Linear Algebra Examples

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}]$

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 $[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] - λ I_{3})$

Step 1.3.2

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

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 13 - λ & 3 + 0 & 1 + 0 \\ - 56 + 0 & - 13 - λ & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 13 - λ & 3 & 1 + 0 \\ - 56 + 0 & - 13 - λ & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 - λ \end{array}]$

Step 1.4.3.2

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 13 - λ & 3 & 1 \\ - 56 + 0 & - 13 - λ & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 - λ \end{array}]$

Step 1.4.3.3

Add $- 56$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 13 - λ & 3 & 1 \\ - 56 & - 13 - λ & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 - λ \end{array}]$

Step 1.4.3.4

Add $- 4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 13 - λ & 3 & 1 \\ - 56 & - 13 - λ & - 4 \\ - 14 + 0 & - 3 + 0 & - 2 - λ \end{array}]$

Step 1.4.3.5

Add $- 14$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 13 - λ & 3 & 1 \\ - 56 & - 13 - λ & - 4 \\ - 14 & - 3 + 0 & - 2 - λ \end{array}]$

Step 1.4.3.6

Add $- 3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 13 - λ & 3 & 1 \\ - 56 & - 13 - λ & - 4 \\ - 14 & - 3 & - 2 - λ \end{array}]$

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.

$| \begin{array}{c} - 13 - λ & - 4 \\ - 3 & - 2 - λ \end{array} |$

Step 1.5.1.4

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

$(13 - λ) | \begin{array}{c} - 13 - λ & - 4 \\ - 3 & - 2 - λ \end{array} |$

Step 1.5.1.5

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

$| \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} |$

Step 1.5.1.6

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

$- 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} |$

Step 1.5.1.7

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

$| \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.1.8

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

$1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.1.9

Add the terms together.

$p (λ) = (13 - λ) | \begin{array}{c} - 13 - λ & - 4 \\ - 3 & - 2 - λ \end{array} | - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2

Evaluate $| \begin{array}{c} - 13 - λ & - 4 \\ - 3 & - 2 - λ \end{array} |$ .

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Step 1.5.2.1

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

$p (λ) = (13 - λ) ((- 13 - λ) (- 2 - λ) - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2

Simplify the determinant.

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Step 1.5.2.2.1

Simplify each term.

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Step 1.5.2.2.1.1

Expand $(- 13 - λ) (- 2 - λ)$ using the FOIL Method.

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Step 1.5.2.2.1.1.1

Apply the distributive property.

$p (λ) = (13 - λ) (- 13 (- 2 - λ) - λ (- 2 - λ) - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.1.2

Apply the distributive property.

$p (λ) = (13 - λ) (- 13 \cdot - 2 - 13 (- λ) - λ (- 2 - λ) - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.1.3

Apply the distributive property.

$p (λ) = (13 - λ) (- 13 \cdot - 2 - 13 (- λ) - λ \cdot - 2 - λ (- λ) - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.2

Simplify and combine like terms.

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Step 1.5.2.2.1.2.1

Simplify each term.

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Step 1.5.2.2.1.2.1.1

Multiply $- 13$ by $- 2$ .

$p (λ) = (13 - λ) (26 - 13 (- λ) - λ \cdot - 2 - λ (- λ) - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.2.1.2

Multiply $- 1$ by $- 13$ .

$p (λ) = (13 - λ) (26 + 13 λ - λ \cdot - 2 - λ (- λ) - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.2.1.3

Multiply $- 2$ by $- 1$ .

$p (λ) = (13 - λ) (26 + 13 λ + 2 λ - λ (- λ) - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (13 - λ) (26 + 13 λ + 2 λ - 1 \cdot - 1 λ \cdot λ - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.2.1.5

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

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Step 1.5.2.2.1.2.1.5.1

Move $λ$ .

$p (λ) = (13 - λ) (26 + 13 λ + 2 λ - 1 \cdot - 1 (λ \cdot λ) - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (13 - λ) (26 + 13 λ + 2 λ - 1 \cdot - 1 λ^{2} - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = (13 - λ) (26 + 13 λ + 2 λ + 1 λ^{2} - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = (13 - λ) (26 + 13 λ + 2 λ + λ^{2} - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.2.2

Add $13 λ$ and $2 λ$ .

$p (λ) = (13 - λ) (26 + 15 λ + λ^{2} - (- 3 \cdot - 4)) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.3

Multiply $- (- 3 \cdot - 4)$ .

