Linear Algebra Examples

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

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

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

Step 1.3.2

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

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

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.

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

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.5

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.6.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.9

Multiply $- 1$ by $1$ .

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

Step 1.4.2

Add the corresponding elements.

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

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $3$ and $0$ .

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

Step 1.4.3.2

Add $0$ and $0$ .

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

Step 1.4.3.3

Add $3$ and $0$ .

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

Step 1.4.3.4

Subtract $λ$ from $0$ .

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

Step 1.4.3.5

Add $1$ and $0$ .

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

Step 1.4.3.6

Add $1$ and $0$ .

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

Step 1.4.3.7

Add $1$ and $0$ .

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

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.

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

Step 1.5.1.4

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

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

Step 1.5.1.5

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

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

Step 1.5.1.6

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

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

Step 1.5.1.7

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

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

Step 1.5.1.8

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

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

Step 1.5.1.9

Add the terms together.

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

Step 1.5.2

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

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

Step 1.5.3

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

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Step 1.5.3.1

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

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

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.

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

Step 1.5.3.2.1.2

Multiply $2$ by $- 1$ .

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

Step 1.5.3.2.1.3

Rewrite using the commutative property of multiplication.

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

Step 1.5.3.2.1.4

Simplify each term.

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Step 1.5.3.2.1.4.1

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

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Step 1.5.3.2.1.4.1.1

Move $λ$ .

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

Step 1.5.3.2.1.4.1.2

Multiply $λ$ by $λ$ .

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

Step 1.5.3.2.1.4.2

Multiply $- 1$ by $- 1$ .

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

Step 1.5.3.2.1.4.3

Multiply $λ^{2}$ by $1$ .

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

Step 1.5.3.2.1.5

Multiply $- 1$ by $1$ .

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

Step 1.5.3.2.2

Reorder $- 2 λ$ and $λ^{2}$ .

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

Step 1.5.4

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

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Step 1.5.4.1

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

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

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

Apply the distributive property.

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

Step 1.5.4.2.1.2

Multiply $3$ by $2$ .

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

Step 1.5.4.2.1.3

Multiply $- 1$ by $3$ .

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

Step 1.5.4.2.1.4

Multiply $- 1$ by $1$ .

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

Step 1.5.4.2.2

Subtract $1$ from $6$ .

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

Step 1.5.5

Simplify the determinant.

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Step 1.5.5.1

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

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

Step 1.5.5.2

Simplify each term.

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Step 1.5.5.2.1

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

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

Step 1.5.5.2.2

Simplify each term.

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Step 1.5.5.2.2.1

Multiply $- 2$ by $- 2$ .

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

Step 1.5.5.2.2.2

Multiply $- 2$ by $- 1$ .

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

Step 1.5.5.2.2.3

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

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Step 1.5.5.2.2.3.1

Move $λ^{2}$ .

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

Step 1.5.5.2.2.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.5.2.2.3.2.1

Raise $λ$ to the power of $1$ .

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

Step 1.5.5.2.2.3.2.2

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

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

Step 1.5.5.2.2.3.3

Add $2$ and $1$ .

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

Step 1.5.5.2.2.4

Rewrite using the commutative property of multiplication.

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

Step 1.5.5.2.2.5

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

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Step 1.5.5.2.2.5.1

Move $λ$ .

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

Step 1.5.5.2.2.5.2

Multiply $λ$ by $λ$ .

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

Step 1.5.5.2.2.6

Multiply $- 1$ by $- 2$ .

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

Step 1.5.5.2.2.7

Multiply $- λ \cdot - 1$ .

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Step 1.5.5.2.2.7.1

Multiply $- 1$ by $- 1$ .

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

Step 1.5.5.2.2.7.2

Multiply $λ$ by $1$ .

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

Step 1.5.5.2.3

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

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Step 1.5.5.2.3.1

Add $- 2 λ^{2}$ and $2 λ^{2}$ .

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

Step 1.5.5.2.3.2

Add $4 λ + 2 - λ^{3}$ and $0$ .

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

Step 1.5.5.2.4

Add $4 λ$ and $λ$ .

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

Step 1.5.5.2.5

Apply the distributive property.

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

Step 1.5.5.2.6

Multiply $- 3$ by $- 3$ .

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

Step 1.5.5.2.7

Multiply $- 3$ by $5$ .

$p (λ) = 5 λ + 2 - λ^{3} + 9 λ - 15$

Step 1.5.5.3

Add $5 λ$ and $9 λ$ .

$p (λ) = 14 λ + 2 - λ^{3} - 15$

Step 1.5.5.4

Subtract $15$ from $2$ .

