 6 (8 λ + 8)$

Step 1.5.5.1.6

Multiply $8$ by $8$ .

$p (λ) = 12 λ^{2} + 60 λ - 16 - λ^{3} + 64 λ + 8 \cdot - 16 - 6 (8 λ + 8)$

Step 1.5.5.1.7

Multiply $8$ by $- 16$ .

$p (λ) = 12 λ^{2} + 60 λ - 16 - λ^{3} + 64 λ - 128 - 6 (8 λ + 8)$

Step 1.5.5.1.8

Apply the distributive property.

$p (λ) = 12 λ^{2} + 60 λ - 16 - λ^{3} + 64 λ - 128 - 6 (8 λ) - 6 \cdot 8$

Step 1.5.5.1.9

Multiply $8$ by $- 6$ .

$p (λ) = 12 λ^{2} + 60 λ - 16 - λ^{3} + 64 λ - 128 - 48 λ - 6 \cdot 8$

Step 1.5.5.1.10

Multiply $- 6$ by $8$ .

$p (λ) = 12 λ^{2} + 60 λ - 16 - λ^{3} + 64 λ - 128 - 48 λ - 48$

Step 1.5.5.2

Add $60 λ$ and $64 λ$ .

$p (λ) = 12 λ^{2} + 124 λ - 16 - λ^{3} - 128 - 48 λ - 48$

Step 1.5.5.3

Subtract $48 λ$ from $124 λ$ .

$p (λ) = 12 λ^{2} + 76 λ - 16 - λ^{3} - 128 - 48$

Step 1.5.5.4

Subtract $128$ from $- 16$ .

$p (λ) = 12 λ^{2} + 76 λ - λ^{3} - 144 - 48$

Step 1.5.5.5

Subtract $48$ from $- 144$ .

$p (λ) = 12 λ^{2} + 76 λ - λ^{3} - 192$

Step 1.5.5.6

Move $76 λ$ .

$p (λ) = 12 λ^{2} - λ^{3} + 76 λ - 192$

Step 1.5.5.7

Reorder $12 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 12 λ^{2} + 76 λ - 192$

Step 1.6

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

$- λ^{3} + 12 λ^{2} + 76 λ - 192 = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Factor the left side of the equation.

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Step 1.7.1.1

Factor $- λ^{3} + 12 λ^{2} + 76 λ - 192$ using the rational roots test.

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Step 1.7.1.1.1

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.

$p = \pm 1, \pm 192, \pm 2, \pm 96, \pm 3, \pm 64, \pm 4, \pm 48, \pm 6, \pm 32, \pm 8, \pm 24, \pm 12, \pm 16$

$q = \pm 1$

Step 1.7.1.1.2

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

$\pm 1, \pm 192, \pm 2, \pm 96, \pm 3, \pm 64, \pm 4, \pm 48, \pm 6, \pm 32, \pm 8, \pm 24, \pm 12, \pm 16$

Step 1.7.1.1.3

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 1.7.1.1.3.1

Substitute $2$ into the polynomial.

$- 2^{3} + 12 \cdot 2^{2} + 76 \cdot 2 - 192$

Step 1.7.1.1.3.2

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

$- 1 \cdot 8 + 12 \cdot 2^{2} + 76 \cdot 2 - 192$

Step 1.7.1.1.3.3

Multiply $- 1$ by $8$ .

$- 8 + 12 \cdot 2^{2} + 76 \cdot 2 - 192$

Step 1.7.1.1.3.4

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

$- 8 + 12 \cdot 4 + 76 \cdot 2 - 192$

Step 1.7.1.1.3.5

Multiply $12$ by $4$ .

$- 8 + 48 + 76 \cdot 2 - 192$

Step 1.7.1.1.3.6

Add $- 8$ and $48$ .

$40 + 76 \cdot 2 - 192$

Step 1.7.1.1.3.7

Multiply $76$ by $2$ .

$40 + 152 - 192$

Step 1.7.1.1.3.8

Add $40$ and $152$ .

$192 - 192$

Step 1.7.1.1.3.9

Subtract $192$ from $192$ .

$0$

Step 1.7.1.1.4

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.

$\frac{- λ^{3} + 12 λ^{2} + 76 λ - 192}{λ - 2}$

Step 1.7.1.1.5

Divide $- λ^{3} + 12 λ^{2} + 76 λ - 192$ by $λ - 2$ .