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Step 1.5.2.2.1.3.1

Multiply $- 3$ by $- 4$ .

$p (λ) = (13 - λ) (26 + 15 λ + λ^{2} - 1 \cdot 12) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.1.3.2

Multiply $- 1$ by $12$ .

$p (λ) = (13 - λ) (26 + 15 λ + λ^{2} - 12) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.2

Subtract $12$ from $26$ .

$p (λ) = (13 - λ) (15 λ + λ^{2} + 14) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.2.2.3

Reorder $15 λ$ and $λ^{2}$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 | \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} | + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.3

Evaluate $| \begin{array}{c} - 56 & - 4 \\ - 14 & - 2 - λ \end{array} |$ .

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Step 1.5.3.1

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

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (- 56 (- 2 - λ) - (- 14 \cdot - 4)) + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.3.2

Simplify the determinant.

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Step 1.5.3.2.1

Simplify each term.

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Step 1.5.3.2.1.1

Apply the distributive property.

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (- 56 \cdot - 2 - 56 (- λ) - (- 14 \cdot - 4)) + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.3.2.1.2

Multiply $- 56$ by $- 2$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (112 - 56 (- λ) - (- 14 \cdot - 4)) + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.3.2.1.3

Multiply $- 1$ by $- 56$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (112 + 56 λ - (- 14 \cdot - 4)) + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.3.2.1.4

Multiply $- (- 14 \cdot - 4)$ .

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Step 1.5.3.2.1.4.1

Multiply $- 14$ by $- 4$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (112 + 56 λ - 1 \cdot 56) + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.3.2.1.4.2

Multiply $- 1$ by $56$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (112 + 56 λ - 56) + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.3.2.2

Subtract $56$ from $112$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 | \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$

Step 1.5.4

Evaluate $| \begin{array}{c} - 56 & - 13 - λ \\ - 14 & - 3 \end{array} |$ .

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Step 1.5.4.1

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

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (- 56 \cdot - 3 - (- 14 (- 13 - λ)))$

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 $- 56$ by $- 3$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (168 - (- 14 (- 13 - λ)))$

Step 1.5.4.2.1.2

Apply the distributive property.

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (168 - (- 14 \cdot - 13 - 14 (- λ)))$

Step 1.5.4.2.1.3

Multiply $- 14$ by $- 13$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (168 - (182 - 14 (- λ)))$

Step 1.5.4.2.1.4

Multiply $- 1$ by $- 14$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (168 - (182 + 14 λ))$

Step 1.5.4.2.1.5

Apply the distributive property.

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (168 - 1 \cdot 182 - (14 λ))$

Step 1.5.4.2.1.6

Multiply $- 1$ by $182$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (168 - 182 - (14 λ))$

Step 1.5.4.2.1.7

Multiply $14$ by $- 1$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (168 - 182 - 14 λ)$

Step 1.5.4.2.2

Subtract $182$ from $168$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (- 14 - 14 λ)$

Step 1.5.4.2.3

Reorder $- 14$ and $- 14 λ$ .

$p (λ) = (13 - λ) (λ^{2} + 15 λ + 14) - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5

Simplify the determinant.

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Step 1.5.5.1

Simplify each term.

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Step 1.5.5.1.1

Expand $(13 - λ) (λ^{2} + 15 λ + 14)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 13 λ^{2} + 13 (15 λ) + 13 \cdot 14 - λ \cdot λ^{2} - λ (15 λ) - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2

Simplify each term.

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Step 1.5.5.1.2.1

Multiply $15$ by $13$ .

$p (λ) = 13 λ^{2} + 195 λ + 13 \cdot 14 - λ \cdot λ^{2} - λ (15 λ) - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.2

Multiply $13$ by $14$ .

$p (λ) = 13 λ^{2} + 195 λ + 182 - λ \cdot λ^{2} - λ (15 λ) - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.3

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

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Step 1.5.5.1.2.3.1

Move $λ^{2}$ .

$p (λ) = 13 λ^{2} + 195 λ + 182 - (λ^{2} λ) - λ (15 λ) - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.5.1.2.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 13 λ^{2} + 195 λ + 182 - (λ^{2} λ^{1}) - λ (15 λ) - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.3.2.2

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

$p (λ) = 13 λ^{2} + 195 λ + 182 - λ^{2 + 1} - λ (15 λ) - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.3.3

Add $2$ and $1$ .

