Linear Algebra Examples

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}]$

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 $[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}]$ for $A$ .

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

Step 1.3.2

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

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

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.

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

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.5

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.6.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.9

Multiply $- 1$ by $1$ .

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

Step 1.4.2

Add the corresponding elements.

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

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $0$ and $0$ .

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

Step 1.4.3.2

Add $- 1$ and $0$ .

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

Step 1.4.3.3

Add $2$ and $0$ .

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

Step 1.4.3.4

Add $1$ and $0$ .

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

Step 1.4.3.5

Add $- 3$ and $0$ .

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

Step 1.4.3.6

Add $4$ and $0$ .

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

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.

$| \begin{array}{c} 3 - λ & 1 \\ 4 & 5 - λ \end{array} |$

Step 1.5.1.4

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

$(3 - λ) | \begin{array}{c} 3 - λ & 1 \\ 4 & 5 - λ \end{array} |$

Step 1.5.1.5

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

$| \begin{array}{c} 2 & 1 \\ - 3 & 5 - λ \end{array} |$

Step 1.5.1.6

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

$0 | \begin{array}{c} 2 & 1 \\ - 3 & 5 - λ \end{array} |$

Step 1.5.1.7

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

$| \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.1.8

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

$- 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.1.9

Add the terms together.

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

Step 1.5.2

Multiply $0$ by $| \begin{array}{c} 2 & 1 \\ - 3 & 5 - λ \end{array} |$ .

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

Step 1.5.3

Evaluate $| \begin{array}{c} 3 - λ & 1 \\ 4 & 5 - λ \end{array} |$ .

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Step 1.5.3.1

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

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

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

Expand $(3 - λ) (5 - λ)$ using the FOIL Method.

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Step 1.5.3.2.1.1.1

Apply the distributive property.

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

Step 1.5.3.2.1.1.2

Apply the distributive property.

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

Step 1.5.3.2.1.1.3

Apply the distributive property.

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

Step 1.5.3.2.1.2

Simplify and combine like terms.

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Step 1.5.3.2.1.2.1

Simplify each term.

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Step 1.5.3.2.1.2.1.1

Multiply $3$ by $5$ .

$p (λ) = (3 - λ) (15 + 3 (- λ) - λ \cdot 5 - λ (- λ) - 4 \cdot 1) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.1.2.1.2

Multiply $- 1$ by $3$ .

$p (λ) = (3 - λ) (15 - 3 λ - λ \cdot 5 - λ (- λ) - 4 \cdot 1) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.1.2.1.3

Multiply $5$ by $- 1$ .

$p (λ) = (3 - λ) (15 - 3 λ - 5 λ - λ (- λ) - 4 \cdot 1) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (3 - λ) (15 - 3 λ - 5 λ - 1 \cdot - 1 λ \cdot λ - 4 \cdot 1) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.1.2.1.5

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

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Step 1.5.3.2.1.2.1.5.1

Move $λ$ .

$p (λ) = (3 - λ) (15 - 3 λ - 5 λ - 1 \cdot - 1 (λ \cdot λ) - 4 \cdot 1) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (3 - λ) (15 - 3 λ - 5 λ - 1 \cdot - 1 λ^{2} - 4 \cdot 1) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = (3 - λ) (15 - 3 λ - 5 λ + 1 λ^{2} - 4 \cdot 1) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = (3 - λ) (15 - 3 λ - 5 λ + λ^{2} - 4 \cdot 1) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.1.2.2

Subtract $5 λ$ from $- 3 λ$ .

$p (λ) = (3 - λ) (15 - 8 λ + λ^{2} - 4 \cdot 1) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.1.3

Multiply $- 4$ by $1$ .

$p (λ) = (3 - λ) (15 - 8 λ + λ^{2} - 4) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.2

Subtract $4$ from $15$ .

$p (λ) = (3 - λ) (- 8 λ + λ^{2} + 11) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.3.2.3

Reorder $- 8 λ$ and $λ^{2}$ .

$p (λ) = (3 - λ) (λ^{2} - 8 λ + 11) + 0 - 1 | \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$

Step 1.5.4

Evaluate $| \begin{array}{c} 2 & 3 - λ \\ - 3 & 4 \end{array} |$ .

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Step 1.5.4.1

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

$p (λ) = (3 - λ) (λ^{2} - 8 λ + 11) + 0 - 1 (2 \cdot 4 - (- 3 (3 - λ)))$

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 $2$ by $4$ .

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

Step 1.5.4.2.1.2

Apply the distributive property.

$p (λ) = (3 - λ) (λ^{2} - 8 λ + 11) + 0 - 1 (8 - (- 3 \cdot 3 - 3 (- λ)))$

Step 1.5.4.2.1.3

Multiply $- 3$ by $3$ .

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

Step 1.5.4.2.1.4

Multiply $- 1$ by $- 3$ .

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

Step 1.5.4.2.1.5

Apply the distributive property.

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

Step 1.5.4.2.1.6

Multiply $- 1$ by $- 9$ .

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

Step 1.5.4.2.1.7

Multiply $3$ by $- 1$ .

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

Step 1.5.4.2.2

Add $8$ and $9$ .

$p (λ) = (3 - λ) (λ^{2} - 8 λ + 11) + 0 - 1 (17 - 3 λ)$

Step 1.5.4.2.3

Reorder $17$ and $- 3 λ$ .

$p (λ) = (3 - λ) (λ^{2} - 8 λ + 11) + 0 - 1 (- 3 λ + 17)$

Step 1.5.5

Simplify the determinant.

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Step 1.5.5.1

Add $(3 - λ) (λ^{2} - 8 λ + 11)$ and $0$ .

$p (λ) = (3 - λ) (λ^{2} - 8 λ + 11) - 1 (- 3 λ + 17)$

Step 1.5.5.2

Simplify each term.