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Step 1.7.1.1.5.1

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

$λ$

$2$

$λ^{3}$

$12 λ^{2}$

$76 λ$

$192$

Step 1.7.1.1.5.2

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

				-	$λ^{2}$
	$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$

Step 1.7.1.1.5.3

Multiply the new quotient term by the divisor.

			-	$λ^{2}$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			-	$λ^{3}$	+	$2 λ^{2}$

Step 1.7.1.1.5.4

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

			-	$λ^{2}$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$

Step 1.7.1.1.5.5

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

			-	$λ^{2}$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$

Step 1.7.1.1.5.6

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

			-	$λ^{2}$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$

Step 1.7.1.1.5.7

Divide the highest order term in the dividend $10 λ^{2}$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$10 λ$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$

Step 1.7.1.1.5.8

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$10 λ$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$
					+	$10 λ^{2}$	-	$20 λ$

Step 1.7.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $10 λ^{2} - 20 λ$

			-	$λ^{2}$	+	$10 λ$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$
					-	$10 λ^{2}$	+	$20 λ$

Step 1.7.1.1.5.10

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

			-	$λ^{2}$	+	$10 λ$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$
					-	$10 λ^{2}$	+	$20 λ$
							+	$96 λ$

Step 1.7.1.1.5.11

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

			-	$λ^{2}$	+	$10 λ$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$
					-	$10 λ^{2}$	+	$20 λ$
							+	$96 λ$	-	$192$

Step 1.7.1.1.5.12

Divide the highest order term in the dividend $96 λ$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$10 λ$	+	$96$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$
					-	$10 λ^{2}$	+	$20 λ$
							+	$96 λ$	-	$192$

Step 1.7.1.1.5.13

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$10 λ$	+	$96$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$
					-	$10 λ^{2}$	+	$20 λ$
							+	$96 λ$	-	$192$
							+	$96 λ$	-	$192$

Step 1.7.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $96 λ - 192$

			-	$λ^{2}$	+	$10 λ$	+	$96$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$
					-	$10 λ^{2}$	+	$20 λ$
							+	$96 λ$	-	$192$
							-	$96 λ$	+	$192$

Step 1.7.1.1.5.15

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

			-	$λ^{2}$	+	$10 λ$	+	$96$
$λ$	-	$2$	-	$λ^{3}$	+	$12 λ^{2}$	+	$76 λ$	-	$192$
			+	$λ^{3}$	-	$2 λ^{2}$
					+	$10 λ^{2}$	+	$76 λ$
					-	$10 λ^{2}$	+	$20 λ$
							+	$96 λ$	-	$192$
							-	$96 λ$	+	$192$
										$0$

Step 1.7.1.1.5.16

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

$- λ^{2} + 10 λ + 96$

Step 1.7.1.1.6

Write $- λ^{3} + 12 λ^{2} + 76 λ - 192$ as a set of factors.

$(λ - 2) (- λ^{2} + 10 λ + 96) = 0$

Step 1.7.1.2

Factor by grouping.

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Step 1.7.1.2.1

Factor by grouping.

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Step 1.7.1.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 96 = - 96$ and whose sum is $b = 10$ .

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Step 1.7.1.2.1.1.1

Factor $10$ out of $10 λ$ .

$(λ - 2) (- λ^{2} + 10 (λ) + 96) = 0$

Step 1.7.1.2.1.1.2

Rewrite $10$ as $- 6$ plus $16$

$(λ - 2) (- λ^{2} + (- 6 + 16) λ + 96) = 0$

Step 1.7.1.2.1.1.3

Apply the distributive property.

$(λ - 2) (- λ^{2} - 6 λ + 16 λ + 96) = 0$

Step 1.7.1.2.1.2

Factor out the greatest common factor from each group.

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Step 1.7.1.2.1.2.1

Group the first two terms and the last two terms.

$(λ - 2) ((- λ^{2} - 6 λ) + 16 λ + 96) = 0$

Step 1.7.1.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$(λ - 2) (λ (- λ - 6) - 16 (- λ - 6)) = 0$

Step 1.7.1.2.1.3

Factor the polynomial by factoring out the greatest common factor, $- λ - 6$ .

$(λ - 2) ((- λ - 6) (λ - 16)) = 0$

Step 1.7.1.2.2

Remove unnecessary parentheses.

