Examples

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}]$

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

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}]$ for

$A$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] - λ I_{3})$

Step 1.3.2

Substitute

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

$I_{3}$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8

Multiply

$- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} - 13 - λ & - 8 + 0 & - 4 + 0 \\ 12 + 0 & 7 - λ & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add

$- 8$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 13 - λ & - 8 & - 4 + 0 \\ 12 + 0 & 7 - λ & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - λ \end{array}]$

Step 1.4.3.2

Add

$- 4$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 13 - λ & - 8 & - 4 \\ 12 + 0 & 7 - λ & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - λ \end{array}]$

Step 1.4.3.3

Add

$12$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 13 - λ & - 8 & - 4 \\ 12 & 7 - λ & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - λ \end{array}]$

Step 1.4.3.4

Add

$4$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 13 - λ & - 8 & - 4 \\ 12 & 7 - λ & 4 \\ 24 + 0 & 16 + 0 & 7 - λ \end{array}]$

Step 1.4.3.5

Add

$24$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 13 - λ & - 8 & - 4 \\ 12 & 7 - λ & 4 \\ 24 & 16 + 0 & 7 - λ \end{array}]$

Step 1.4.3.6

Add

$16$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} - 13 - λ & - 8 & - 4 \\ 12 & 7 - λ & 4 \\ 24 & 16 & 7 - λ \end{array}]$

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.

$| \begin{array}{c} 7 - λ & 4 \\ 16 & 7 - λ \end{array} |$

Step 1.5.1.4

Multiply element

$a_{11}$ by its cofactor.

$(- 13 - λ) | \begin{array}{c} 7 - λ & 4 \\ 16 & 7 - λ \end{array} |$

Step 1.5.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} |$

Step 1.5.1.6

Multiply element

$a_{12}$ by its cofactor.

$8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} |$

Step 1.5.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

$| \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.1.8

Multiply element

$a_{13}$ by its cofactor.

$- 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.1.9

Add the terms together.

$p (λ) = (- 13 - λ) | \begin{array}{c} 7 - λ & 4 \\ 16 & 7 - λ \end{array} | + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2

Evaluate

$| \begin{array}{c} 7 - λ & 4 \\ 16 & 7 - λ \end{array} |$ .

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Step 1.5.2.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 13 - λ) ((7 - λ) (7 - λ) - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

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

$(7 - λ) (7 - λ)$ using the FOIL Method.

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Step 1.5.2.2.1.1.1

Apply the distributive property.

$p (λ) = (- 13 - λ) (7 (7 - λ) - λ (7 - λ) - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.1.2

Apply the distributive property.

$p (λ) = (- 13 - λ) (7 \cdot 7 + 7 (- λ) - λ (7 - λ) - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.1.3

Apply the distributive property.

$p (λ) = (- 13 - λ) (7 \cdot 7 + 7 (- λ) - λ \cdot 7 - λ (- λ) - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

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

$7$ by

$7$ .

$p (λ) = (- 13 - λ) (49 + 7 (- λ) - λ \cdot 7 - λ (- λ) - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.2.1.2

Multiply

$- 1$ by

$7$ .

$p (λ) = (- 13 - λ) (49 - 7 λ - λ \cdot 7 - λ (- λ) - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.2.1.3

Multiply

$7$ by

$- 1$ .

$p (λ) = (- 13 - λ) (49 - 7 λ - 7 λ - λ (- λ) - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (- 13 - λ) (49 - 7 λ - 7 λ - 1 \cdot - 1 λ \cdot λ - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

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

$λ$ .

$p (λ) = (- 13 - λ) (49 - 7 λ - 7 λ - 1 \cdot - 1 (λ \cdot λ) - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.2.1.5.2

Multiply

$λ$ by

$λ$ .

$p (λ) = (- 13 - λ) (49 - 7 λ - 7 λ - 1 \cdot - 1 λ^{2} - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.2.1.6

Multiply

$- 1$ by

$- 1$ .