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Step 1.5.5.2.1

Expand $(3 - λ) (λ^{2} - 8 λ + 11)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 3 λ^{2} + 3 (- 8 λ) + 3 \cdot 11 - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot 11 - 1 (- 3 λ + 17)$

Step 1.5.5.2.2

Simplify each term.

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Step 1.5.5.2.2.1

Multiply $- 8$ by $3$ .

$p (λ) = 3 λ^{2} - 24 λ + 3 \cdot 11 - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot 11 - 1 (- 3 λ + 17)$

Step 1.5.5.2.2.2

Multiply $3$ by $11$ .

$p (λ) = 3 λ^{2} - 24 λ + 33 - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot 11 - 1 (- 3 λ + 17)$

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 $λ^{2}$ .

$p (λ) = 3 λ^{2} - 24 λ + 33 - (λ^{2} λ) - λ (- 8 λ) - λ \cdot 11 - 1 (- 3 λ + 17)$

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 $1$ .

$p (λ) = 3 λ^{2} - 24 λ + 33 - (λ^{2} λ^{1}) - λ (- 8 λ) - λ \cdot 11 - 1 (- 3 λ + 17)$

Step 1.5.5.2.2.3.2.2

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

$p (λ) = 3 λ^{2} - 24 λ + 33 - λ^{2 + 1} - λ (- 8 λ) - λ \cdot 11 - 1 (- 3 λ + 17)$

Step 1.5.5.2.2.3.3

Add $2$ and $1$ .

$p (λ) = 3 λ^{2} - 24 λ + 33 - λ^{3} - λ (- 8 λ) - λ \cdot 11 - 1 (- 3 λ + 17)$

Step 1.5.5.2.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = 3 λ^{2} - 24 λ + 33 - λ^{3} - 1 \cdot - 8 λ \cdot λ - λ \cdot 11 - 1 (- 3 λ + 17)$

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 $λ$ .

$p (λ) = 3 λ^{2} - 24 λ + 33 - λ^{3} - 1 \cdot - 8 (λ \cdot λ) - λ \cdot 11 - 1 (- 3 λ + 17)$

Step 1.5.5.2.2.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 3 λ^{2} - 24 λ + 33 - λ^{3} - 1 \cdot - 8 λ^{2} - λ \cdot 11 - 1 (- 3 λ + 17)$

Step 1.5.5.2.2.6

Multiply $- 1$ by $- 8$ .

$p (λ) = 3 λ^{2} - 24 λ + 33 - λ^{3} + 8 λ^{2} - λ \cdot 11 - 1 (- 3 λ + 17)$

Step 1.5.5.2.2.7

Multiply $11$ by $- 1$ .

$p (λ) = 3 λ^{2} - 24 λ + 33 - λ^{3} + 8 λ^{2} - 11 λ - 1 (- 3 λ + 17)$

Step 1.5.5.2.3

Add $3 λ^{2}$ and $8 λ^{2}$ .

$p (λ) = 11 λ^{2} - 24 λ + 33 - λ^{3} - 11 λ - 1 (- 3 λ + 17)$

Step 1.5.5.2.4

Subtract $11 λ$ from $- 24 λ$ .

$p (λ) = 11 λ^{2} - 35 λ + 33 - λ^{3} - 1 (- 3 λ + 17)$

Step 1.5.5.2.5

Apply the distributive property.

$p (λ) = 11 λ^{2} - 35 λ + 33 - λ^{3} - 1 (- 3 λ) - 1 \cdot 17$

Step 1.5.5.2.6

Multiply $- 3$ by $- 1$ .

$p (λ) = 11 λ^{2} - 35 λ + 33 - λ^{3} + 3 λ - 1 \cdot 17$

Step 1.5.5.2.7

Multiply $- 1$ by $17$ .

$p (λ) = 11 λ^{2} - 35 λ + 33 - λ^{3} + 3 λ - 17$

Step 1.5.5.3

Add $- 35 λ$ and $3 λ$ .

$p (λ) = 11 λ^{2} - 32 λ + 33 - λ^{3} - 17$

Step 1.5.5.4

Subtract $17$ from $33$ .

$p (λ) = 11 λ^{2} - 32 λ - λ^{3} + 16$

Step 1.5.5.5

Move $- 32 λ$ .

$p (λ) = 11 λ^{2} - λ^{3} - 32 λ + 16$

Step 1.5.5.6

Reorder $11 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 11 λ^{2} - 32 λ + 16$

Step 1.6

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

$- λ^{3} + 11 λ^{2} - 32 λ + 16 = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Factor $- λ^{3} + 11 λ^{2} - 32 λ + 16$ using the rational roots test.

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Step 1.7.1.1

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.

$p = \pm 1, \pm 16, \pm 2, \pm 8, \pm 4$

$q = \pm 1$

Step 1.7.1.2

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

$\pm 1, \pm 16, \pm 2, \pm 8, \pm 4$

Step 1.7.1.3

Substitute $4$ and simplify the expression. In this case, the expression is equal to $0$ so $4$ is a root of the polynomial.

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Step 1.7.1.3.1

Substitute $4$ into the polynomial.

$- 4^{3} + 11 \cdot 4^{2} - 32 \cdot 4 + 16$

Step 1.7.1.3.2

Raise $4$ to the power of $3$ .

$- 1 \cdot 64 + 11 \cdot 4^{2} - 32 \cdot 4 + 16$

Step 1.7.1.3.3

Multiply $- 1$ by $64$ .

$- 64 + 11 \cdot 4^{2} - 32 \cdot 4 + 16$

Step 1.7.1.3.4

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

$- 64 + 11 \cdot 16 - 32 \cdot 4 + 16$

Step 1.7.1.3.5

Multiply $11$ by $16$ .

$- 64 + 176 - 32 \cdot 4 + 16$

Step 1.7.1.3.6

Add $- 64$ and $176$ .

$112 - 32 \cdot 4 + 16$

Step 1.7.1.3.7

Multiply $- 32$ by $4$ .