$(λ - 2) (- λ - 6) (λ - 16) = 0$

Step 1.7.2

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

$λ - 2 = 0$

$- λ - 6 = 0$

$λ - 16 = 0$

Step 1.7.3

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

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Step 1.7.3.1

Set $λ - 2$ equal to $0$ .

$λ - 2 = 0$

Step 1.7.3.2

Add $2$ to both sides of the equation.

$λ = 2$

Step 1.7.4

Set $- λ - 6$ equal to $0$ and solve for $λ$ .

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Step 1.7.4.1

Set $- λ - 6$ equal to $0$ .

$- λ - 6 = 0$

Step 1.7.4.2

Solve $- λ - 6 = 0$ for $λ$ .

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Step 1.7.4.2.1

Add $6$ to both sides of the equation.

$- λ = 6$

Step 1.7.4.2.2

Divide each term in $- λ = 6$ by $- 1$ and simplify.

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Step 1.7.4.2.2.1

Divide each term in $- λ = 6$ by $- 1$ .

$\frac{- λ}{- 1} = \frac{6}{- 1}$

Step 1.7.4.2.2.2

Simplify the left side.

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Step 1.7.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{λ}{1} = \frac{6}{- 1}$

Step 1.7.4.2.2.2.2

Divide $λ$ by $1$ .

$λ = \frac{6}{- 1}$

Step 1.7.4.2.2.3

Simplify the right side.

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Step 1.7.4.2.2.3.1

Divide $6$ by $- 1$ .

$λ = - 6$

Step 1.7.5

Set $λ - 16$ equal to $0$ and solve for $λ$ .

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Step 1.7.5.1

Set $λ - 16$ equal to $0$ .

$λ - 16 = 0$

Step 1.7.5.2

Add $16$ to both sides of the equation.

$λ = 16$

Step 1.7.6

The final solution is all the values that make $(λ - 2) (- λ - 6) (λ - 16) = 0$ true.

$λ = 2, - 6, 16$

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 $λ = 2$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] - 2 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $- 2$ by each element of the matrix.

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 \cdot 1 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $- 2$ by $1$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 & - 2 \cdot 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 & 0 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 \\ - 2 \cdot 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 \cdot 1 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.5

Multiply $- 2$ by $1$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & - 2 \cdot 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.6

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ - 2 \cdot 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.7

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & - 2 \cdot 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.8

Multiply $- 2$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - 2 \cdot 1 \end{array}]$

Step 3.2.1.2.9

Multiply $- 2$ by $1$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 2 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - 2 \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} - 4 - 2 & - 8 + 0 & - 6 + 0 \\ - 8 + 0 & 6 - 2 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 2 \end{array}]$

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Subtract $2$ from $- 4$ .

$[\begin{array}{c} - 6 & - 8 + 0 & - 6 + 0 \\ - 8 + 0 & 6 - 2 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 2 \end{array}]$

Step 3.2.3.2

Add $- 8$ and $0$ .

$[\begin{array}{c} - 6 & - 8 & - 6 + 0 \\ - 8 + 0 & 6 - 2 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 2 \end{array}]$

Step 3.2.3.3

Add $- 6$ and $0$ .

$[\begin{array}{c} - 6 & - 8 & - 6 \\ - 8 + 0 & 6 - 2 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 2 \end{array}]$

Step 3.2.3.4

Add $- 8$ and $0$ .

$[\begin{array}{c} - 6 & - 8 & - 6 \\ - 8 & 6 - 2 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 2 \end{array}]$

Step 3.2.3.5

Subtract $2$ from $6$ .

$[\begin{array}{c} - 6 & - 8 & - 6 \\ - 8 & 4 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 2 \end{array}]$

Step 3.2.3.6

Add $- 8$ and $0$ .

$[\begin{array}{c} - 6 & - 8 & - 6 \\ - 8 & 4 & - 8 \\ 8 + 0 & - 7 + 0 & 10 - 2 \end{array}]$

Step 3.2.3.7

Add $8$ and $0$ .

$[\begin{array}{c} - 6 & - 8 & - 6 \\ - 8 & 4 & - 8 \\ 8 & - 7 + 0 & 10 - 2 \end{array}]$

Step 3.2.3.8

Add $- 7$ and $0$ .