$p (λ) = 13 λ^{2} + 195 λ + 182 - λ^{3} - λ (15 λ) - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = 13 λ^{2} + 195 λ + 182 - λ^{3} - 1 \cdot 15 λ \cdot λ - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.5

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

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Step 1.5.5.1.2.5.1

Move $λ$ .

$p (λ) = 13 λ^{2} + 195 λ + 182 - λ^{3} - 1 \cdot 15 (λ \cdot λ) - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 13 λ^{2} + 195 λ + 182 - λ^{3} - 1 \cdot 15 λ^{2} - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.6

Multiply $- 1$ by $15$ .

$p (λ) = 13 λ^{2} + 195 λ + 182 - λ^{3} - 15 λ^{2} - λ \cdot 14 - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.2.7

Multiply $14$ by $- 1$ .

$p (λ) = 13 λ^{2} + 195 λ + 182 - λ^{3} - 15 λ^{2} - 14 λ - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.3

Subtract $15 λ^{2}$ from $13 λ^{2}$ .

$p (λ) = - 2 λ^{2} + 195 λ + 182 - λ^{3} - 14 λ - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.4

Subtract $14 λ$ from $195 λ$ .

$p (λ) = - 2 λ^{2} + 181 λ + 182 - λ^{3} - 3 (56 λ + 56) + 1 (- 14 λ - 14)$

Step 1.5.5.1.5

Apply the distributive property.

$p (λ) = - 2 λ^{2} + 181 λ + 182 - λ^{3} - 3 (56 λ) - 3 \cdot 56 + 1 (- 14 λ - 14)$

Step 1.5.5.1.6

Multiply $56$ by $- 3$ .

$p (λ) = - 2 λ^{2} + 181 λ + 182 - λ^{3} - 168 λ - 3 \cdot 56 + 1 (- 14 λ - 14)$

Step 1.5.5.1.7

Multiply $- 3$ by $56$ .

$p (λ) = - 2 λ^{2} + 181 λ + 182 - λ^{3} - 168 λ - 168 + 1 (- 14 λ - 14)$

Step 1.5.5.1.8

Multiply $- 14 λ - 14$ by $1$ .

$p (λ) = - 2 λ^{2} + 181 λ + 182 - λ^{3} - 168 λ - 168 - 14 λ - 14$

Step 1.5.5.2

Subtract $168 λ$ from $181 λ$ .

$p (λ) = - 2 λ^{2} + 13 λ + 182 - λ^{3} - 168 - 14 λ - 14$

Step 1.5.5.3

Subtract $14 λ$ from $13 λ$ .

$p (λ) = - 2 λ^{2} - λ + 182 - λ^{3} - 168 - 14$

Step 1.5.5.4

Subtract $168$ from $182$ .

$p (λ) = - 2 λ^{2} - λ - λ^{3} + 14 - 14$

Step 1.5.5.5

Combine the opposite terms in $- 2 λ^{2} - λ - λ^{3} + 14 - 14$ .

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Step 1.5.5.5.1

Subtract $14$ from $14$ .

$p (λ) = - 2 λ^{2} - λ - λ^{3} + 0$

Step 1.5.5.5.2

Add $- 2 λ^{2} - λ - λ^{3}$ and $0$ .

$p (λ) = - 2 λ^{2} - λ - λ^{3}$

Step 1.5.5.6

Move $- λ$ .

$p (λ) = - 2 λ^{2} - λ^{3} - λ$

Step 1.5.5.7

Reorder $- 2 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} - 2 λ^{2} - λ$

Step 1.6

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

$- λ^{3} - 2 λ^{2} - λ = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Factor the left side of the equation.

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Step 1.7.1.1

Factor $- λ$ out of $- λ^{3} - 2 λ^{2} - λ$ .

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Step 1.7.1.1.1

Factor $- λ$ out of $- λ^{3}$ .

$- λ \cdot λ^{2} - 2 λ^{2} - λ = 0$

Step 1.7.1.1.2

Factor $- λ$ out of $- 2 λ^{2}$ .

$- λ \cdot λ^{2} - λ (2 λ) - λ = 0$

Step 1.7.1.1.3

Factor $- λ$ out of $- λ$ .

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

Step 1.7.1.1.4

Factor $- λ$ out of $- λ (λ^{2}) - λ (2 λ)$ .