$p (λ) = 14 λ - λ^{3} - 13$

Step 1.5.5.5

Reorder $14 λ$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 14 λ - 13$

Step 1.6

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

$- λ^{3} + 14 λ - 13 = 0$

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 $- 1$ out of $- λ^{3} + 14 λ - 13$ .

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Step 1.7.1.1.1

Factor $- 1$ out of $- λ^{3}$ .

$- (λ^{3}) + 14 λ - 13 = 0$

Step 1.7.1.1.2

Factor $- 1$ out of $14 λ$ .

$- (λ^{3}) - (- 14 λ) - 13 = 0$

Step 1.7.1.1.3

Rewrite $- 13$ as $- 1 (13)$ .

$- (λ^{3}) - (- 14 λ) - 1 \cdot 13 = 0$

Step 1.7.1.1.4

Factor $- 1$ out of $- (λ^{3}) - (- 14 λ)$ .

$- (λ^{3} - 14 λ) - 1 \cdot 13 = 0$

Step 1.7.1.1.5

Factor $- 1$ out of $- (λ^{3} - 14 λ) - 1 (13)$ .

$- (λ^{3} - 14 λ + 13) = 0$

Step 1.7.1.2

Factor.

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Step 1.7.1.2.1

Factor $λ^{3} - 14 λ + 13$ using the rational roots test.

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Step 1.7.1.2.1.1

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.

$p = \pm 1, \pm 13$

$q = \pm 1$

Step 1.7.1.2.1.2

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

$\pm 1, \pm 13$

Step 1.7.1.2.1.3

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

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Step 1.7.1.2.1.3.1

Substitute $1$ into the polynomial.

$1^{3} - 14 \cdot 1 + 13$

Step 1.7.1.2.1.3.2

Raise $1$ to the power of $3$ .

$1 - 14 \cdot 1 + 13$

Step 1.7.1.2.1.3.3

Multiply $- 14$ by $1$ .

$1 - 14 + 13$

Step 1.7.1.2.1.3.4

Subtract $14$ from $1$ .

$- 13 + 13$

Step 1.7.1.2.1.3.5

Add $- 13$ and $13$ .

$0$

Step 1.7.1.2.1.4

Since $1$ is a known root, divide the polynomial by $λ - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{λ^{3} - 14 λ + 13}{λ - 1}$

Step 1.7.1.2.1.5

Divide $λ^{3} - 14 λ + 13$ by $λ - 1$ .

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Step 1.7.1.2.1.5.1

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

$λ$

$1$

$λ^{3}$

$0 λ^{2}$

$14 λ$

$13$

Step 1.7.1.2.1.5.2

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

					$λ^{2}$
	$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$

Step 1.7.1.2.1.5.3

Multiply the new quotient term by the divisor.

				$λ^{2}$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			+	$λ^{3}$	-	$λ^{2}$

Step 1.7.1.2.1.5.4

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

				$λ^{2}$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$

Step 1.7.1.2.1.5.5

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

				$λ^{2}$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$

Step 1.7.1.2.1.5.6

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

				$λ^{2}$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$

Step 1.7.1.2.1.5.7

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

				$λ^{2}$	+	$λ$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$

Step 1.7.1.2.1.5.8

Multiply the new quotient term by the divisor.

				$λ^{2}$	+	$λ$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$
					+	$λ^{2}$	-	$λ$

Step 1.7.1.2.1.5.9

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

				$λ^{2}$	+	$λ$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$
					-	$λ^{2}$	+	$λ$

Step 1.7.1.2.1.5.10

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

				$λ^{2}$	+	$λ$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$
					-	$λ^{2}$	+	$λ$
							-	$13 λ$

Step 1.7.1.2.1.5.11

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

				$λ^{2}$	+	$λ$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$
					-	$λ^{2}$	+	$λ$
							-	$13 λ$	+	$13$

Step 1.7.1.2.1.5.12

Divide the highest order term in the dividend $- 13 λ$ by the highest order term in divisor $λ$ .

				$λ^{2}$	+	$λ$	-	$13$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$
					-	$λ^{2}$	+	$λ$
							-	$13 λ$	+	$13$

Step 1.7.1.2.1.5.13

Multiply the new quotient term by the divisor.