$[\begin{array}{c} - 6 & - 8 & - 6 \\ - 8 & 4 & - 8 \\ 8 & - 7 & 10 - 2 \end{array}]$

Step 3.2.3.9

Subtract $2$ from $10$ .

$[\begin{array}{c} - 6 & - 8 & - 6 \\ - 8 & 4 & - 8 \\ 8 & - 7 & 8 \end{array}]$

Step 3.3

Find the null space when $λ = 2$ .

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Step 3.3.1

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

$[\begin{array}{c} - 6 & - 8 & - 6 & 0 \\ - 8 & 4 & - 8 & 0 \\ 8 & - 7 & 8 & 0 \end{array}]$

Step 3.3.2

Find the reduced row echelon form.

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Step 3.3.2.1

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

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Step 3.3.2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{6}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot - 8 & - \frac{1}{6} \cdot - 6 & - \frac{1}{6} \cdot 0 \\ - 8 & 4 & - 8 & 0 \\ 8 & - 7 & 8 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & 1 & 0 \\ - 8 & 4 & - 8 & 0 \\ 8 & - 7 & 8 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{4}{3} & 1 & 0 \\ - 8 + 8 \cdot 1 & 4 + 8 (\frac{4}{3}) & - 8 + 8 \cdot 1 & 0 + 8 \cdot 0 \\ 8 & - 7 & 8 & 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & 1 & 0 \\ 0 & \frac{44}{3} & 0 & 0 \\ 8 & - 7 & 8 & 0 \end{array}]$

Step 3.3.2.3

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.2.3.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{4}{3} & 1 & 0 \\ 0 & \frac{44}{3} & 0 & 0 \\ 8 - 8 \cdot 1 & - 7 - 8 (\frac{4}{3}) & 8 - 8 \cdot 1 & 0 - 8 \cdot 0 \end{array}]$

Step 3.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & 1 & 0 \\ 0 & \frac{44}{3} & 0 & 0 \\ 0 & - \frac{53}{3} & 0 & 0 \end{array}]$

Step 3.3.2.4

Multiply each element of $R_{2}$ by $\frac{3}{44}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.2.4.1

Multiply each element of $R_{2}$ by $\frac{3}{44}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{4}{3} & 1 & 0 \\ \frac{3}{44} \cdot 0 & \frac{3}{44} \cdot \frac{44}{3} & \frac{3}{44} \cdot 0 & \frac{3}{44} \cdot 0 \\ 0 & - \frac{53}{3} & 0 & 0 \end{array}]$

Step 3.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & - \frac{53}{3} & 0 & 0 \end{array}]$

Step 3.3.2.5

Perform the row operation $R_{3} = R_{3} + \frac{53}{3} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.3.2.5.1

Perform the row operation $R_{3} = R_{3} + \frac{53}{3} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{4}{3} & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 + \frac{53}{3} \cdot 0 & - \frac{53}{3} + \frac{53}{3} \cdot 1 & 0 + \frac{53}{3} \cdot 0 & 0 + \frac{53}{3} \cdot 0 \end{array}]$

Step 3.3.2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{4}{3} & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3.3.2.6

Perform the row operation $R_{1} = R_{1} - \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.3.2.6.1

Perform the row operation $R_{1} = R_{1} - \frac{4}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{4}{3} \cdot 0 & \frac{4}{3} - \frac{4}{3} \cdot 1 & 1 - \frac{4}{3} \cdot 0 & 0 - \frac{4}{3} \cdot 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3.3.2.6.2

Simplify $R_{1}$ .

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

Step 3.3.3

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

$x + z = 0$

$y = 0$

$0 = 0$

Step 3.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - z \\ 0 \\ z \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} - 1 \\ 0 \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${z [\begin{array}{c} - 1 \\ 0 \\ 1 \end{array}] | z \in R}$

Step 3.3.7

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

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

Step 4

Find the eigenvector using the eigenvalue $λ = - 6$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + 6 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 4.2

Simplify.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $6$ by each element of the matrix.