$p (λ) = (- 13 - λ) (49 - 7 λ - 7 λ + 1 λ^{2} - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.2.1.7

Multiply

$λ^{2}$ by

$1$ .

$p (λ) = (- 13 - λ) (49 - 7 λ - 7 λ + λ^{2} - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.2.2

Subtract

$7 λ$ from

$- 7 λ$ .

$p (λ) = (- 13 - λ) (49 - 14 λ + λ^{2} - 16 \cdot 4) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.1.3

Multiply

$- 16$ by

$4$ .

$p (λ) = (- 13 - λ) (49 - 14 λ + λ^{2} - 64) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.2

Subtract

$64$ from

$49$ .

$p (λ) = (- 13 - λ) (- 14 λ + λ^{2} - 15) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.2.2.3

Reorder

$- 14 λ$ and

$λ^{2}$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 | \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} | - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.3

Evaluate

$| \begin{array}{c} 12 & 4 \\ 24 & 7 - λ \end{array} |$ .

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Step 1.5.3.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (12 (7 - λ) - 24 \cdot 4) - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

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.

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (12 \cdot 7 + 12 (- λ) - 24 \cdot 4) - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.3.2.1.2

Multiply

$12$ by

$7$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (84 + 12 (- λ) - 24 \cdot 4) - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.3.2.1.3

Multiply

$- 1$ by

$12$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (84 - 12 λ - 24 \cdot 4) - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.3.2.1.4

Multiply

$- 24$ by

$4$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (84 - 12 λ - 96) - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.3.2.2

Subtract

$96$ from

$84$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (- 12 λ - 12) - 4 | \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$

Step 1.5.4

Evaluate

$| \begin{array}{c} 12 & 7 - λ \\ 24 & 16 \end{array} |$ .

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Step 1.5.4.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (- 12 λ - 12) - 4 (12 \cdot 16 - 24 (7 - λ))$

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

$12$ by

$16$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (- 12 λ - 12) - 4 (192 - 24 (7 - λ))$

Step 1.5.4.2.1.2

Apply the distributive property.

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (- 12 λ - 12) - 4 (192 - 24 \cdot 7 - 24 (- λ))$

Step 1.5.4.2.1.3

Multiply

$- 24$ by

$7$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (- 12 λ - 12) - 4 (192 - 168 - 24 (- λ))$

Step 1.5.4.2.1.4

Multiply

$- 1$ by

$- 24$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (- 12 λ - 12) - 4 (192 - 168 + 24 λ)$

Step 1.5.4.2.2

Subtract

$168$ from

$192$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (- 12 λ - 12) - 4 (24 + 24 λ)$

Step 1.5.4.2.3

Reorder

$24$ and

$24 λ$ .

$p (λ) = (- 13 - λ) (λ^{2} - 14 λ - 15) + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

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

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

$p (λ) = - 13 λ^{2} - 13 (- 14 λ) - 13 \cdot - 15 - λ \cdot λ^{2} - λ (- 14 λ) - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.2

Simplify each term.

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Step 1.5.5.1.2.1

Multiply

$- 14$ by

$- 13$ .

$p (λ) = - 13 λ^{2} + 182 λ - 13 \cdot - 15 - λ \cdot λ^{2} - λ (- 14 λ) - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.2.2

Multiply

$- 13$ by

$- 15$ .

$p (λ) = - 13 λ^{2} + 182 λ + 195 - λ \cdot λ^{2} - λ (- 14 λ) - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

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

$λ^{2}$ .

$p (λ) = - 13 λ^{2} + 182 λ + 195 - (λ^{2} λ) - λ (- 14 λ) - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

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

$1$ .