				$λ^{2}$	+	$λ$	-	$13$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$
					-	$λ^{2}$	+	$λ$
							-	$13 λ$	+	$13$
							-	$13 λ$	+	$13$

Step 1.7.1.2.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 13 λ + 13$

				$λ^{2}$	+	$λ$	-	$13$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$
					-	$λ^{2}$	+	$λ$
							-	$13 λ$	+	$13$
							+	$13 λ$	-	$13$

Step 1.7.1.2.1.5.15

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

				$λ^{2}$	+	$λ$	-	$13$
$λ$	-	$1$		$λ^{3}$	+	$0 λ^{2}$	-	$14 λ$	+	$13$
			-	$λ^{3}$	+	$λ^{2}$
					+	$λ^{2}$	-	$14 λ$
					-	$λ^{2}$	+	$λ$
							-	$13 λ$	+	$13$
							+	$13 λ$	-	$13$
										$0$

Step 1.7.1.2.1.5.16

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

$λ^{2} + λ - 13$

Step 1.7.1.2.1.6

Write $λ^{3} - 14 λ + 13$ as a set of factors.

$- ((λ - 1) (λ^{2} + λ - 13)) = 0$

Step 1.7.1.2.2

Remove unnecessary parentheses.

$- (λ - 1) (λ^{2} + λ - 13) = 0$

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 $0$ .

$λ - 1 = 0$

$λ^{2} + λ - 13 = 0$

Step 1.7.3

Set $λ - 1$ equal to $0$ and solve for $λ$ .

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Step 1.7.3.1

Set $λ - 1$ equal to $0$ .

$λ - 1 = 0$

Step 1.7.3.2

Add $1$ to both sides of the equation.

$λ = 1$

Step 1.7.4

Set $λ^{2} + λ - 13$ equal to $0$ and solve for $λ$ .

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Step 1.7.4.1

Set $λ^{2} + λ - 13$ equal to $0$ .

$λ^{2} + λ - 13 = 0$

Step 1.7.4.2

Solve $λ^{2} + λ - 13 = 0$ for $λ$ .

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Step 1.7.4.2.1

Use the quadratic formula to find the solutions.

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

Step 1.7.4.2.2

Substitute the values $a = 1$ , $b = 1$ , and $c = - 13$ into the quadratic formula and solve for $λ$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 13)}}{2 \cdot 1}$

Step 1.7.4.2.3

Simplify.

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Step 1.7.4.2.3.1

Simplify the numerator.

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Step 1.7.4.2.3.1.1

One to any power is one.

$λ = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 13}}{2 \cdot 1}$

Step 1.7.4.2.3.1.2

Multiply $- 4 \cdot 1 \cdot - 13$ .

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Step 1.7.4.2.3.1.2.1

Multiply $- 4$ by $1$ .

$λ = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 13}}{2 \cdot 1}$

Step 1.7.4.2.3.1.2.2

Multiply $- 4$ by $- 13$ .

$λ = \frac{- 1 \pm \sqrt{1 + 52}}{2 \cdot 1}$

Step 1.7.4.2.3.1.3

Add $1$ and $52$ .

$λ = \frac{- 1 \pm \sqrt{53}}{2 \cdot 1}$

Step 1.7.4.2.3.2

Multiply $2$ by $1$ .

$λ = \frac{- 1 \pm \sqrt{53}}{2}$

Step 1.7.4.2.4

The final answer is the combination of both solutions.

$λ = - \frac{1 - \sqrt{53}}{2}, - \frac{1 + \sqrt{53}}{2}$

Step 1.7.5

The final solution is all the values that make $- (λ - 1) (λ^{2} + λ - 13) = 0$ true.

$λ = 1, - \frac{1 - \sqrt{53}}{2}, - \frac{1 + \sqrt{53}}{2}$

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 $λ = 1$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] - [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Subtract the corresponding elements.

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

Step 3.2.2

Simplify each element.

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Step 3.2.2.1

Subtract $1$ from $- 2$ .

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

Step 3.2.2.2

Subtract $0$ from $3$ .

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

Step 3.2.2.3

Subtract $0$ from $0$ .

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

Step 3.2.2.4

Subtract $0$ from $3$ .

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

Step 3.2.2.5

Subtract $1$ from $0$ .

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

Step 3.2.2.6

Subtract $0$ from $1$ .

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

Step 3.2.2.7

Subtract $0$ from $1$ .

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

Step 3.2.2.8

Subtract $0$ from $1$ .

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

Step 3.2.2.9

Subtract $1$ from $2$ .

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

Step 3.3

Find the null space when $λ = 1$ .

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Step 3.3.1

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

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

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 $- \frac{1}{3}$ 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 $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

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

Step 3.3.2.1.2

Simplify $R_{1}$ .

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

Step 3.3.2.2

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

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Step 3.3.2.2.1

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

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

Step 3.3.2.2.2

Simplify $R_{2}$ .