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 \cdot 1 & 6 \cdot 0 & 6 \cdot 0 \\ 6 \cdot 0 & 6 \cdot 1 & 6 \cdot 0 \\ 6 \cdot 0 & 6 \cdot 0 & 6 \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $6$ by $1$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 & 6 \cdot 0 & 6 \cdot 0 \\ 6 \cdot 0 & 6 \cdot 1 & 6 \cdot 0 \\ 6 \cdot 0 & 6 \cdot 0 & 6 \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $6$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 & 0 & 6 \cdot 0 \\ 6 \cdot 0 & 6 \cdot 1 & 6 \cdot 0 \\ 6 \cdot 0 & 6 \cdot 0 & 6 \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $6$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 & 0 & 0 \\ 6 \cdot 0 & 6 \cdot 1 & 6 \cdot 0 \\ 6 \cdot 0 & 6 \cdot 0 & 6 \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $6$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 & 0 & 0 \\ 0 & 6 \cdot 1 & 6 \cdot 0 \\ 6 \cdot 0 & 6 \cdot 0 & 6 \cdot 1 \end{array}]$

Step 4.2.1.2.5

Multiply $6$ by $1$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 & 0 & 0 \\ 0 & 6 & 6 \cdot 0 \\ 6 \cdot 0 & 6 \cdot 0 & 6 \cdot 1 \end{array}]$

Step 4.2.1.2.6

Multiply $6$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 6 \cdot 0 & 6 \cdot 0 & 6 \cdot 1 \end{array}]$

Step 4.2.1.2.7

Multiply $6$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 6 \cdot 0 & 6 \cdot 1 \end{array}]$

Step 4.2.1.2.8

Multiply $6$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \cdot 1 \end{array}]$

Step 4.2.1.2.9

Multiply $6$ by $1$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} - 4 + 6 & - 8 + 0 & - 6 + 0 \\ - 8 + 0 & 6 + 6 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 + 6 \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Add $- 4$ and $6$ .

$[\begin{array}{c} 2 & - 8 + 0 & - 6 + 0 \\ - 8 + 0 & 6 + 6 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 + 6 \end{array}]$

Step 4.2.3.2

Add $- 8$ and $0$ .

$[\begin{array}{c} 2 & - 8 & - 6 + 0 \\ - 8 + 0 & 6 + 6 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 + 6 \end{array}]$

Step 4.2.3.3

Add $- 6$ and $0$ .

$[\begin{array}{c} 2 & - 8 & - 6 \\ - 8 + 0 & 6 + 6 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 + 6 \end{array}]$

Step 4.2.3.4

Add $- 8$ and $0$ .

$[\begin{array}{c} 2 & - 8 & - 6 \\ - 8 & 6 + 6 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 + 6 \end{array}]$

Step 4.2.3.5

Add $6$ and $6$ .

$[\begin{array}{c} 2 & - 8 & - 6 \\ - 8 & 12 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 + 6 \end{array}]$

Step 4.2.3.6

Add $- 8$ and $0$ .

$[\begin{array}{c} 2 & - 8 & - 6 \\ - 8 & 12 & - 8 \\ 8 + 0 & - 7 + 0 & 10 + 6 \end{array}]$

Step 4.2.3.7

Add $8$ and $0$ .

$[\begin{array}{c} 2 & - 8 & - 6 \\ - 8 & 12 & - 8 \\ 8 & - 7 + 0 & 10 + 6 \end{array}]$

Step 4.2.3.8

Add $- 7$ and $0$ .

$[\begin{array}{c} 2 & - 8 & - 6 \\ - 8 & 12 & - 8 \\ 8 & - 7 & 10 + 6 \end{array}]$

Step 4.2.3.9

Add $10$ and $6$ .

$[\begin{array}{c} 2 & - 8 & - 6 \\ - 8 & 12 & - 8 \\ 8 & - 7 & 16 \end{array}]$

Step 4.3

Find the null space when $λ = - 6$ .

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Step 4.3.1

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

$[\begin{array}{c} 2 & - 8 & - 6 & 0 \\ - 8 & 12 & - 8 & 0 \\ 8 & - 7 & 16 & 0 \end{array}]$

Step 4.3.2

Find the reduced row echelon form.