$- λ (λ^{2} + 2 λ) - λ \cdot 1 = 0$

Step 1.7.1.1.5

Factor $- λ$ out of $- λ (λ^{2} + 2 λ) - λ (1)$ .

$- λ (λ^{2} + 2 λ + 1) = 0$

Step 1.7.1.2

Factor using the perfect square rule.

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Step 1.7.1.2.1

Rewrite $1$ as $1^{2}$ .

$- λ (λ^{2} + 2 λ + 1^{2}) = 0$

Step 1.7.1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 λ = 2 \cdot λ \cdot 1$

Step 1.7.1.2.3

Rewrite the polynomial.

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

Step 1.7.1.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = λ$ and $b = 1$ .

$- λ {(λ + 1)}^{2} = 0$

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 $0$ .

$λ = 0$

${(λ + 1)}^{2} = 0$

Step 1.7.3

Set $λ$ equal to $0$ .

$λ = 0$

Step 1.7.4

Set ${(λ + 1)}^{2}$ equal to $0$ and solve for $λ$ .

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Step 1.7.4.1

Set ${(λ + 1)}^{2}$ equal to $0$ .

${(λ + 1)}^{2} = 0$

Step 1.7.4.2

Solve ${(λ + 1)}^{2} = 0$ for $λ$ .

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Step 1.7.4.2.1

Set the $λ + 1$ equal to $0$ .

$λ + 1 = 0$

Step 1.7.4.2.2

Subtract $1$ from both sides of the equation.

$λ = - 1$

Step 1.7.5

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

$λ = 0, - 1$

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 $λ = 0$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + 0 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $0$ by each element of the matrix.

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 \cdot 1 & 0 \cdot 0 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $0$ by $1$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 & 0 \cdot 0 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $0$ by $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $0$ by $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 \cdot 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $0$ by $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.5

Multiply $0$ by $1$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.6

Multiply $0$ by $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.7

Multiply $0$ by $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.8

Multiply $0$ by $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.9

Multiply $0$ by $1$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

Step 3.2.2

Adding any matrix to a null matrix is the matrix itself.

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Step 3.2.2.1

Add the corresponding elements.

$[\begin{array}{c} 13 + 0 & 3 + 0 & 1 + 0 \\ - 56 + 0 & - 13 + 0 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 0 \end{array}]$

Step 3.2.2.2

Simplify each element.

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Step 3.2.2.2.1

Add $13$ and $0$ .

$[\begin{array}{c} 13 & 3 + 0 & 1 + 0 \\ - 56 + 0 & - 13 + 0 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 0 \end{array}]$

Step 3.2.2.2.2

Add $3$ and $0$ .

$[\begin{array}{c} 13 & 3 & 1 + 0 \\ - 56 + 0 & - 13 + 0 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 0 \end{array}]$

Step 3.2.2.2.3

Add $1$ and $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 + 0 & - 13 + 0 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 0 \end{array}]$

Step 3.2.2.2.4

Add $- 56$ and $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 + 0 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 0 \end{array}]$

Step 3.2.2.2.5

Add $- 13$ and $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 0 \end{array}]$

Step 3.2.2.2.6

Add $- 4$ and $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 + 0 & - 3 + 0 & - 2 + 0 \end{array}]$

Step 3.2.2.2.7

Add $- 14$ and $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 + 0 & - 2 + 0 \end{array}]$

Step 3.2.2.2.8

Add $- 3$ and $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 + 0 \end{array}]$

Step 3.2.2.2.9

Add $- 2$ and $0$ .

$[\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}]$

Step 3.3

Find the null space when $λ = 0$ .

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Step 3.3.1

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

$[\begin{array}{c} 13 & 3 & 1 & 0 \\ - 56 & - 13 & - 4 & 0 \\ - 14 & - 3 & - 2 & 0 \end{array}]$

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 $\frac{1}{13}$ 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 $\frac{1}{13}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{13}{13} & \frac{3}{13} & \frac{1}{13} & \frac{0}{13} \\ - 56 & - 13 & - 4 & 0 \\ - 14 & - 3 & - 2 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{13} & \frac{1}{13} & 0 \\ - 56 & - 13 & - 4 & 0 \\ - 14 & - 3 & - 2 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation $R_{2} = R_{2} + 56 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 56 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{13} & \frac{1}{13} & 0 \\ - 56 + 56 \cdot 1 & - 13 + 56 (\frac{3}{13}) & - 4 + 56 (\frac{1}{13}) & 0 + 56 \cdot 0 \\ - 14 & - 3 & - 2 & 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{13} & \frac{1}{13} & 0 \\ 0 & - \frac{1}{13} & \frac{4}{13} & 0 \\ - 14 & - 3 & - 2 & 0 \end{array}]$