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

Step 3.3.2.3

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

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Step 3.3.2.3.1

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

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

Step 3.3.2.3.2

Simplify $R_{3}$ .

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

Step 3.3.2.4

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

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

Step 3.3.2.4.2

Simplify $R_{2}$ .

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

Step 3.3.2.5

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

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Step 3.3.2.5.1

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

$[\begin{array}{c} 1 & - 1 & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 - 2 \cdot 0 & 2 - 2 \cdot 1 & 1 - 2 (\frac{1}{2}) & 0 - 2 \cdot 0 \end{array}]$

Step 3.3.2.5.2

Simplify $R_{3}$ .

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

Step 3.3.2.6

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

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Step 3.3.2.6.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + \frac{1}{2} & 0 + 0 \\ 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3.3.2.6.2

Simplify $R_{1}$ .

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

Step 3.3.3

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

$x + \frac{1}{2} z = 0$

$y + \frac{1}{2} z = 0$

$0 = 0$

Step 3.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{z}{2} \\ - \frac{z}{2} \\ z \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} - \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${z [\begin{array}{c} - \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{array}] | z \in R}$

Step 3.3.7

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

${[\begin{array}{c} - \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{array}]}$

Step 4

Find the eigenvector using the eigenvalue $λ = - \frac{1 - \sqrt{53}}{2}$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + \frac{1 - \sqrt{53}}{2} [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $\frac{1 - \sqrt{53}}{2}$ by each element of the matrix.

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} \cdot 1 & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 & \frac{1 - \sqrt{53}}{2} \cdot 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $\frac{1 - \sqrt{53}}{2}$ by $1$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 & \frac{1 - \sqrt{53}}{2} \cdot 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $\frac{1 - \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} & 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 & \frac{1 - \sqrt{53}}{2} \cdot 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $\frac{1 - \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} & 0 & 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 & \frac{1 - \sqrt{53}}{2} \cdot 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $\frac{1 - \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 & \frac{1 - \sqrt{53}}{2} \cdot 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.5

Multiply $\frac{1 - \sqrt{53}}{2}$ by $1$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 - \sqrt{53}}{2} & \frac{1 - \sqrt{53}}{2} \cdot 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.6

Multiply $\frac{1 - \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 - \sqrt{53}}{2} & 0 \\ \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.7

Multiply $\frac{1 - \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 - \sqrt{53}}{2} & 0 \\ 0 & \frac{1 - \sqrt{53}}{2} \cdot 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.8

Multiply $\frac{1 - \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 - \sqrt{53}}{2} & 0 \\ 0 & 0 & \frac{1 - \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.9

Multiply $\frac{1 - \sqrt{53}}{2}$ by $1$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 - \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 - \sqrt{53}}{2} & 0 \\ 0 & 0 & \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} - 2 + \frac{1 - \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - 2 \cdot \frac{2}{2} + \frac{1 - \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.2

Combine $- 2$ and $\frac{2}{2}$ .

$[\begin{array}{c} \frac{- 2 \cdot 2}{2} + \frac{1 - \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.3

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{- 2 \cdot 2 + 1 - \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.4

Simplify the numerator.

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Step 4.2.3.4.1

Multiply $- 2$ by $2$ .

$[\begin{array}{c} \frac{- 4 + 1 - \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.4.2

Add $- 4$ and $1$ .

$[\begin{array}{c} \frac{- 3 - \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.5

Rewrite $- 3$ as $- 1 (3)$ .

$[\begin{array}{c} \frac{- 1 (3) - \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.6

Factor $- 1$ out of $- \sqrt{53}$ .

$[\begin{array}{c} \frac{- 1 (3) - (\sqrt{53})}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.7

Factor $- 1$ out of $- 1 (3) - (\sqrt{53})$ .

$[\begin{array}{c} \frac{- 1 (3 + \sqrt{53})}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.8

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.9

Add $3$ and $0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.10

Add $0$ and $0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 + 0 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.11

Add $3$ and $0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & 0 + \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.12

Subtract $- \frac{1 - \sqrt{53}}{2}$ from $0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.13

Add $1$ and $0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.14

Add $1$ and $0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 \\ 1 & 1 + 0 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.15

Add $1$ and $0$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 \\ 1 & 1 & 2 + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.16

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 \\ 1 & 1 & 2 \cdot \frac{2}{2} + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.17

Combine $2$ and $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 \\ 1 & 1 & \frac{2 \cdot 2}{2} + \frac{1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.18

Combine the numerators over the common denominator.