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Step 4.3.2.1

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

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Step 4.3.2.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{- 8}{2} & \frac{- 6}{2} & \frac{0}{2} \\ - 8 & 12 & - 8 & 0 \\ 8 & - 7 & 16 & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ - 8 & 12 & - 8 & 0 \\ 8 & - 7 & 16 & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ - 8 + 8 \cdot 1 & 12 + 8 \cdot - 4 & - 8 + 8 \cdot - 3 & 0 + 8 \cdot 0 \\ 8 & - 7 & 16 & 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ 0 & - 20 & - 32 & 0 \\ 8 & - 7 & 16 & 0 \end{array}]$

Step 4.3.2.3

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.3.2.3.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ 0 & - 20 & - 32 & 0 \\ 8 - 8 \cdot 1 & - 7 - 8 \cdot - 4 & 16 - 8 \cdot - 3 & 0 - 8 \cdot 0 \end{array}]$

Step 4.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ 0 & - 20 & - 32 & 0 \\ 0 & 25 & 40 & 0 \end{array}]$

Step 4.3.2.4

Multiply each element of $R_{2}$ by $- \frac{1}{20}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.2.4.1

Multiply each element of $R_{2}$ by $- \frac{1}{20}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ - \frac{1}{20} \cdot 0 & - \frac{1}{20} \cdot - 20 & - \frac{1}{20} \cdot - 32 & - \frac{1}{20} \cdot 0 \\ 0 & 25 & 40 & 0 \end{array}]$

Step 4.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ 0 & 1 & \frac{8}{5} & 0 \\ 0 & 25 & 40 & 0 \end{array}]$

Step 4.3.2.5

Perform the row operation $R_{3} = R_{3} - 25 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.3.2.5.1

Perform the row operation $R_{3} = R_{3} - 25 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ 0 & 1 & \frac{8}{5} & 0 \\ 0 - 25 \cdot 0 & 25 - 25 \cdot 1 & 40 - 25 (\frac{8}{5}) & 0 - 25 \cdot 0 \end{array}]$

Step 4.3.2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 4 & - 3 & 0 \\ 0 & 1 & \frac{8}{5} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.6

Perform the row operation $R_{1} = R_{1} + 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 4.3.2.6.1

Perform the row operation $R_{1} = R_{1} + 4 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 4 \cdot 0 & - 4 + 4 \cdot 1 & - 3 + 4 (\frac{8}{5}) & 0 + 4 \cdot 0 \\ 0 & 1 & \frac{8}{5} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{17}{5} & 0 \\ 0 & 1 & \frac{8}{5} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.3

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

$x + \frac{17}{5} z = 0$

$y + \frac{8}{5} z = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{17 z}{5} \\ - \frac{8 z}{5} \\ z \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} - \frac{17}{5} \\ - \frac{8}{5} \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${z [\begin{array}{c} - \frac{17}{5} \\ - \frac{8}{5} \\ 1 \end{array}] | z \in R}$

Step 4.3.7

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

${[\begin{array}{c} - \frac{17}{5} \\ - \frac{8}{5} \\ 1 \end{array}]}$

Step 5

Find the eigenvector using the eigenvalue $λ = 16$ .

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Step 5.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] - 16 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 5.2

Simplify.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Multiply $- 16$ by each element of the matrix.

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 \cdot 1 & - 16 \cdot 0 & - 16 \cdot 0 \\ - 16 \cdot 0 & - 16 \cdot 1 & - 16 \cdot 0 \\ - 16 \cdot 0 & - 16 \cdot 0 & - 16 \cdot 1 \end{array}]$

Step 5.2.1.2

Simplify each element in the matrix.

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Step 5.2.1.2.1

Multiply $- 16$ by $1$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 & - 16 \cdot 0 & - 16 \cdot 0 \\ - 16 \cdot 0 & - 16 \cdot 1 & - 16 \cdot 0 \\ - 16 \cdot 0 & - 16 \cdot 0 & - 16 \cdot 1 \end{array}]$

Step 5.2.1.2.2

Multiply $- 16$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 & 0 & - 16 \cdot 0 \\ - 16 \cdot 0 & - 16 \cdot 1 & - 16 \cdot 0 \\ - 16 \cdot 0 & - 16 \cdot 0 & - 16 \cdot 1 \end{array}]$

Step 5.2.1.2.3

Multiply $- 16$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 & 0 & 0 \\ - 16 \cdot 0 & - 16 \cdot 1 & - 16 \cdot 0 \\ - 16 \cdot 0 & - 16 \cdot 0 & - 16 \cdot 1 \end{array}]$

