Linear Algebra Examples

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

Step 1

Find the eigenvectors.

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Step 1.1

Find the eigenvalues.

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Step 1.1.1

Set up the formula to find the characteristic equation $p (λ)$ .

$p (λ) = determinant (A - λ I_{3})$

Step 1.1.2

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 1.1.3

Substitute the known values into $p (λ) = determinant (A - λ I_{3})$ .

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Step 1.1.3.1

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

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

Step 1.1.3.2

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

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

Step 1.1.4

Simplify.

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Step 1.1.4.1

Simplify each term.

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Step 1.1.4.1.1

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

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

Step 1.1.4.1.2

Simplify each element in the matrix.

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Step 1.1.4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 1.1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.1.4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 1.1.4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 1.1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.1.4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 1.1.4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 1.1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.1.4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 1.1.4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 1.1.4.1.2.5

Multiply $- 1$ by $1$ .

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

Step 1.1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.1.4.1.2.6.1

Multiply $0$ by $- 1$ .

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

Step 1.1.4.1.2.6.2

Multiply $0$ by $λ$ .

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

Step 1.1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.1.4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 1.1.4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 1.1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.1.4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 1.1.4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 1.1.4.1.2.9

Multiply $- 1$ by $1$ .

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

Step 1.1.4.2

Add the corresponding elements.

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

Step 1.1.4.3

Simplify each element.

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Step 1.1.4.3.1

Add $2$ and $0$ .

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

Step 1.1.4.3.2

Add $0$ and $0$ .

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

Step 1.1.4.3.3

Add $2$ and $0$ .

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

Step 1.1.4.3.4

Add $0$ and $0$ .

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

Step 1.1.4.3.5

Add $4$ and $0$ .

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

Step 1.1.4.3.6

Add $- 1$ and $0$ .

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

Step 1.1.5

Find the determinant.

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Step 1.1.5.1

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 column $3$ by its cofactor and add.

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Step 1.1.5.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 1.1.5.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 1.1.5.1.3

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

$| \begin{array}{c} 2 & 5 - λ \\ 4 & - 1 \end{array} |$

Step 1.1.5.1.4

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

$0 | \begin{array}{c} 2 & 5 - λ \\ 4 & - 1 \end{array} |$

Step 1.1.5.1.5

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

$| \begin{array}{c} 5 - λ & 2 \\ 4 & - 1 \end{array} |$

Step 1.1.5.1.6

Multiply element $a_{23}$ by its cofactor.

$0 | \begin{array}{c} 5 - λ & 2 \\ 4 & - 1 \end{array} |$

Step 1.1.5.1.7

The minor for $a_{33}$ is the determinant with row $3$ and column $3$ deleted.

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

Step 1.1.5.1.8

Multiply element $a_{33}$ by its cofactor.

$(4 - λ) | \begin{array}{c} 5 - λ & 2 \\ 2 & 5 - λ \end{array} |$

Step 1.1.5.1.9

Add the terms together.

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

Step 1.1.5.2

Multiply $0$ by $| \begin{array}{c} 2 & 5 - λ \\ 4 & - 1 \end{array} |$ .

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

Step 1.1.5.3

Multiply $0$ by $| \begin{array}{c} 5 - λ & 2 \\ 4 & - 1 \end{array} |$ .

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

Step 1.1.5.4

Evaluate $| \begin{array}{c} 5 - λ & 2 \\ 2 & 5 - λ \end{array} |$ .

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Step 1.1.5.4.1

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

$p (λ) = 0 + 0 + (4 - λ) ((5 - λ) (5 - λ) - 2 \cdot 2)$

Step 1.1.5.4.2

Simplify the determinant.

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Step 1.1.5.4.2.1

Simplify each term.

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Step 1.1.5.4.2.1.1

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

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Step 1.1.5.4.2.1.1.1

Apply the distributive property.

$p (λ) = 0 + 0 + (4 - λ) (5 (5 - λ) - λ (5 - λ) - 2 \cdot 2)$

Step 1.1.5.4.2.1.1.2

Apply the distributive property.

$p (λ) = 0 + 0 + (4 - λ) (5 \cdot 5 + 5 (- λ) - λ (5 - λ) - 2 \cdot 2)$

Step 1.1.5.4.2.1.1.3

Apply the distributive property.