$p (λ) = - 13 λ^{2} + 182 λ + 195 - (λ^{2} λ^{1}) - λ (- 14 λ) - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.2.3.2.2

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$p (λ) = - 13 λ^{2} + 182 λ + 195 - λ^{2 + 1} - λ (- 14 λ) - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.2.3.3

Add

$2$ and

$1$ .

$p (λ) = - 13 λ^{2} + 182 λ + 195 - λ^{3} - λ (- 14 λ) - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = - 13 λ^{2} + 182 λ + 195 - λ^{3} - 1 \cdot - 14 λ \cdot λ - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

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

$λ$ .

$p (λ) = - 13 λ^{2} + 182 λ + 195 - λ^{3} - 1 \cdot - 14 (λ \cdot λ) - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.2.5.2

Multiply

$λ$ by

$λ$ .

$p (λ) = - 13 λ^{2} + 182 λ + 195 - λ^{3} - 1 \cdot - 14 λ^{2} - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.2.6

Multiply

$- 1$ by

$- 14$ .

$p (λ) = - 13 λ^{2} + 182 λ + 195 - λ^{3} + 14 λ^{2} - λ \cdot - 15 + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.2.7

Multiply

$- 15$ by

$- 1$ .

$p (λ) = - 13 λ^{2} + 182 λ + 195 - λ^{3} + 14 λ^{2} + 15 λ + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.3

Add

$- 13 λ^{2}$ and

$14 λ^{2}$ .

$p (λ) = λ^{2} + 182 λ + 195 - λ^{3} + 15 λ + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.4

Add

$182 λ$ and

$15 λ$ .

$p (λ) = λ^{2} + 197 λ + 195 - λ^{3} + 8 (- 12 λ - 12) - 4 (24 λ + 24)$

Step 1.5.5.1.5

Apply the distributive property.

$p (λ) = λ^{2} + 197 λ + 195 - λ^{3} + 8 (- 12 λ) + 8 \cdot - 12 - 4 (24 λ + 24)$

Step 1.5.5.1.6

Multiply

$- 12$ by

$8$ .

$p (λ) = λ^{2} + 197 λ + 195 - λ^{3} - 96 λ + 8 \cdot - 12 - 4 (24 λ + 24)$

Step 1.5.5.1.7

Multiply

$8$ by

$- 12$ .

$p (λ) = λ^{2} + 197 λ + 195 - λ^{3} - 96 λ - 96 - 4 (24 λ + 24)$

Step 1.5.5.1.8

Apply the distributive property.

$p (λ) = λ^{2} + 197 λ + 195 - λ^{3} - 96 λ - 96 - 4 (24 λ) - 4 \cdot 24$

Step 1.5.5.1.9

Multiply

$24$ by

$- 4$ .

$p (λ) = λ^{2} + 197 λ + 195 - λ^{3} - 96 λ - 96 - 96 λ - 4 \cdot 24$

Step 1.5.5.1.10

Multiply

$- 4$ by

$24$ .

$p (λ) = λ^{2} + 197 λ + 195 - λ^{3} - 96 λ - 96 - 96 λ - 96$

Step 1.5.5.2

Subtract

$96 λ$ from

$197 λ$ .

$p (λ) = λ^{2} + 101 λ + 195 - λ^{3} - 96 - 96 λ - 96$

Step 1.5.5.3

Subtract

$96 λ$ from

$101 λ$ .

$p (λ) = λ^{2} + 5 λ + 195 - λ^{3} - 96 - 96$

Step 1.5.5.4

Subtract

$96$ from

$195$ .

$p (λ) = λ^{2} + 5 λ - λ^{3} + 99 - 96$

Step 1.5.5.5

Subtract

$96$ from

$99$ .

$p (λ) = λ^{2} + 5 λ - λ^{3} + 3$

Step 1.5.5.6

Move

$5 λ$ .

$p (λ) = λ^{2} - λ^{3} + 5 λ + 3$

Step 1.5.5.7

Reorder

$λ^{2}$ and

$- λ^{3}$ .