$112 - 128 + 16$

Step 1.7.1.3.8

Subtract $128$ from $112$ .

$- 16 + 16$

Step 1.7.1.3.9

Add $- 16$ and $16$ .

$0$

Step 1.7.1.4

Since $4$ is a known root, divide the polynomial by $λ - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- λ^{3} + 11 λ^{2} - 32 λ + 16}{λ - 4}$

Step 1.7.1.5

Divide $- λ^{3} + 11 λ^{2} - 32 λ + 16$ by $λ - 4$ .

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Step 1.7.1.5.1

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

$λ$

$4$

$λ^{3}$

$11 λ^{2}$

$32 λ$

$16$

Step 1.7.1.5.2

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

				-	$λ^{2}$
	$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$

Step 1.7.1.5.3

Multiply the new quotient term by the divisor.

			-	$λ^{2}$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			-	$λ^{3}$	+	$4 λ^{2}$

Step 1.7.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $- λ^{3} + 4 λ^{2}$

			-	$λ^{2}$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$

Step 1.7.1.5.5

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

			-	$λ^{2}$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$

Step 1.7.1.5.6

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

			-	$λ^{2}$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$

Step 1.7.1.5.7

Divide the highest order term in the dividend $7 λ^{2}$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$7 λ$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$

Step 1.7.1.5.8

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$7 λ$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$
					+	$7 λ^{2}$	-	$28 λ$

Step 1.7.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $7 λ^{2} - 28 λ$

			-	$λ^{2}$	+	$7 λ$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$
					-	$7 λ^{2}$	+	$28 λ$

Step 1.7.1.5.10

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

			-	$λ^{2}$	+	$7 λ$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$
					-	$7 λ^{2}$	+	$28 λ$
							-	$4 λ$

Step 1.7.1.5.11

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

			-	$λ^{2}$	+	$7 λ$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$
					-	$7 λ^{2}$	+	$28 λ$
							-	$4 λ$	+	$16$

Step 1.7.1.5.12

Divide the highest order term in the dividend $- 4 λ$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$7 λ$	-	$4$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$
					-	$7 λ^{2}$	+	$28 λ$
							-	$4 λ$	+	$16$

Step 1.7.1.5.13

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$7 λ$	-	$4$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$
					-	$7 λ^{2}$	+	$28 λ$
							-	$4 λ$	+	$16$
							-	$4 λ$	+	$16$

Step 1.7.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 λ + 16$

			-	$λ^{2}$	+	$7 λ$	-	$4$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$
					-	$7 λ^{2}$	+	$28 λ$
							-	$4 λ$	+	$16$
							+	$4 λ$	-	$16$

Step 1.7.1.5.15

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

			-	$λ^{2}$	+	$7 λ$	-	$4$
$λ$	-	$4$	-	$λ^{3}$	+	$11 λ^{2}$	-	$32 λ$	+	$16$
			+	$λ^{3}$	-	$4 λ^{2}$
					+	$7 λ^{2}$	-	$32 λ$
					-	$7 λ^{2}$	+	$28 λ$
							-	$4 λ$	+	$16$
							+	$4 λ$	-	$16$
										$0$

Step 1.7.1.5.16

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

$- λ^{2} + 7 λ - 4$

Step 1.7.1.6

Write $- λ^{3} + 11 λ^{2} - 32 λ + 16$ as a set of factors.

$(λ - 4) (- λ^{2} + 7 λ - 4) = 0$

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 $0$ .

$λ - 4 = 0$

$- λ^{2} + 7 λ - 4 = 0$

Step 1.7.3

Set $λ - 4$ equal to $0$ and solve for $λ$ .

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Step 1.7.3.1

Set $λ - 4$ equal to $0$ .

$λ - 4 = 0$

Step 1.7.3.2

Add $4$ to both sides of the equation.

$λ = 4$

Step 1.7.4

Set $- λ^{2} + 7 λ - 4$ equal to $0$ and solve for $λ$ .

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Step 1.7.4.1

Set $- λ^{2} + 7 λ - 4$ equal to $0$ .

$- λ^{2} + 7 λ - 4 = 0$

Step 1.7.4.2

Solve $- λ^{2} + 7 λ - 4 = 0$ 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 $a = - 1$ , $b = 7$ , and $c = - 4$ into the quadratic formula and solve for $λ$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (- 1 \cdot - 4)}}{2 \cdot - 1}$

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

Raise $7$ to the power of $2$ .

$λ = \frac{- 7 \pm \sqrt{49 - 4 \cdot - 1 \cdot - 4}}{2 \cdot - 1}$

Step 1.7.4.2.3.1.2

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

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Step 1.7.4.2.3.1.2.1

Multiply $- 4$ by $- 1$ .

$λ = \frac{- 7 \pm \sqrt{49 + 4 \cdot - 4}}{2 \cdot - 1}$

Step 1.7.4.2.3.1.2.2

Multiply $4$ by $- 4$ .

$λ = \frac{- 7 \pm \sqrt{49 - 16}}{2 \cdot - 1}$

Step 1.7.4.2.3.1.3

Subtract $16$ from $49$ .

$λ = \frac{- 7 \pm \sqrt{33}}{2 \cdot - 1}$

Step 1.7.4.2.3.2

Multiply $2$ by $- 1$ .

$λ = \frac{- 7 \pm \sqrt{33}}{- 2}$

Step 1.7.4.2.3.3

Simplify $\frac{- 7 \pm \sqrt{33}}{- 2}$ .

$λ = \frac{7 \pm \sqrt{33}}{2}$

Step 1.7.4.2.4

The final answer is the combination of both solutions.

$λ = \frac{7 + \sqrt{33}}{2}, \frac{7 - \sqrt{33}}{2}$

Step 1.7.5

The final solution is all the values that make $(λ - 4) (- λ^{2} + 7 λ - 4) = 0$ true.