Step 5.2.1.2.4

Multiply $- 16$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 & 0 & 0 \\ 0 & - 16 \cdot 1 & - 16 \cdot 0 \\ - 16 \cdot 0 & - 16 \cdot 0 & - 16 \cdot 1 \end{array}]$

Step 5.2.1.2.5

Multiply $- 16$ by $1$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 & 0 & 0 \\ 0 & - 16 & - 16 \cdot 0 \\ - 16 \cdot 0 & - 16 \cdot 0 & - 16 \cdot 1 \end{array}]$

Step 5.2.1.2.6

Multiply $- 16$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 & 0 & 0 \\ 0 & - 16 & 0 \\ - 16 \cdot 0 & - 16 \cdot 0 & - 16 \cdot 1 \end{array}]$

Step 5.2.1.2.7

Multiply $- 16$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & - 16 \cdot 0 & - 16 \cdot 1 \end{array}]$

Step 5.2.1.2.8

Multiply $- 16$ by $0$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & 0 & - 16 \cdot 1 \end{array}]$

Step 5.2.1.2.9

Multiply $- 16$ by $1$ .

$[\begin{array}{c} - 4 & - 8 & - 6 \\ - 8 & 6 & - 8 \\ 8 & - 7 & 10 \end{array}] + [\begin{array}{c} - 16 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & 0 & - 16 \end{array}]$

Step 5.2.2

Add the corresponding elements.

$[\begin{array}{c} - 4 - 16 & - 8 + 0 & - 6 + 0 \\ - 8 + 0 & 6 - 16 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 16 \end{array}]$

Step 5.2.3

Simplify each element.

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Step 5.2.3.1

Subtract $16$ from $- 4$ .

$[\begin{array}{c} - 20 & - 8 + 0 & - 6 + 0 \\ - 8 + 0 & 6 - 16 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 16 \end{array}]$

Step 5.2.3.2

Add $- 8$ and $0$ .

$[\begin{array}{c} - 20 & - 8 & - 6 + 0 \\ - 8 + 0 & 6 - 16 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 16 \end{array}]$

Step 5.2.3.3

Add $- 6$ and $0$ .

$[\begin{array}{c} - 20 & - 8 & - 6 \\ - 8 + 0 & 6 - 16 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 16 \end{array}]$

Step 5.2.3.4

Add $- 8$ and $0$ .

$[\begin{array}{c} - 20 & - 8 & - 6 \\ - 8 & 6 - 16 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 16 \end{array}]$

Step 5.2.3.5

Subtract $16$ from $6$ .

$[\begin{array}{c} - 20 & - 8 & - 6 \\ - 8 & - 10 & - 8 + 0 \\ 8 + 0 & - 7 + 0 & 10 - 16 \end{array}]$

Step 5.2.3.6

Add $- 8$ and $0$ .

$[\begin{array}{c} - 20 & - 8 & - 6 \\ - 8 & - 10 & - 8 \\ 8 + 0 & - 7 + 0 & 10 - 16 \end{array}]$

Step 5.2.3.7

Add $8$ and $0$ .

$[\begin{array}{c} - 20 & - 8 & - 6 \\ - 8 & - 10 & - 8 \\ 8 & - 7 + 0 & 10 - 16 \end{array}]$

Step 5.2.3.8

Add $- 7$ and $0$ .

$[\begin{array}{c} - 20 & - 8 & - 6 \\ - 8 & - 10 & - 8 \\ 8 & - 7 & 10 - 16 \end{array}]$

Step 5.2.3.9

Subtract $16$ from $10$ .

$[\begin{array}{c} - 20 & - 8 & - 6 \\ - 8 & - 10 & - 8 \\ 8 & - 7 & - 6 \end{array}]$

Step 5.3

Find the null space when $λ = 16$ .

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Step 5.3.1

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

$[\begin{array}{c} - 20 & - 8 & - 6 & 0 \\ - 8 & - 10 & - 8 & 0 \\ 8 & - 7 & - 6 & 0 \end{array}]$

Step 5.3.2

Find the reduced row echelon form.

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Step 5.3.2.1

Multiply each element of $R_{1}$ by $- \frac{1}{20}$ to make the entry at $1, 1$ a $1$ .