$p (λ) = - λ^{3} + λ^{2} + 5 λ + 3$

Step 1.6

Set the characteristic polynomial equal to

$0$ to find the eigenvalues

$λ$ .

$- λ^{3} + λ^{2} + 5 λ + 3 = 0$

Step 1.7

Solve for

$λ$ .

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Step 1.7.1

Factor the left side of the equation.

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Step 1.7.1.1

Factor

$- λ^{3} + λ^{2} + 5 λ + 3$ using the rational roots test.

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Step 1.7.1.1.1

If a polynomial function has integer coefficients, then every rational zero will have the form

$\frac{p}{q}$ where

$p$ is a factor of the constant and

$q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 3$

$q = \pm 1$

Step 1.7.1.1.2

Find every combination of

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

$\pm 1, \pm 3$

Step 1.7.1.1.3

Substitute

$- 1$ and simplify the expression. In this case, the expression is equal to

$0$ so

$- 1$ is a root of the polynomial.

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Step 1.7.1.1.3.1

Substitute

$- 1$ into the polynomial.

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

Step 1.7.1.1.3.2

Raise

$- 1$ to the power of

$3$ .

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

Step 1.7.1.1.3.3

Multiply

$- 1$ by

$- 1$ .

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

Step 1.7.1.1.3.4

Raise

$- 1$ to the power of

$2$ .

$1 + 1 + 5 \cdot - 1 + 3$

Step 1.7.1.1.3.5

Add

$1$ and

$1$ .

$2 + 5 \cdot - 1 + 3$

Step 1.7.1.1.3.6

Multiply

$5$ by

$- 1$ .

$2 - 5 + 3$

Step 1.7.1.1.3.7

Subtract

$5$ from

$2$ .

$- 3 + 3$

Step 1.7.1.1.3.8

Add

$- 3$ and

$3$ .

$0$

Step 1.7.1.1.4

Since

$- 1$ is a known root, divide the polynomial by

$λ + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- λ^{3} + λ^{2} + 5 λ + 3}{λ + 1}$

Step 1.7.1.1.5

Divide

$- λ^{3} + λ^{2} + 5 λ + 3$ by

$λ + 1$ .

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Step 1.7.1.1.5.1

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

$0$ .

$λ$

$1$

$λ^{3}$

$λ^{2}$

$5 λ$

$3$

Step 1.7.1.1.5.2

Divide the highest order term in the dividend

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

$λ$ .

				-	$λ^{2}$
	$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$

Step 1.7.1.1.5.3

Multiply the new quotient term by the divisor.

			-	$λ^{2}$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			-	$λ^{3}$	-	$λ^{2}$

Step 1.7.1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in

$- λ^{3} - λ^{2}$

			-	$λ^{2}$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$

Step 1.7.1.1.5.5

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

			-	$λ^{2}$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$

Step 1.7.1.1.5.6

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

			-	$λ^{2}$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$

Step 1.7.1.1.5.7

Divide the highest order term in the dividend

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

$λ$ .

			-	$λ^{2}$	+	$2 λ$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$

Step 1.7.1.1.5.8

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$2 λ$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$
					+	$2 λ^{2}$	+	$2 λ$

Step 1.7.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in

$2 λ^{2} + 2 λ$

			-	$λ^{2}$	+	$2 λ$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$
					-	$2 λ^{2}$	-	$2 λ$

Step 1.7.1.1.5.10

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

			-	$λ^{2}$	+	$2 λ$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$
					-	$2 λ^{2}$	-	$2 λ$
							+	$3 λ$

Step 1.7.1.1.5.11

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

			-	$λ^{2}$	+	$2 λ$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$
					-	$2 λ^{2}$	-	$2 λ$
							+	$3 λ$	+	$3$

Step 1.7.1.1.5.12

Divide the highest order term in the dividend

$3 λ$ by the highest order term in divisor

$λ$ .