$λ = 4, \frac{7 + \sqrt{33}}{2}, \frac{7 - \sqrt{33}}{2}$

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 $λ = 4$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] - 4 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $- 4$ by each element of the matrix.

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 \cdot 1 & - 4 \cdot 0 & - 4 \cdot 0 \\ - 4 \cdot 0 & - 4 \cdot 1 & - 4 \cdot 0 \\ - 4 \cdot 0 & - 4 \cdot 0 & - 4 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $- 4$ by $1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 & - 4 \cdot 0 & - 4 \cdot 0 \\ - 4 \cdot 0 & - 4 \cdot 1 & - 4 \cdot 0 \\ - 4 \cdot 0 & - 4 \cdot 0 & - 4 \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $- 4$ by $0$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 & 0 & - 4 \cdot 0 \\ - 4 \cdot 0 & - 4 \cdot 1 & - 4 \cdot 0 \\ - 4 \cdot 0 & - 4 \cdot 0 & - 4 \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $- 4$ by $0$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 & 0 & 0 \\ - 4 \cdot 0 & - 4 \cdot 1 & - 4 \cdot 0 \\ - 4 \cdot 0 & - 4 \cdot 0 & - 4 \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $- 4$ by $0$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 & 0 & 0 \\ 0 & - 4 \cdot 1 & - 4 \cdot 0 \\ - 4 \cdot 0 & - 4 \cdot 0 & - 4 \cdot 1 \end{array}]$

Step 3.2.1.2.5

Multiply $- 4$ by $1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 & 0 & 0 \\ 0 & - 4 & - 4 \cdot 0 \\ - 4 \cdot 0 & - 4 \cdot 0 & - 4 \cdot 1 \end{array}]$

Step 3.2.1.2.6

Multiply $- 4$ by $0$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 & 0 & 0 \\ 0 & - 4 & 0 \\ - 4 \cdot 0 & - 4 \cdot 0 & - 4 \cdot 1 \end{array}]$

Step 3.2.1.2.7

Multiply $- 4$ by $0$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & - 4 \cdot 0 & - 4 \cdot 1 \end{array}]$

Step 3.2.1.2.8

Multiply $- 4$ by $0$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & - 4 \cdot 1 \end{array}]$

Step 3.2.1.2.9

Multiply $- 4$ by $1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - 4 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & - 4 \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} 3 - 4 & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - 4 & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - 4 \end{array}]$

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Subtract $4$ from $3$ .

$[\begin{array}{c} - 1 & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - 4 & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - 4 \end{array}]$

Step 3.2.3.2

Add $0$ and $0$ .

$[\begin{array}{c} - 1 & 0 & - 1 + 0 \\ 2 + 0 & 3 - 4 & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - 4 \end{array}]$

Step 3.2.3.3

Add $- 1$ and $0$ .

$[\begin{array}{c} - 1 & 0 & - 1 \\ 2 + 0 & 3 - 4 & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - 4 \end{array}]$

Step 3.2.3.4

Add $2$ and $0$ .

$[\begin{array}{c} - 1 & 0 & - 1 \\ 2 & 3 - 4 & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - 4 \end{array}]$

Step 3.2.3.5

Subtract $4$ from $3$ .

$[\begin{array}{c} - 1 & 0 & - 1 \\ 2 & - 1 & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - 4 \end{array}]$

Step 3.2.3.6

Add $1$ and $0$ .

$[\begin{array}{c} - 1 & 0 & - 1 \\ 2 & - 1 & 1 \\ - 3 + 0 & 4 + 0 & 5 - 4 \end{array}]$

Step 3.2.3.7

Add $- 3$ and $0$ .

$[\begin{array}{c} - 1 & 0 & - 1 \\ 2 & - 1 & 1 \\ - 3 & 4 + 0 & 5 - 4 \end{array}]$

Step 3.2.3.8

Add $4$ and $0$ .

$[\begin{array}{c} - 1 & 0 & - 1 \\ 2 & - 1 & 1 \\ - 3 & 4 & 5 - 4 \end{array}]$

Step 3.2.3.9

Subtract $4$ from $5$ .

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

Step 3.3

Find the null space when $λ = 4$ .

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Step 3.3.1

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

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

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 $- 1$ 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 $- 1$ to make the entry at $1, 1$ a $1$ .

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

Step 3.3.2.1.2

Simplify $R_{1}$ .

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

Step 3.3.2.2

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

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Step 3.3.2.2.1

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

$[\begin{array}{c} 1 & 0 & 1 & 0 \\ 2 - 2 \cdot 1 & - 1 - 2 \cdot 0 & 1 - 2 \cdot 1 & 0 - 2 \cdot 0 \\ - 3 & 4 & 1 & 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{2}$ .

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

Step 3.3.2.3

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

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Step 3.3.2.3.1

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

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

Step 3.3.2.3.2

Simplify $R_{3}$ .

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

Step 3.3.2.4

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

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

Step 3.3.2.4.2

Simplify $R_{2}$ .

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

Step 3.3.2.5

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

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Step 3.3.2.5.1

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

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

Step 3.3.2.5.2

Simplify $R_{3}$ .

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

Step 3.3.3

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

$x + z = 0$

$y + z = 0$

$0 = 0$

Step 3.3.4

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

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

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

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

Step 3.3.7

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

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

Step 4

Find the eigenvector using the eigenvalue $λ = \frac{7 + \sqrt{33}}{2}$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] - \frac{7 + \sqrt{33}}{2} [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $- \frac{7 + \sqrt{33}}{2}$ by each element of the matrix.

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} \cdot 1 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $- \frac{7 + \sqrt{33}}{2} \cdot 0$ .

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Step 4.2.1.2.2.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 \frac{7 + \sqrt{33}}{2} & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.2.2

Multiply $0$ by $\frac{7 + \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $- \frac{7 + \sqrt{33}}{2} \cdot 0$ .