Step 3.3.2.3

Perform the row operation $R_{3} = R_{3} + 14 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.2.3.1

Perform the row operation $R_{3} = R_{3} + 14 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{13} & \frac{1}{13} & 0 \\ 0 & - \frac{1}{13} & \frac{4}{13} & 0 \\ - 14 + 14 \cdot 1 & - 3 + 14 (\frac{3}{13}) & - 2 + 14 (\frac{1}{13}) & 0 + 14 \cdot 0 \end{array}]$

Step 3.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{13} & \frac{1}{13} & 0 \\ 0 & - \frac{1}{13} & \frac{4}{13} & 0 \\ 0 & \frac{3}{13} & - \frac{12}{13} & 0 \end{array}]$

Step 3.3.2.4

Multiply each element of $R_{2}$ by $- 13$ 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 $- 13$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{13} & \frac{1}{13} & 0 \\ - 13 \cdot 0 & - 13 (- \frac{1}{13}) & - 13 (\frac{4}{13}) & - 13 \cdot 0 \\ 0 & \frac{3}{13} & - \frac{12}{13} & 0 \end{array}]$

Step 3.3.2.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{13} & \frac{1}{13} & 0 \\ 0 & 1 & - 4 & 0 \\ 0 & \frac{3}{13} & - \frac{12}{13} & 0 \end{array}]$

Step 3.3.2.5

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

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Step 3.3.2.5.1

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

$[\begin{array}{c} 1 & \frac{3}{13} & \frac{1}{13} & 0 \\ 0 & 1 & - 4 & 0 \\ 0 - \frac{3}{13} \cdot 0 & \frac{3}{13} - \frac{3}{13} \cdot 1 & - \frac{12}{13} - \frac{3}{13} \cdot - 4 & 0 - \frac{3}{13} \cdot 0 \end{array}]$

Step 3.3.2.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{13} & \frac{1}{13} & 0 \\ 0 & 1 & - 4 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3.3.2.6

Perform the row operation $R_{1} = R_{1} - \frac{3}{13} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.3.2.6.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{13} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{3}{13} \cdot 0 & \frac{3}{13} - \frac{3}{13} \cdot 1 & \frac{1}{13} - \frac{3}{13} \cdot - 4 & 0 - \frac{3}{13} \cdot 0 \\ 0 & 1 & - 4 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 3.3.2.6.2

Simplify $R_{1}$ .

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

Step 3.3.3

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

$x + z = 0$

$y - 4 z = 0$

$0 = 0$

Step 3.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - z \\ 4 z \\ z \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} - 1 \\ 4 \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${z [\begin{array}{c} - 1 \\ 4 \\ 1 \end{array}] | z \in R}$

Step 3.3.7

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

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

Step 4

Find the eigenvector using the eigenvalue $λ = - 1$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 13 & 3 & 1 \\ - 56 & - 13 & - 4 \\ - 14 & - 3 & - 2 \end{array}] + [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 4.2

Simplify.

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Step 4.2.1

Add the corresponding elements.

$[\begin{array}{c} 13 + 1 & 3 + 0 & 1 + 0 \\ - 56 + 0 & - 13 + 1 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 1 \end{array}]$

Step 4.2.2

Simplify each element.

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Step 4.2.2.1

Add $13$ and $1$ .

$[\begin{array}{c} 14 & 3 + 0 & 1 + 0 \\ - 56 + 0 & - 13 + 1 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 1 \end{array}]$

Step 4.2.2.2

Add $3$ and $0$ .

$[\begin{array}{c} 14 & 3 & 1 + 0 \\ - 56 + 0 & - 13 + 1 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 1 \end{array}]$

Step 4.2.2.3

Add $1$ and $0$ .

$[\begin{array}{c} 14 & 3 & 1 \\ - 56 + 0 & - 13 + 1 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 1 \end{array}]$

Step 4.2.2.4

Add $- 56$ and $0$ .

$[\begin{array}{c} 14 & 3 & 1 \\ - 56 & - 13 + 1 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 1 \end{array}]$

Step 4.2.2.5

Add $- 13$ and $1$ .