			-	$λ^{2}$	+	$2 λ$	+	$3$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$
					-	$2 λ^{2}$	-	$2 λ$
							+	$3 λ$	+	$3$

Step 1.7.1.1.5.13

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$2 λ$	+	$3$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$
					-	$2 λ^{2}$	-	$2 λ$
							+	$3 λ$	+	$3$
							+	$3 λ$	+	$3$

Step 1.7.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in

$3 λ + 3$

			-	$λ^{2}$	+	$2 λ$	+	$3$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$
					-	$2 λ^{2}$	-	$2 λ$
							+	$3 λ$	+	$3$
							-	$3 λ$	-	$3$

Step 1.7.1.1.5.15

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

			-	$λ^{2}$	+	$2 λ$	+	$3$
$λ$	+	$1$	-	$λ^{3}$	+	$λ^{2}$	+	$5 λ$	+	$3$
			+	$λ^{3}$	+	$λ^{2}$
					+	$2 λ^{2}$	+	$5 λ$
					-	$2 λ^{2}$	-	$2 λ$
							+	$3 λ$	+	$3$
							-	$3 λ$	-	$3$
										$0$

Step 1.7.1.1.5.16

Since the remander is

$0$ , the final answer is the quotient.

$- λ^{2} + 2 λ + 3$

Step 1.7.1.1.6

Write

$- λ^{3} + λ^{2} + 5 λ + 3$ as a set of factors.

$(λ + 1) (- λ^{2} + 2 λ + 3) = 0$

Step 1.7.1.2

Factor by grouping.

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Step 1.7.1.2.1

Factor by grouping.

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Step 1.7.1.2.1.1

For a polynomial of the form

$a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is

$a \cdot c = - 1 \cdot 3 = - 3$ and whose sum is

$b = 2$ .

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Step 1.7.1.2.1.1.1

Factor

$2$ out of

$2 λ$ .

$(λ + 1) (- λ^{2} + 2 (λ) + 3) = 0$

Step 1.7.1.2.1.1.2

Rewrite

$2$ as

$- 1$ plus

$3$

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

Step 1.7.1.2.1.1.3

Apply the distributive property.

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

Step 1.7.1.2.1.2

Factor out the greatest common factor from each group.

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Step 1.7.1.2.1.2.1

Group the first two terms and the last two terms.

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

Step 1.7.1.2.1.2.2

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

$(λ + 1) (λ (- λ - 1) - 3 (- λ - 1)) = 0$

Step 1.7.1.2.1.3

Factor the polynomial by factoring out the greatest common factor,

$- λ - 1$ .

$(λ + 1) ((- λ - 1) (λ - 3)) = 0$

Step 1.7.1.2.2

Remove unnecessary parentheses.

$(λ + 1) (- λ - 1) (λ - 3) = 0$

Step 1.7.2

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$λ + 1 = 0$

$- λ - 1 = 0$

$λ - 3 = 0$

Step 1.7.3

Set

$λ + 1$ equal to

$0$ and solve for

$λ$ .

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Step 1.7.3.1

Set

$λ + 1$ equal to

$0$ .

$λ + 1 = 0$

Step 1.7.3.2

Subtract

$1$ from both sides of the equation.

$λ = - 1$

Step 1.7.4

Set

$λ - 3$ equal to

$0$ and solve for

$λ$ .

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Step 1.7.4.1

Set

$λ - 3$ equal to

$0$ .

$λ - 3 = 0$

Step 1.7.4.2

Add

$3$ to both sides of the equation.

$λ = 3$

Step 1.7.5

The final solution is all the values that make

$(λ + 1) (- λ - 1) (λ - 3) = 0$ true.

$λ = - 1, 3$

Step 2

The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where

$N$ is the null space and

$I$ is the identity matrix.

$ε_{A} = N (A - λ I_{3})$

Step 3

Find the eigenvector using the eigenvalue

$λ = - 1$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 3.2

Simplify.