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Step 4.2.1.2.3.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \frac{7 + \sqrt{33}}{2} \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.3.2

Multiply $0$ by $\frac{7 + \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $- \frac{7 + \sqrt{33}}{2} \cdot 0$ .

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Step 4.2.1.2.4.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ 0 \frac{7 + \sqrt{33}}{2} & - \frac{7 + \sqrt{33}}{2} \cdot 1 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.4.2

Multiply $0$ by $\frac{7 + \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.5

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 + \sqrt{33}}{2} & - \frac{7 + \sqrt{33}}{2} \cdot 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.6

Multiply $- \frac{7 + \sqrt{33}}{2} \cdot 0$ .

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Step 4.2.1.2.6.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 + \sqrt{33}}{2} & 0 \frac{7 + \sqrt{33}}{2} \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.6.2

Multiply $0$ by $\frac{7 + \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 + \sqrt{33}}{2} & 0 \\ - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.7

Multiply $- \frac{7 + \sqrt{33}}{2} \cdot 0$ .

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Step 4.2.1.2.7.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 + \sqrt{33}}{2} & 0 \\ 0 \frac{7 + \sqrt{33}}{2} & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.7.2

Multiply $0$ by $\frac{7 + \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 + \sqrt{33}}{2} & 0 \\ 0 & - \frac{7 + \sqrt{33}}{2} \cdot 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.8

Multiply $- \frac{7 + \sqrt{33}}{2} \cdot 0$ .

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Step 4.2.1.2.8.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 + \sqrt{33}}{2} & 0 \\ 0 & 0 \frac{7 + \sqrt{33}}{2} & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.8.2

Multiply $0$ by $\frac{7 + \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 + \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 + \sqrt{33}}{2} & 0 \\ 0 & 0 & - \frac{7 + \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 4.2.1.2.9

Multiply $- 1$ by $1$ .

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} 3 - \frac{7 + \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} 3 \cdot \frac{2}{2} - \frac{7 + \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.2

Combine $3$ and $\frac{2}{2}$ .

$[\begin{array}{c} \frac{3 \cdot 2}{2} - \frac{7 + \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.3

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{3 \cdot 2 - (7 + \sqrt{33})}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4

Simplify the numerator.

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Step 4.2.3.4.1

Multiply $3$ by $2$ .

$[\begin{array}{c} \frac{6 - (7 + \sqrt{33})}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4.2

Apply the distributive property.

$[\begin{array}{c} \frac{6 - 1 \cdot 7 - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4.3

Multiply $- 1$ by $7$ .

$[\begin{array}{c} \frac{6 - 7 - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.4.4

Subtract $7$ from $6$ .

$[\begin{array}{c} \frac{- 1 - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.5

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

$[\begin{array}{c} \frac{- 1 (1) - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.6

Factor $- 1$ out of $- \sqrt{33}$ .

$[\begin{array}{c} \frac{- 1 (1) - (\sqrt{33})}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.7

Factor $- 1$ out of $- 1 (1) - (\sqrt{33})$ .

$[\begin{array}{c} \frac{- 1 (1 + \sqrt{33})}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.8

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.9

Add $0$ and $0$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.10

Add $- 1$ and $0$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 + 0 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.11

Add $2$ and $0$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & 3 - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.12

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & 3 \cdot \frac{2}{2} - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.13

Combine $3$ and $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{3 \cdot 2}{2} - \frac{7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.14

Combine the numerators over the common denominator.

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{3 \cdot 2 - (7 + \sqrt{33})}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.15

Simplify the numerator.

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Step 4.2.3.15.1

Multiply $3$ by $2$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{6 - (7 + \sqrt{33})}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.15.2

Apply the distributive property.

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{6 - 1 \cdot 7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.15.3

Multiply $- 1$ by $7$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{6 - 7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.15.4

Subtract $7$ from $6$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{- 1 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.16

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

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{- 1 (1) - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.17

Factor $- 1$ out of $- \sqrt{33}$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{- 1 (1) - (\sqrt{33})}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.18

Factor $- 1$ out of $- 1 (1) - (\sqrt{33})$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{- 1 (1 + \sqrt{33})}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.19

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.20

Add $1$ and $0$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.21

Add $- 3$ and $0$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 & 4 + 0 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.22

Add $4$ and $0$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 & 4 & 5 - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.23

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 & 4 & 5 \cdot \frac{2}{2} - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.24

Combine $5$ and $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{5 \cdot 2}{2} - \frac{7 + \sqrt{33}}{2} \end{array}]$

Step 4.2.3.25

Combine the numerators over the common denominator.

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{5 \cdot 2 - (7 + \sqrt{33})}{2} \end{array}]$

Step 4.2.3.26

Simplify the numerator.

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Step 4.2.3.26.1

Multiply $5$ by $2$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{10 - (7 + \sqrt{33})}{2} \end{array}]$

Step 4.2.3.26.2

Apply the distributive property.

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{10 - 1 \cdot 7 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.26.3

Multiply $- 1$ by $7$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{10 - 7 - \sqrt{33}}{2} \end{array}]$

Step 4.2.3.26.4

Subtract $7$ from $10$ .

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{3 - \sqrt{33}}{2} \end{array}]$

Step 4.3

Find the null space when $λ = \frac{7 + \sqrt{33}}{2}$ .