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 \\ 1 & 1 & \frac{2 \cdot 2 + 1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.19

Simplify the numerator.

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Step 4.2.3.19.1

Multiply $2$ by $2$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 \\ 1 & 1 & \frac{4 + 1 - \sqrt{53}}{2} \end{array}]$

Step 4.2.3.19.2

Add $4$ and $1$ .

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 \\ 1 & 1 & \frac{5 - \sqrt{53}}{2} \end{array}]$

Step 4.3

Find the null space when $λ = - \frac{1 - \sqrt{53}}{2}$ .

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Step 4.3.1

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

$[\begin{array}{c} - \frac{3 + \sqrt{53}}{2} & 3 & 0 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 & 0 \\ 1 & 1 & \frac{5 - \sqrt{53}}{2} & 0 \end{array}]$

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 $- \frac{2}{3 + \sqrt{53}}$ 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 $- \frac{2}{3 + \sqrt{53}}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{2}{3 + \sqrt{53}} (- \frac{3 + \sqrt{53}}{2}) & - \frac{2}{3 + \sqrt{53}} \cdot 3 & - \frac{2}{3 + \sqrt{53}} \cdot 0 & - \frac{2}{3 + \sqrt{53}} \cdot 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 & 0 \\ 1 & 1 & \frac{5 - \sqrt{53}}{2} & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3 (3 - \sqrt{53})}{22} & 0 & 0 \\ 3 & \frac{1 - \sqrt{53}}{2} & 1 & 0 \\ 1 & 1 & \frac{5 - \sqrt{53}}{2} & 0 \end{array}]$

Step 4.3.2.2

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

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Step 4.3.2.2.1

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

$[\begin{array}{c} 1 & \frac{3 (3 - \sqrt{53})}{22} & 0 & 0 \\ 3 - 3 \cdot 1 & \frac{1 - \sqrt{53}}{2} - 3 \frac{3 (3 - \sqrt{53})}{22} & 1 - 3 \cdot 0 & 0 - 3 \cdot 0 \\ 1 & 1 & \frac{5 - \sqrt{53}}{2} & 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3 (3 - \sqrt{53})}{22} & 0 & 0 \\ 0 & - \frac{8 + \sqrt{53}}{11} & 1 & 0 \\ 1 & 1 & \frac{5 - \sqrt{53}}{2} & 0 \end{array}]$

Step 4.3.2.3

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

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Step 4.3.2.3.1

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

$[\begin{array}{c} 1 & \frac{3 (3 - \sqrt{53})}{22} & 0 & 0 \\ 0 & - \frac{8 + \sqrt{53}}{11} & 1 & 0 \\ 1 - 1 & 1 - \frac{3 (3 - \sqrt{53})}{22} & \frac{5 - \sqrt{53}}{2} - 0 & 0 - 0 \end{array}]$

Step 4.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3 (3 - \sqrt{53})}{22} & 0 & 0 \\ 0 & - \frac{8 + \sqrt{53}}{11} & 1 & 0 \\ 0 & \frac{13 + 3 \sqrt{53}}{22} & \frac{5 - \sqrt{53}}{2} & 0 \end{array}]$

Step 4.3.2.4

Multiply each element of $R_{2}$ by $- \frac{11}{8 + \sqrt{53}}$ 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 $- \frac{11}{8 + \sqrt{53}}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3 (3 - \sqrt{53})}{22} & 0 & 0 \\ - \frac{11}{8 + \sqrt{53}} \cdot 0 & - \frac{11}{8 + \sqrt{53}} (- \frac{8 + \sqrt{53}}{11}) & - \frac{11}{8 + \sqrt{53}} \cdot 1 & - \frac{11}{8 + \sqrt{53}} \cdot 0 \\ 0 & \frac{13 + 3 \sqrt{53}}{22} & \frac{5 - \sqrt{53}}{2} & 0 \end{array}]$

Step 4.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3 (3 - \sqrt{53})}{22} & 0 & 0 \\ 0 & 1 & - 8 + \sqrt{53} & 0 \\ 0 & \frac{13 + 3 \sqrt{53}}{22} & \frac{5 - \sqrt{53}}{2} & 0 \end{array}]$