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Step 3.2.1

Add the corresponding elements.

$[\begin{array}{c} - 13 + 1 & - 8 + 0 & - 4 + 0 \\ 12 + 0 & 7 + 1 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 + 1 \end{array}]$

Step 3.2.2

Simplify each element.

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Step 3.2.2.1

Add

$- 13$ and

$1$ .

$[\begin{array}{c} - 12 & - 8 + 0 & - 4 + 0 \\ 12 + 0 & 7 + 1 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 + 1 \end{array}]$

Step 3.2.2.2

Add

$- 8$ and

$0$ .

$[\begin{array}{c} - 12 & - 8 & - 4 + 0 \\ 12 + 0 & 7 + 1 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 + 1 \end{array}]$

Step 3.2.2.3

Add

$- 4$ and

$0$ .

$[\begin{array}{c} - 12 & - 8 & - 4 \\ 12 + 0 & 7 + 1 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 + 1 \end{array}]$

Step 3.2.2.4

Add

$12$ and

$0$ .

$[\begin{array}{c} - 12 & - 8 & - 4 \\ 12 & 7 + 1 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 + 1 \end{array}]$

Step 3.2.2.5

Add

$7$ and

$1$ .

$[\begin{array}{c} - 12 & - 8 & - 4 \\ 12 & 8 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 + 1 \end{array}]$

Step 3.2.2.6

Add

$4$ and

$0$ .

$[\begin{array}{c} - 12 & - 8 & - 4 \\ 12 & 8 & 4 \\ 24 + 0 & 16 + 0 & 7 + 1 \end{array}]$

Step 3.2.2.7

Add

$24$ and

$0$ .

$[\begin{array}{c} - 12 & - 8 & - 4 \\ 12 & 8 & 4 \\ 24 & 16 + 0 & 7 + 1 \end{array}]$

Step 3.2.2.8

Add

$16$ and

$0$ .

$[\begin{array}{c} - 12 & - 8 & - 4 \\ 12 & 8 & 4 \\ 24 & 16 & 7 + 1 \end{array}]$

Step 3.2.2.9

Add

$7$ and

$1$ .

$[\begin{array}{c} - 12 & - 8 & - 4 \\ 12 & 8 & 4 \\ 24 & 16 & 8 \end{array}]$

Step 3.3

Find the null space when

$λ = - 1$ .

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Step 3.3.1

Write as an augmented matrix for

$A x = 0$ .

$[\begin{array}{c} - 12 & - 8 & - 4 & 0 \\ 12 & 8 & 4 & 0 \\ 24 & 16 & 8 & 0 \end{array}]$

Step 3.3.2

Find the reduced row echelon form.

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Step 3.3.2.1

Multiply each element of

$R_{1}$ by

$- \frac{1}{12}$ 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

$- \frac{1}{12}$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} - \frac{1}{12} \cdot - 12 & - \frac{1}{12} \cdot - 8 & - \frac{1}{12} \cdot - 4 & - \frac{1}{12} \cdot 0 \\ 12 & 8 & 4 & 0 \\ 24 & 16 & 8 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{1}{3} & 0 \\ 12 & 8 & 4 & 0 \\ 24 & 16 & 8 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation

$R_{2} = R_{2} - 12 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 3.3.2.2.1

Perform the row operation

$R_{2} = R_{2} - 12 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{1}{3} & 0 \\ 12 - 12 \cdot 1 & 8 - 12 (\frac{2}{3}) & 4 - 12 (\frac{1}{3}) & 0 - 12 \cdot 0 \\ 24 & 16 & 8 & 0 \end{array}]$

Step 3.3.2.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 0 & 0 \\ 24 & 16 & 8 & 0 \end{array}]$

Step 3.3.2.3

Perform the row operation

$R_{3} = R_{3} - 24 R_{1}$ to make the entry at

$3, 1$ a

$0$ .