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Step 4.3.1

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

$[\begin{array}{c} - \frac{1 + \sqrt{33}}{2} & 0 & - 1 & 0 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 & 0 \\ - 3 & 4 & \frac{3 - \sqrt{33}}{2} & 0 \end{array}]$

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 $- \frac{2}{1 + \sqrt{33}}$ 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 $- \frac{2}{1 + \sqrt{33}}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{2}{1 + \sqrt{33}} (- \frac{1 + \sqrt{33}}{2}) & - \frac{2}{1 + \sqrt{33}} \cdot 0 & - \frac{2}{1 + \sqrt{33}} \cdot - 1 & - \frac{2}{1 + \sqrt{33}} \cdot 0 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 & 0 \\ - 3 & 4 & \frac{3 - \sqrt{33}}{2} & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 - \sqrt{33}}{16} & 0 \\ 2 & - \frac{1 + \sqrt{33}}{2} & 1 & 0 \\ - 3 & 4 & \frac{3 - \sqrt{33}}{2} & 0 \end{array}]$

Step 4.3.2.2

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

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Step 4.3.2.2.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1 - \sqrt{33}}{16} & 0 \\ 2 - 2 \cdot 1 & - \frac{1 + \sqrt{33}}{2} - 2 \cdot 0 & 1 - 2 (- \frac{1 - \sqrt{33}}{16}) & 0 - 2 \cdot 0 \\ - 3 & 4 & \frac{3 - \sqrt{33}}{2} & 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 - \sqrt{33}}{16} & 0 \\ 0 & - \frac{1 + \sqrt{33}}{2} & \frac{9 - \sqrt{33}}{8} & 0 \\ - 3 & 4 & \frac{3 - \sqrt{33}}{2} & 0 \end{array}]$

Step 4.3.2.3

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

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Step 4.3.2.3.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1 - \sqrt{33}}{16} & 0 \\ 0 & - \frac{1 + \sqrt{33}}{2} & \frac{9 - \sqrt{33}}{8} & 0 \\ - 3 + 3 \cdot 1 & 4 + 3 \cdot 0 & \frac{3 - \sqrt{33}}{2} + 3 (- \frac{1 - \sqrt{33}}{16}) & 0 + 3 \cdot 0 \end{array}]$

Step 4.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 - \sqrt{33}}{16} & 0 \\ 0 & - \frac{1 + \sqrt{33}}{2} & \frac{9 - \sqrt{33}}{8} & 0 \\ 0 & 4 & \frac{21 - 5 \sqrt{33}}{16} & 0 \end{array}]$

Step 4.3.2.4

Multiply each element of $R_{2}$ by $- \frac{2}{1 + \sqrt{33}}$ 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 $- \frac{2}{1 + \sqrt{33}}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 - \sqrt{33}}{16} & 0 \\ - \frac{2}{1 + \sqrt{33}} \cdot 0 & - \frac{2}{1 + \sqrt{33}} (- \frac{1 + \sqrt{33}}{2}) & - \frac{2}{1 + \sqrt{33}} \cdot \frac{9 - \sqrt{33}}{8} & - \frac{2}{1 + \sqrt{33}} \cdot 0 \\ 0 & 4 & \frac{21 - 5 \sqrt{33}}{16} & 0 \end{array}]$

Step 4.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 - \sqrt{33}}{16} & 0 \\ 0 & 1 & \frac{21 - 5 \sqrt{33}}{64} & 0 \\ 0 & 4 & \frac{21 - 5 \sqrt{33}}{16} & 0 \end{array}]$

Step 4.3.2.5

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

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Step 4.3.2.5.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1 - \sqrt{33}}{16} & 0 \\ 0 & 1 & \frac{21 - 5 \sqrt{33}}{64} & 0 \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & \frac{21 - 5 \sqrt{33}}{16} - 4 \frac{21 - 5 \sqrt{33}}{64} & 0 - 4 \cdot 0 \end{array}]$

Step 4.3.2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 - \sqrt{33}}{16} & 0 \\ 0 & 1 & \frac{21 - 5 \sqrt{33}}{64} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.3

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

$x - \frac{1 - \sqrt{33}}{16} z = 0$

$y + \frac{21 - 5 \sqrt{33}}{64} z = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{z}{16} - \frac{z \sqrt{33}}{16} \\ - \frac{21 z}{64} + \frac{5 \sqrt{33} z}{64} \\ z \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} \frac{1}{16} - \frac{\sqrt{33}}{16} \\ - \frac{21}{64} + \frac{5 \sqrt{33}}{64} \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${z [\begin{array}{c} \frac{1}{16} - \frac{\sqrt{33}}{16} \\ - \frac{21}{64} + \frac{5 \sqrt{33}}{64} \\ 1 \end{array}] | z \in R}$

Step 4.3.7

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

${[\begin{array}{c} \frac{1}{16} - \frac{\sqrt{33}}{16} \\ - \frac{21}{64} + \frac{5 \sqrt{33}}{64} \\ 1 \end{array}]}$

Step 5

Find the eigenvector using the eigenvalue $λ = \frac{7 - \sqrt{33}}{2}$ .

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Step 5.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] - \frac{7 - \sqrt{33}}{2} [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $- \frac{7 - \sqrt{33}}{2}$ by each element of the matrix.

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} \cdot 1 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2

Simplify each element in the matrix.

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Step 5.2.1.2.1

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.2

Multiply $- \frac{7 - \sqrt{33}}{2} \cdot 0$ .

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Step 5.2.1.2.2.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 \frac{7 - \sqrt{33}}{2} & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.2.2

Multiply $0$ by $\frac{7 - \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.3

Multiply $- \frac{7 - \sqrt{33}}{2} \cdot 0$ .

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Step 5.2.1.2.3.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \frac{7 - \sqrt{33}}{2} \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.3.2

Multiply $0$ by $\frac{7 - \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.4

Multiply $- \frac{7 - \sqrt{33}}{2} \cdot 0$ .

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Step 5.2.1.2.4.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ 0 \frac{7 - \sqrt{33}}{2} & - \frac{7 - \sqrt{33}}{2} \cdot 1 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.4.2

Multiply $0$ by $\frac{7 - \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.5

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 - \sqrt{33}}{2} & - \frac{7 - \sqrt{33}}{2} \cdot 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.6

Multiply $- \frac{7 - \sqrt{33}}{2} \cdot 0$ .

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Step 5.2.1.2.6.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 - \sqrt{33}}{2} & 0 \frac{7 - \sqrt{33}}{2} \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.6.2

Multiply $0$ by $\frac{7 - \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 - \sqrt{33}}{2} & 0 \\ - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.7

Multiply $- \frac{7 - \sqrt{33}}{2} \cdot 0$ .