$[\begin{array}{c} 14 & 3 & 1 \\ - 56 & - 12 & - 4 + 0 \\ - 14 + 0 & - 3 + 0 & - 2 + 1 \end{array}]$

Step 4.2.2.6

Add $- 4$ and $0$ .

$[\begin{array}{c} 14 & 3 & 1 \\ - 56 & - 12 & - 4 \\ - 14 + 0 & - 3 + 0 & - 2 + 1 \end{array}]$

Step 4.2.2.7

Add $- 14$ and $0$ .

$[\begin{array}{c} 14 & 3 & 1 \\ - 56 & - 12 & - 4 \\ - 14 & - 3 + 0 & - 2 + 1 \end{array}]$

Step 4.2.2.8

Add $- 3$ and $0$ .

$[\begin{array}{c} 14 & 3 & 1 \\ - 56 & - 12 & - 4 \\ - 14 & - 3 & - 2 + 1 \end{array}]$

Step 4.2.2.9

Add $- 2$ and $1$ .

$[\begin{array}{c} 14 & 3 & 1 \\ - 56 & - 12 & - 4 \\ - 14 & - 3 & - 1 \end{array}]$

Step 4.3

Find the null space when $λ = - 1$ .

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Step 4.3.1

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

$[\begin{array}{c} 14 & 3 & 1 & 0 \\ - 56 & - 12 & - 4 & 0 \\ - 14 & - 3 & - 1 & 0 \end{array}]$

Step 4.3.2

Find the reduced row echelon form.

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Step 4.3.2.1

Multiply each element of $R_{1}$ by $\frac{1}{14}$ to make the entry at $1, 1$ a $1$ .

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Step 4.3.2.1.1

Multiply each element of $R_{1}$ by $\frac{1}{14}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{14}{14} & \frac{3}{14} & \frac{1}{14} & \frac{0}{14} \\ - 56 & - 12 & - 4 & 0 \\ - 14 & - 3 & - 1 & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{14} & \frac{1}{14} & 0 \\ - 56 & - 12 & - 4 & 0 \\ - 14 & - 3 & - 1 & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation $R_{2} = R_{2} + 56 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.3.2.2.1

Perform the row operation $R_{2} = R_{2} + 56 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{14} & \frac{1}{14} & 0 \\ - 56 + 56 \cdot 1 & - 12 + 56 (\frac{3}{14}) & - 4 + 56 (\frac{1}{14}) & 0 + 56 \cdot 0 \\ - 14 & - 3 & - 1 & 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{14} & \frac{1}{14} & 0 \\ 0 & 0 & 0 & 0 \\ - 14 & - 3 & - 1 & 0 \end{array}]$

Step 4.3.2.3

Perform the row operation $R_{3} = R_{3} + 14 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.3.2.3.1

Perform the row operation $R_{3} = R_{3} + 14 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{14} & \frac{1}{14} & 0 \\ 0 & 0 & 0 & 0 \\ - 14 + 14 \cdot 1 & - 3 + 14 (\frac{3}{14}) & - 1 + 14 (\frac{1}{14}) & 0 + 14 \cdot 0 \end{array}]$

Step 4.3.2.3.2

Simplify $R_{3}$ .

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

Step 4.3.3

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

$x + \frac{3}{14} y + \frac{1}{14} z = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{3 y}{14} - \frac{z}{14} \\ y \\ z \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = y [\begin{array}{c} - \frac{3}{14} \\ 1 \\ 0 \end{array}] + z [\begin{array}{c} - \frac{1}{14} \\ 0 \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${y [\begin{array}{c} - \frac{3}{14} \\ 1 \\ 0 \end{array}] + z [\begin{array}{c} - \frac{1}{14} \\ 0 \\ 1 \end{array}] | y, z \in R}$

Step 4.3.7

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

${[\begin{array}{c} - \frac{3}{14} \\ 1 \\ 0 \end{array}], [\begin{array}{c} - \frac{1}{14} \\ 0 \\ 1 \end{array}]}$

Step 5

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

${[\begin{array}{c} - 1 \\ 4 \\ 1 \end{array}], [\begin{array}{c} - \frac{3}{14} \\ 1 \\ 0 \end{array}], [\begin{array}{c} - \frac{1}{14} \\ 0 \\ 1 \end{array}]}$