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Step 5.3.2.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{20}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{20} \cdot - 20 & - \frac{1}{20} \cdot - 8 & - \frac{1}{20} \cdot - 6 & - \frac{1}{20} \cdot 0 \\ - 8 & - 10 & - 8 & 0 \\ 8 & - 7 & - 6 & 0 \end{array}]$

Step 5.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{3}{10} & 0 \\ - 8 & - 10 & - 8 & 0 \\ 8 & - 7 & - 6 & 0 \end{array}]$

Step 5.3.2.2

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 5.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 8 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{3}{10} & 0 \\ - 8 + 8 \cdot 1 & - 10 + 8 (\frac{2}{5}) & - 8 + 8 (\frac{3}{10}) & 0 + 8 \cdot 0 \\ 8 & - 7 & - 6 & 0 \end{array}]$

Step 5.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{3}{10} & 0 \\ 0 & - \frac{34}{5} & - \frac{28}{5} & 0 \\ 8 & - 7 & - 6 & 0 \end{array}]$

Step 5.3.2.3

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 5.3.2.3.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{3}{10} & 0 \\ 0 & - \frac{34}{5} & - \frac{28}{5} & 0 \\ 8 - 8 \cdot 1 & - 7 - 8 (\frac{2}{5}) & - 6 - 8 (\frac{3}{10}) & 0 - 8 \cdot 0 \end{array}]$

Step 5.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{3}{10} & 0 \\ 0 & - \frac{34}{5} & - \frac{28}{5} & 0 \\ 0 & - \frac{51}{5} & - \frac{42}{5} & 0 \end{array}]$

Step 5.3.2.4

Multiply each element of $R_{2}$ by $- \frac{5}{34}$ to make the entry at $2, 2$ a $1$ .

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Step 5.3.2.4.1

Multiply each element of $R_{2}$ by $- \frac{5}{34}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{3}{10} & 0 \\ - \frac{5}{34} \cdot 0 & - \frac{5}{34} (- \frac{34}{5}) & - \frac{5}{34} (- \frac{28}{5}) & - \frac{5}{34} \cdot 0 \\ 0 & - \frac{51}{5} & - \frac{42}{5} & 0 \end{array}]$

Step 5.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{3}{10} & 0 \\ 0 & 1 & \frac{14}{17} & 0 \\ 0 & - \frac{51}{5} & - \frac{42}{5} & 0 \end{array}]$

Step 5.3.2.5

Perform the row operation $R_{3} = R_{3} + \frac{51}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 5.3.2.5.1

Perform the row operation $R_{3} = R_{3} + \frac{51}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{3}{10} & 0 \\ 0 & 1 & \frac{14}{17} & 0 \\ 0 + \frac{51}{5} \cdot 0 & - \frac{51}{5} + \frac{51}{5} \cdot 1 & - \frac{42}{5} + \frac{51}{5} \cdot \frac{14}{17} & 0 + \frac{51}{5} \cdot 0 \end{array}]$

Step 5.3.2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{2}{5} & \frac{3}{10} & 0 \\ 0 & 1 & \frac{14}{17} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.2.6

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

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Step 5.3.2.6.1

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

$[\begin{array}{c} 1 - \frac{2}{5} \cdot 0 & \frac{2}{5} - \frac{2}{5} \cdot 1 & \frac{3}{10} - \frac{2}{5} \cdot \frac{14}{17} & 0 - \frac{2}{5} \cdot 0 \\ 0 & 1 & \frac{14}{17} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{34} & 0 \\ 0 & 1 & \frac{14}{17} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.3

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

$x - \frac{1}{34} z = 0$

$y + \frac{14}{17} z = 0$

$0 = 0$

Step 5.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{z}{34} \\ - \frac{14 z}{17} \\ z \end{array}]$

Step 5.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} \frac{1}{34} \\ - \frac{14}{17} \\ 1 \end{array}]$

Step 5.3.6

Write as a solution set.

${z [\begin{array}{c} \frac{1}{34} \\ - \frac{14}{17} \\ 1 \end{array}] | z \in R}$

Step 5.3.7

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

${[\begin{array}{c} \frac{1}{34} \\ - \frac{14}{17} \\ 1 \end{array}]}$

Step 6

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

${[\begin{array}{c} - 1 \\ 0 \\ 1 \end{array}], [\begin{array}{c} - \frac{17}{5} \\ - \frac{8}{5} \\ 1 \end{array}], [\begin{array}{c} \frac{1}{34} \\ - \frac{14}{17} \\ 1 \end{array}]}$