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Step 5.2.1.2.7.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 - \sqrt{33}}{2} & 0 \\ 0 \frac{7 - \sqrt{33}}{2} & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.7.2

Multiply $0$ by $\frac{7 - \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 - \sqrt{33}}{2} & 0 \\ 0 & - \frac{7 - \sqrt{33}}{2} \cdot 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.8

Multiply $- \frac{7 - \sqrt{33}}{2} \cdot 0$ .

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Step 5.2.1.2.8.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 - \sqrt{33}}{2} & 0 \\ 0 & 0 \frac{7 - \sqrt{33}}{2} & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.8.2

Multiply $0$ by $\frac{7 - \sqrt{33}}{2}$ .

$[\begin{array}{c} 3 & 0 & - 1 \\ 2 & 3 & 1 \\ - 3 & 4 & 5 \end{array}] + [\begin{array}{c} - \frac{7 - \sqrt{33}}{2} & 0 & 0 \\ 0 & - \frac{7 - \sqrt{33}}{2} & 0 \\ 0 & 0 & - \frac{7 - \sqrt{33}}{2} \cdot 1 \end{array}]$

Step 5.2.1.2.9

Multiply $- 1$ by $1$ .

Step 5.2.2

Add the corresponding elements.

$[\begin{array}{c} 3 - \frac{7 - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3

Simplify each element.

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Step 5.2.3.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} 3 \cdot \frac{2}{2} - \frac{7 - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.2

Combine $3$ and $\frac{2}{2}$ .

$[\begin{array}{c} \frac{3 \cdot 2}{2} - \frac{7 - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.3

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{3 \cdot 2 - (7 - \sqrt{33})}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.4

Simplify the numerator.

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Step 5.2.3.4.1

Multiply $3$ by $2$ .

$[\begin{array}{c} \frac{6 - (7 - \sqrt{33})}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.4.2

Apply the distributive property.

$[\begin{array}{c} \frac{6 - 1 \cdot 7 - - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.4.3

Multiply $- 1$ by $7$ .

$[\begin{array}{c} \frac{6 - 7 - - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.4.4

Multiply $- - \sqrt{33}$ .

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Step 5.2.3.4.4.1

Multiply $- 1$ by $- 1$ .

$[\begin{array}{c} \frac{6 - 7 + 1 \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.4.4.2

Multiply $\sqrt{33}$ by $1$ .

$[\begin{array}{c} \frac{6 - 7 + \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.4.5

Subtract $7$ from $6$ .

$[\begin{array}{c} \frac{- 1 + \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.5

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

$[\begin{array}{c} \frac{- 1 (1) + \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.6

Factor $- 1$ out of $\sqrt{33}$ .

$[\begin{array}{c} \frac{- 1 (1) - 1 (- \sqrt{33})}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.7

Factor $- 1$ out of $- 1 (1) - 1 (- \sqrt{33})$ .

$[\begin{array}{c} \frac{- 1 (1 - \sqrt{33})}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.8

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 + 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.9

Add $0$ and $0$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 + 0 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.10

Add $- 1$ and $0$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 + 0 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.11

Add $2$ and $0$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & 3 - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.12

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & 3 \cdot \frac{2}{2} - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.13

Combine $3$ and $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{3 \cdot 2}{2} - \frac{7 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.14

Combine the numerators over the common denominator.

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{3 \cdot 2 - (7 - \sqrt{33})}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.15

Simplify the numerator.

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Step 5.2.3.15.1

Multiply $3$ by $2$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{6 - (7 - \sqrt{33})}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.15.2

Apply the distributive property.

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{6 - 1 \cdot 7 - - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.15.3

Multiply $- 1$ by $7$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{6 - 7 - - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.15.4

Multiply $- - \sqrt{33}$ .

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Step 5.2.3.15.4.1

Multiply $- 1$ by $- 1$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{6 - 7 + 1 \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.15.4.2

Multiply $\sqrt{33}$ by $1$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{6 - 7 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.15.5

Subtract $7$ from $6$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{- 1 + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.16

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

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{- 1 (1) + \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.17

Factor $- 1$ out of $\sqrt{33}$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{- 1 (1) - 1 (- \sqrt{33})}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.18

Factor $- 1$ out of $- 1 (1) - 1 (- \sqrt{33})$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & \frac{- 1 (1 - \sqrt{33})}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.19

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 + 0 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.20

Add $1$ and $0$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 + 0 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.21

Add $- 3$ and $0$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 + 0 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.22

Add $4$ and $0$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & 5 - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.23

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & 5 \cdot \frac{2}{2} - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.24

Combine $5$ and $\frac{2}{2}$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{5 \cdot 2}{2} - \frac{7 - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.25

Combine the numerators over the common denominator.

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{5 \cdot 2 - (7 - \sqrt{33})}{2} \end{array}]$

Step 5.2.3.26

Simplify the numerator.

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Step 5.2.3.26.1

Multiply $5$ by $2$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{10 - (7 - \sqrt{33})}{2} \end{array}]$

Step 5.2.3.26.2

Apply the distributive property.

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{10 - 1 \cdot 7 - - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.26.3

Multiply $- 1$ by $7$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{10 - 7 - - \sqrt{33}}{2} \end{array}]$

Step 5.2.3.26.4

Multiply $- - \sqrt{33}$ .

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Step 5.2.3.26.4.1

Multiply $- 1$ by $- 1$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{10 - 7 + 1 \sqrt{33}}{2} \end{array}]$

Step 5.2.3.26.4.2

Multiply $\sqrt{33}$ by $1$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{10 - 7 + \sqrt{33}}{2} \end{array}]$

Step 5.2.3.26.5

Subtract $7$ from $10$ .