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Step 3.3.2.3.1

Perform the row operation

$R_{3} = R_{3} - 24 R_{1}$ to make the entry at

$3, 1$ a

$0$ .

$[\begin{array}{c} 1 & \frac{2}{3} & \frac{1}{3} & 0 \\ 0 & 0 & 0 & 0 \\ 24 - 24 \cdot 1 & 16 - 24 (\frac{2}{3}) & 8 - 24 (\frac{1}{3}) & 0 - 24 \cdot 0 \end{array}]$

Step 3.3.2.3.2

Simplify

$R_{3}$ .

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

Step 3.3.3

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

$x + \frac{2}{3} y + \frac{1}{3} z = 0$

$0 = 0$

Step 3.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} - \frac{2 y}{3} - \frac{z}{3} \\ y \\ z \end{array}]$

Step 3.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = y [\begin{array}{c} - \frac{2}{3} \\ 1 \\ 0 \end{array}] + z [\begin{array}{c} - \frac{1}{3} \\ 0 \\ 1 \end{array}]$

Step 3.3.6

Write as a solution set.

${y [\begin{array}{c} - \frac{2}{3} \\ 1 \\ 0 \end{array}] + z [\begin{array}{c} - \frac{1}{3} \\ 0 \\ 1 \end{array}] | y, z \in R}$

Step 3.3.7

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

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

Step 4

Find the eigenvector using the eigenvalue

$λ = 3$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] - 3 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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

$- 3$ by each element of the matrix.

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 \cdot 1 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply

$- 3$ by

$1$ .

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 & - 3 \cdot 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply

$- 3$ by

$0$ .

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 & 0 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply

$- 3$ by

$0$ .

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 \\ - 3 \cdot 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply

$- 3$ by

$0$ .

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 \\ 0 & - 3 \cdot 1 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.5

Multiply

$- 3$ by

$1$ .

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 \\ 0 & - 3 & - 3 \cdot 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.6

Multiply

$- 3$ by

$0$ .

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 \\ 0 & - 3 & 0 \\ - 3 \cdot 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.7

Multiply

$- 3$ by

$0$ .

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & - 3 \cdot 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.8

Multiply

$- 3$ by

$0$ .

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & - 3 \cdot 1 \end{array}]$

Step 4.2.1.2.9

Multiply

$- 3$ by

$1$ .

$[\begin{array}{c} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{array}] + [\begin{array}{c} - 3 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & - 3 \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} - 13 - 3 & - 8 + 0 & - 4 + 0 \\ 12 + 0 & 7 - 3 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - 3 \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Subtract

$3$ from

$- 13$ .

$[\begin{array}{c} - 16 & - 8 + 0 & - 4 + 0 \\ 12 + 0 & 7 - 3 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - 3 \end{array}]$

Step 4.2.3.2

Add

$- 8$ and

$0$ .

$[\begin{array}{c} - 16 & - 8 & - 4 + 0 \\ 12 + 0 & 7 - 3 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - 3 \end{array}]$

Step 4.2.3.3

Add

$- 4$ and

$0$ .

$[\begin{array}{c} - 16 & - 8 & - 4 \\ 12 + 0 & 7 - 3 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - 3 \end{array}]$

Step 4.2.3.4

Add

$12$ and

$0$ .

$[\begin{array}{c} - 16 & - 8 & - 4 \\ 12 & 7 - 3 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - 3 \end{array}]$

Step 4.2.3.5

Subtract

$3$ from

$7$ .

$[\begin{array}{c} - 16 & - 8 & - 4 \\ 12 & 4 & 4 + 0 \\ 24 + 0 & 16 + 0 & 7 - 3 \end{array}]$

Step 4.2.3.6

Add

$4$ and

$0$ .