Step 4.3.2.5

Perform the row operation $R_{3} = R_{3} - \frac{13 + 3 \sqrt{53}}{22} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.3.2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{13 + 3 \sqrt{53}}{22} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3 (3 - \sqrt{53})}{22} & 0 & 0 \\ 0 & 1 & - 8 + \sqrt{53} & 0 \\ 0 - \frac{13 + 3 \sqrt{53}}{22} \cdot 0 & \frac{13 + 3 \sqrt{53}}{22} - \frac{13 + 3 \sqrt{53}}{22} \cdot 1 & \frac{5 - \sqrt{53}}{2} - \frac{13 + 3 \sqrt{53}}{22} (- 8 + \sqrt{53}) & 0 - \frac{13 + 3 \sqrt{53}}{22} \cdot 0 \end{array}]$

Step 4.3.2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3 (3 - \sqrt{53})}{22} & 0 & 0 \\ 0 & 1 & - 8 + \sqrt{53} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.6

Perform the row operation $R_{1} = R_{1} - \frac{3 (3 - \sqrt{53})}{22} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 4.3.2.6.1

Perform the row operation $R_{1} = R_{1} - \frac{3 (3 - \sqrt{53})}{22} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3 (3 - \sqrt{53})}{22} \cdot 0 & \frac{3 (3 - \sqrt{53})}{22} - \frac{3 (3 - \sqrt{53})}{22} \cdot 1 & 0 - \frac{3 (3 - \sqrt{53})}{22} (- 8 + \sqrt{53}) & 0 - \frac{3 (3 - \sqrt{53})}{22} \cdot 0 \\ 0 & 1 & - 8 + \sqrt{53} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{21 - 3 \sqrt{53}}{2} & 0 \\ 0 & 1 & - 8 + \sqrt{53} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.3

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

$x + \frac{21 - 3 \sqrt{53}}{2} z = 0$

$y + (- 8 + \sqrt{53}) z = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{21 z}{2} + \frac{3 \sqrt{53} z}{2} \\ 8 z - \sqrt{53} z \\ z \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} - \frac{21}{2} + \frac{3 \sqrt{53}}{2} \\ 8 - \sqrt{53} \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${z [\begin{array}{c} - \frac{21}{2} + \frac{3 \sqrt{53}}{2} \\ 8 - \sqrt{53} \\ 1 \end{array}] | z \in R}$

Step 4.3.7

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

${[\begin{array}{c} - \frac{21}{2} + \frac{3 \sqrt{53}}{2} \\ 8 - \sqrt{53} \\ 1 \end{array}]}$

Step 5

Find the eigenvector using the eigenvalue $λ = - \frac{1 + \sqrt{53}}{2}$ .

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Step 5.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + \frac{1 + \sqrt{53}}{2} [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $\frac{1 + \sqrt{53}}{2}$ by each element of the matrix.

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} \cdot 1 & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 & \frac{1 + \sqrt{53}}{2} \cdot 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2

Simplify each element in the matrix.

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Step 5.2.1.2.1

Multiply $\frac{1 + \sqrt{53}}{2}$ by $1$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 & \frac{1 + \sqrt{53}}{2} \cdot 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.2

Multiply $\frac{1 + \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} & 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 & \frac{1 + \sqrt{53}}{2} \cdot 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.3

Multiply $\frac{1 + \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} & 0 & 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 & \frac{1 + \sqrt{53}}{2} \cdot 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.4

Multiply $\frac{1 + \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 & \frac{1 + \sqrt{53}}{2} \cdot 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.5

Multiply $\frac{1 + \sqrt{53}}{2}$ by $1$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 + \sqrt{53}}{2} & \frac{1 + \sqrt{53}}{2} \cdot 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.6

Multiply $\frac{1 + \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 + \sqrt{53}}{2} & 0 \\ \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.7

Multiply $\frac{1 + \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 + \sqrt{53}}{2} & 0 \\ 0 & \frac{1 + \sqrt{53}}{2} \cdot 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.8

Multiply $\frac{1 + \sqrt{53}}{2}$ by $0$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 + \sqrt{53}}{2} & 0 \\ 0 & 0 & \frac{1 + \sqrt{53}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.9

Multiply $\frac{1 + \sqrt{53}}{2}$ by $1$ .

$[\begin{array}{c} - 2 & 3 & 0 \\ 3 & 0 & 1 \\ 1 & 1 & 2 \end{array}] + [\begin{array}{c} \frac{1 + \sqrt{53}}{2} & 0 & 0 \\ 0 & \frac{1 + \sqrt{53}}{2} & 0 \\ 0 & 0 & \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.2

Add the corresponding elements.

$[\begin{array}{c} - 2 + \frac{1 + \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3

Simplify each element.

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Step 5.2.3.1

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - 2 \cdot \frac{2}{2} + \frac{1 + \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.2

Combine $- 2$ and $\frac{2}{2}$ .