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 \\ - 3 & 4 & \frac{3 + \sqrt{33}}{2} \end{array}]$

Step 5.3

Find the null space when $λ = \frac{7 - \sqrt{33}}{2}$ .

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Step 5.3.1

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

$[\begin{array}{c} - \frac{1 - \sqrt{33}}{2} & 0 & - 1 & 0 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 & 0 \\ - 3 & 4 & \frac{3 + \sqrt{33}}{2} & 0 \end{array}]$

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 $- \frac{2}{1 - \sqrt{33}}$ 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 $- \frac{2}{1 - \sqrt{33}}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{2}{1 - \sqrt{33}} (- \frac{1 - \sqrt{33}}{2}) & - \frac{2}{1 - \sqrt{33}} \cdot 0 & - \frac{2}{1 - \sqrt{33}} \cdot - 1 & - \frac{2}{1 - \sqrt{33}} \cdot 0 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 & 0 \\ - 3 & 4 & \frac{3 + \sqrt{33}}{2} & 0 \end{array}]$

Step 5.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 + \sqrt{33}}{16} & 0 \\ 2 & - \frac{1 - \sqrt{33}}{2} & 1 & 0 \\ - 3 & 4 & \frac{3 + \sqrt{33}}{2} & 0 \end{array}]$

Step 5.3.2.2

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

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Step 5.3.2.2.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1 + \sqrt{33}}{16} & 0 \\ 2 - 2 \cdot 1 & - \frac{1 - \sqrt{33}}{2} - 2 \cdot 0 & 1 - 2 (- \frac{1 + \sqrt{33}}{16}) & 0 - 2 \cdot 0 \\ - 3 & 4 & \frac{3 + \sqrt{33}}{2} & 0 \end{array}]$

Step 5.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 + \sqrt{33}}{16} & 0 \\ 0 & - \frac{1 - \sqrt{33}}{2} & \frac{9 + \sqrt{33}}{8} & 0 \\ - 3 & 4 & \frac{3 + \sqrt{33}}{2} & 0 \end{array}]$

Step 5.3.2.3

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

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Step 5.3.2.3.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1 + \sqrt{33}}{16} & 0 \\ 0 & - \frac{1 - \sqrt{33}}{2} & \frac{9 + \sqrt{33}}{8} & 0 \\ - 3 + 3 \cdot 1 & 4 + 3 \cdot 0 & \frac{3 + \sqrt{33}}{2} + 3 (- \frac{1 + \sqrt{33}}{16}) & 0 + 3 \cdot 0 \end{array}]$

Step 5.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 + \sqrt{33}}{16} & 0 \\ 0 & - \frac{1 - \sqrt{33}}{2} & \frac{9 + \sqrt{33}}{8} & 0 \\ 0 & 4 & \frac{21 + 5 \sqrt{33}}{16} & 0 \end{array}]$

Step 5.3.2.4

Multiply each element of $R_{2}$ by $- \frac{2}{1 - \sqrt{33}}$ 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 $- \frac{2}{1 - \sqrt{33}}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 + \sqrt{33}}{16} & 0 \\ - \frac{2}{1 - \sqrt{33}} \cdot 0 & - \frac{2}{1 - \sqrt{33}} (- \frac{1 - \sqrt{33}}{2}) & - \frac{2}{1 - \sqrt{33}} \cdot \frac{9 + \sqrt{33}}{8} & - \frac{2}{1 - \sqrt{33}} \cdot 0 \\ 0 & 4 & \frac{21 + 5 \sqrt{33}}{16} & 0 \end{array}]$

Step 5.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 + \sqrt{33}}{16} & 0 \\ 0 & 1 & \frac{21 + 5 \sqrt{33}}{64} & 0 \\ 0 & 4 & \frac{21 + 5 \sqrt{33}}{16} & 0 \end{array}]$

Step 5.3.2.5

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

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Step 5.3.2.5.1

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

$[\begin{array}{c} 1 & 0 & - \frac{1 + \sqrt{33}}{16} & 0 \\ 0 & 1 & \frac{21 + 5 \sqrt{33}}{64} & 0 \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & \frac{21 + 5 \sqrt{33}}{16} - 4 \frac{21 + 5 \sqrt{33}}{64} & 0 - 4 \cdot 0 \end{array}]$

Step 5.3.2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1 + \sqrt{33}}{16} & 0 \\ 0 & 1 & \frac{21 + 5 \sqrt{33}}{64} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.3

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

$x - \frac{1 + \sqrt{33}}{16} z = 0$

$y + \frac{21 + 5 \sqrt{33}}{64} z = 0$

$0 = 0$

Step 5.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{z}{16} + \frac{z \sqrt{33}}{16} \\ - \frac{21 z}{64} - \frac{5 \sqrt{33} z}{64} \\ z \end{array}]$

Step 5.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} \frac{1}{16} + \frac{\sqrt{33}}{16} \\ - \frac{21}{64} - \frac{5 \sqrt{33}}{64} \\ 1 \end{array}]$

Step 5.3.6

Write as a solution set.

${z [\begin{array}{c} \frac{1}{16} + \frac{\sqrt{33}}{16} \\ - \frac{21}{64} - \frac{5 \sqrt{33}}{64} \\ 1 \end{array}] | z \in R}$

Step 5.3.7

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

${[\begin{array}{c} \frac{1}{16} + \frac{\sqrt{33}}{16} \\ - \frac{21}{64} - \frac{5 \sqrt{33}}{64} \\ 1 \end{array}]}$

Step 6

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

${[\begin{array}{c} - 1 \\ - 1 \\ 1 \end{array}], [\begin{array}{c} \frac{1}{16} - \frac{\sqrt{33}}{16} \\ - \frac{21}{64} + \frac{5 \sqrt{33}}{64} \\ 1 \end{array}], [\begin{array}{c} \frac{1}{16} + \frac{\sqrt{33}}{16} \\ - \frac{21}{64} - \frac{5 \sqrt{33}}{64} \\ 1 \end{array}]}$