$[\begin{array}{c} - 16 & - 8 & - 4 \\ 12 & 4 & 4 \\ 24 + 0 & 16 + 0 & 7 - 3 \end{array}]$

Step 4.2.3.7

Add

$24$ and

$0$ .

$[\begin{array}{c} - 16 & - 8 & - 4 \\ 12 & 4 & 4 \\ 24 & 16 + 0 & 7 - 3 \end{array}]$

Step 4.2.3.8

Add

$16$ and

$0$ .

$[\begin{array}{c} - 16 & - 8 & - 4 \\ 12 & 4 & 4 \\ 24 & 16 & 7 - 3 \end{array}]$

Step 4.2.3.9

Subtract

$3$ from

$7$ .

$[\begin{array}{c} - 16 & - 8 & - 4 \\ 12 & 4 & 4 \\ 24 & 16 & 4 \end{array}]$

Step 4.3

Find the null space when

$λ = 3$ .

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Step 4.3.1

Write as an augmented matrix for

$A x = 0$ .

$[\begin{array}{c} - 16 & - 8 & - 4 & 0 \\ 12 & 4 & 4 & 0 \\ 24 & 16 & 4 & 0 \end{array}]$

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

$- \frac{1}{16}$ 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

$- \frac{1}{16}$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} - \frac{1}{16} \cdot - 16 & - \frac{1}{16} \cdot - 8 & - \frac{1}{16} \cdot - 4 & - \frac{1}{16} \cdot 0 \\ 12 & 4 & 4 & 0 \\ 24 & 16 & 4 & 0 \end{array}]$

Step 4.3.2.1.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{4} & 0 \\ 12 & 4 & 4 & 0 \\ 24 & 16 & 4 & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation

$R_{2} = R_{2} - 12 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

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Step 4.3.2.2.1

Perform the row operation

$R_{2} = R_{2} - 12 R_{1}$ to make the entry at

$2, 1$ a

$0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{4} & 0 \\ 12 - 12 \cdot 1 & 4 - 12 (\frac{1}{2}) & 4 - 12 (\frac{1}{4}) & 0 - 12 \cdot 0 \\ 24 & 16 & 4 & 0 \end{array}]$

Step 4.3.2.2.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{4} & 0 \\ 0 & - 2 & 1 & 0 \\ 24 & 16 & 4 & 0 \end{array}]$

Step 4.3.2.3

Perform the row operation

$R_{3} = R_{3} - 24 R_{1}$ to make the entry at

$3, 1$ a

$0$ .

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Step 4.3.2.3.1

Perform the row operation

$R_{3} = R_{3} - 24 R_{1}$ to make the entry at

$3, 1$ a

$0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{4} & 0 \\ 0 & - 2 & 1 & 0 \\ 24 - 24 \cdot 1 & 16 - 24 (\frac{1}{2}) & 4 - 24 (\frac{1}{4}) & 0 - 24 \cdot 0 \end{array}]$

Step 4.3.2.3.2

Simplify

$R_{3}$ .

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

Step 4.3.2.4

Multiply each element of

$R_{2}$ by

$- \frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

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Step 4.3.2.4.1

Multiply each element of

$R_{2}$ by

$- \frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

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

Step 4.3.2.4.2

Simplify

$R_{2}$ .

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

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

$0$ .

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

Step 4.3.2.5.2

Simplify

$R_{3}$ .

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

Step 4.3.2.6

Perform the row operation

$R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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Step 4.3.2.6.1

Perform the row operation

$R_{1} = R_{1} - \frac{1}{2} R_{2}$ to make the entry at

$1, 2$ a

$0$ .

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

Step 4.3.2.6.2

Simplify

$R_{1}$ .

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

Step 4.3.3

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

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

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

$0 = 0$

Step 4.3.4

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

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

Step 4.3.5

Write the solution as a linear combination of vectors.

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

Step 4.3.6

Write as a solution set.

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

Step 4.3.7

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

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

Step 5

The eigenspace of

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

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

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