$[\begin{array}{c} \frac{- 2 \cdot 2}{2} + \frac{1 + \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.3

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{- 2 \cdot 2 + 1 + \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.4

Simplify the numerator.

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Step 5.2.3.4.1

Multiply $- 2$ by $2$ .

$[\begin{array}{c} \frac{- 4 + 1 + \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.4.2

Add $- 4$ and $1$ .

$[\begin{array}{c} \frac{- 3 + \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.5

Rewrite $- 3$ as $- 1 (3)$ .

$[\begin{array}{c} \frac{- 1 (3) + \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.6

Factor $- 1$ out of $\sqrt{53}$ .

$[\begin{array}{c} \frac{- 1 (3) - 1 (- \sqrt{53})}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.7

Factor $- 1$ out of $- 1 (3) - 1 (- \sqrt{53})$ .

$[\begin{array}{c} \frac{- 1 (3 - \sqrt{53})}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.8

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 + 0 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.9

Add $3$ and $0$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 + 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.10

Add $0$ and $0$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 + 0 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.11

Add $3$ and $0$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & 0 + \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.12

Add $0$ and $\frac{1 + \sqrt{53}}{2}$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 + 0 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.13

Add $1$ and $0$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 \\ 1 + 0 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.14

Add $1$ and $0$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 \\ 1 & 1 + 0 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.15

Add $1$ and $0$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 \\ 1 & 1 & 2 + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.16

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 \\ 1 & 1 & 2 \cdot \frac{2}{2} + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.17

Combine $2$ and $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 \\ 1 & 1 & \frac{2 \cdot 2}{2} + \frac{1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.18

Combine the numerators over the common denominator.

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 \\ 1 & 1 & \frac{2 \cdot 2 + 1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.19

Simplify the numerator.

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Step 5.2.3.19.1

Multiply $2$ by $2$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 \\ 1 & 1 & \frac{4 + 1 + \sqrt{53}}{2} \end{array}]$

Step 5.2.3.19.2

Add $4$ and $1$ .

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 \\ 1 & 1 & \frac{5 + \sqrt{53}}{2} \end{array}]$

Step 5.3

Find the null space when $λ = - \frac{1 + \sqrt{53}}{2}$ .

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Step 5.3.1

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

$[\begin{array}{c} - \frac{3 - \sqrt{53}}{2} & 3 & 0 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 & 0 \\ 1 & 1 & \frac{5 + \sqrt{53}}{2} & 0 \end{array}]$

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 $- \frac{2}{3 - \sqrt{53}}$ 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 $- \frac{2}{3 - \sqrt{53}}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{2}{3 - \sqrt{53}} (- \frac{3 - \sqrt{53}}{2}) & - \frac{2}{3 - \sqrt{53}} \cdot 3 & - \frac{2}{3 - \sqrt{53}} \cdot 0 & - \frac{2}{3 - \sqrt{53}} \cdot 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 & 0 \\ 1 & 1 & \frac{5 + \sqrt{53}}{2} & 0 \end{array}]$

Step 5.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3 (3 + \sqrt{53})}{22} & 0 & 0 \\ 3 & \frac{1 + \sqrt{53}}{2} & 1 & 0 \\ 1 & 1 & \frac{5 + \sqrt{53}}{2} & 0 \end{array}]$

Step 5.3.2.2

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

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Step 5.3.2.2.1

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

$[\begin{array}{c} 1 & \frac{3 (3 + \sqrt{53})}{22} & 0 & 0 \\ 3 - 3 \cdot 1 & \frac{1 + \sqrt{53}}{2} - 3 \frac{3 (3 + \sqrt{53})}{22} & 1 - 3 \cdot 0 & 0 - 3 \cdot 0 \\ 1 & 1 & \frac{5 + \sqrt{53}}{2} & 0 \end{array}]$

Step 5.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3 (3 + \sqrt{53})}{22} & 0 & 0 \\ 0 & - \frac{8 - \sqrt{53}}{11} & 1 & 0 \\ 1 & 1 & \frac{5 + \sqrt{53}}{2} & 0 \end{array}]$

Step 5.3.2.3

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

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Step 5.3.2.3.1

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

$[\begin{array}{c} 1 & \frac{3 (3 + \sqrt{53})}{22} & 0 & 0 \\ 0 & - \frac{8 - \sqrt{53}}{11} & 1 & 0 \\ 1 - 1 & 1 - \frac{3 (3 + \sqrt{53})}{22} & \frac{5 + \sqrt{53}}{2} - 0 & 0 - 0 \end{array}]$

Step 5.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3 (3 + \sqrt{53})}{22} & 0 & 0 \\ 0 & - \frac{8 - \sqrt{53}}{11} & 1 & 0 \\ 0 & \frac{13 - 3 \sqrt{53}}{22} & \frac{5 + \sqrt{53}}{2} & 0 \end{array}]$

Step 5.3.2.4

Multiply each element of $R_{2}$ by $- \frac{11}{8 - \sqrt{53}}$ 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 $- \frac{11}{8 - \sqrt{53}}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3 (3 + \sqrt{53})}{22} & 0 & 0 \\ - \frac{11}{8 - \sqrt{53}} \cdot 0 & - \frac{11}{8 - \sqrt{53}} (- \frac{8 - \sqrt{53}}{11}) & - \frac{11}{8 - \sqrt{53}} \cdot 1 & - \frac{11}{8 - \sqrt{53}} \cdot 0 \\ 0 & \frac{13 - 3 \sqrt{53}}{22} & \frac{5 + \sqrt{53}}{2} & 0 \end{array}]$

Step 5.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3 (3 + \sqrt{53})}{22} & 0 & 0 \\ 0 & 1 & - 8 - \sqrt{53} & 0 \\ 0 & \frac{13 - 3 \sqrt{53}}{22} & \frac{5 + \sqrt{53}}{2} & 0 \end{array}]$

Step 5.3.2.5

Perform the row operation $R_{3} = R_{3} - \frac{13 - 3 \sqrt{53}}{22} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 5.3.2.5.1

Perform the row operation $R_{3} = R_{3} - \frac{13 - 3 \sqrt{53}}{22} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3 (3 + \sqrt{53})}{22} & 0 & 0 \\ 0 & 1 & - 8 - \sqrt{53} & 0 \\ 0 - \frac{13 - 3 \sqrt{53}}{22} \cdot 0 & \frac{13 - 3 \sqrt{53}}{22} - \frac{13 - 3 \sqrt{53}}{22} \cdot 1 & \frac{5 + \sqrt{53}}{2} - \frac{13 - 3 \sqrt{53}}{22} (- 8 - \sqrt{53}) & 0 - \frac{13 - 3 \sqrt{53}}{22} \cdot 0 \end{array}]$

Step 5.3.2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3 (3 + \sqrt{53})}{22} & 0 & 0 \\ 0 & 1 & - 8 - \sqrt{53} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.2.6

Perform the row operation $R_{1} = R_{1} - \frac{3 (3 + \sqrt{53})}{22} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 5.3.2.6.1

Perform the row operation $R_{1} = R_{1} - \frac{3 (3 + \sqrt{53})}{22} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3 (3 + \sqrt{53})}{22} \cdot 0 & \frac{3 (3 + \sqrt{53})}{22} - \frac{3 (3 + \sqrt{53})}{22} \cdot 1 & 0 - \frac{3 (3 + \sqrt{53})}{22} (- 8 - \sqrt{53}) & 0 - \frac{3 (3 + \sqrt{53})}{22} \cdot 0 \\ 0 & 1 & - 8 - \sqrt{53} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{21 + 3 \sqrt{53}}{2} & 0 \\ 0 & 1 & - 8 - \sqrt{53} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.3

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

$x + \frac{21 + 3 \sqrt{53}}{2} z = 0$

$y + (- 8 - \sqrt{53}) z = 0$

$0 = 0$

Step 5.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{21 z}{2} - \frac{3 \sqrt{53} z}{2} \\ 8 z + \sqrt{53} z \\ z \end{array}]$

Step 5.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} - \frac{21}{2} - \frac{3 \sqrt{53}}{2} \\ 8 + \sqrt{53} \\ 1 \end{array}]$

Step 5.3.6

Write as a solution set.

${z [\begin{array}{c} - \frac{21}{2} - \frac{3 \sqrt{53}}{2} \\ 8 + \sqrt{53} \\ 1 \end{array}] | z \in R}$

Step 5.3.7

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

${[\begin{array}{c} - \frac{21}{2} - \frac{3 \sqrt{53}}{2} \\ 8 + \sqrt{53} \\ 1 \end{array}]}$

Step 6

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

${[\begin{array}{c} - \frac{1}{2} \\ - \frac{1}{2} \\ 1 \end{array}], [\begin{array}{c} - \frac{21}{2} + \frac{3 \sqrt{53}}{2} \\ 8 - \sqrt{53} \\ 1 \end{array}], [\begin{array}{c} - \frac{21}{2} - \frac{3 \sqrt{53}}{2} \\ 8 + \sqrt{53} \\ 1 \end{array}]}$