$p (λ) = 0 + 0 + (4 - λ) (5 \cdot 5 + 5 (- λ) - λ \cdot 5 - λ (- λ) - 2 \cdot 2)$

Step 1.1.5.4.2.1.2

Simplify and combine like terms.

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Step 1.1.5.4.2.1.2.1

Simplify each term.

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Step 1.1.5.4.2.1.2.1.1

Multiply $5$ by $5$ .

$p (λ) = 0 + 0 + (4 - λ) (25 + 5 (- λ) - λ \cdot 5 - λ (- λ) - 2 \cdot 2)$

Step 1.1.5.4.2.1.2.1.2

Multiply $- 1$ by $5$ .

$p (λ) = 0 + 0 + (4 - λ) (25 - 5 λ - λ \cdot 5 - λ (- λ) - 2 \cdot 2)$

Step 1.1.5.4.2.1.2.1.3

Multiply $5$ by $- 1$ .

$p (λ) = 0 + 0 + (4 - λ) (25 - 5 λ - 5 λ - λ (- λ) - 2 \cdot 2)$

Step 1.1.5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0 + 0 + (4 - λ) (25 - 5 λ - 5 λ - 1 \cdot - 1 λ \cdot λ - 2 \cdot 2)$

Step 1.1.5.4.2.1.2.1.5

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

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Step 1.1.5.4.2.1.2.1.5.1

Move $λ$ .

$p (λ) = 0 + 0 + (4 - λ) (25 - 5 λ - 5 λ - 1 \cdot - 1 (λ \cdot λ) - 2 \cdot 2)$

Step 1.1.5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 + 0 + (4 - λ) (25 - 5 λ - 5 λ - 1 \cdot - 1 λ^{2} - 2 \cdot 2)$

Step 1.1.5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 0 + 0 + (4 - λ) (25 - 5 λ - 5 λ + 1 λ^{2} - 2 \cdot 2)$

Step 1.1.5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0 + 0 + (4 - λ) (25 - 5 λ - 5 λ + λ^{2} - 2 \cdot 2)$

Step 1.1.5.4.2.1.2.2

Subtract $5 λ$ from $- 5 λ$ .

$p (λ) = 0 + 0 + (4 - λ) (25 - 10 λ + λ^{2} - 2 \cdot 2)$

Step 1.1.5.4.2.1.3

Multiply $- 2$ by $2$ .

$p (λ) = 0 + 0 + (4 - λ) (25 - 10 λ + λ^{2} - 4)$

Step 1.1.5.4.2.2

Subtract $4$ from $25$ .

$p (λ) = 0 + 0 + (4 - λ) (- 10 λ + λ^{2} + 21)$

Step 1.1.5.4.2.3

Reorder $- 10 λ$ and $λ^{2}$ .

$p (λ) = 0 + 0 + (4 - λ) (λ^{2} - 10 λ + 21)$

Step 1.1.5.5

Simplify the determinant.

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Step 1.1.5.5.1

Combine the opposite terms in $0 + 0 + (4 - λ) (λ^{2} - 10 λ + 21)$ .

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Step 1.1.5.5.1.1

Add $0$ and $0$ .

$p (λ) = 0 + (4 - λ) (λ^{2} - 10 λ + 21)$

Step 1.1.5.5.1.2

Add $0$ and $(4 - λ) (λ^{2} - 10 λ + 21)$ .

$p (λ) = (4 - λ) (λ^{2} - 10 λ + 21)$

Step 1.1.5.5.2

Expand $(4 - λ) (λ^{2} - 10 λ + 21)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 4 λ^{2} + 4 (- 10 λ) + 4 \cdot 21 - λ \cdot λ^{2} - λ (- 10 λ) - λ \cdot 21$

Step 1.1.5.5.3

Simplify each term.

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Step 1.1.5.5.3.1

Multiply $- 10$ by $4$ .

$p (λ) = 4 λ^{2} - 40 λ + 4 \cdot 21 - λ \cdot λ^{2} - λ (- 10 λ) - λ \cdot 21$

Step 1.1.5.5.3.2

Multiply $4$ by $21$ .

$p (λ) = 4 λ^{2} - 40 λ + 84 - λ \cdot λ^{2} - λ (- 10 λ) - λ \cdot 21$

Step 1.1.5.5.3.3

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

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Step 1.1.5.5.3.3.1

Move $λ^{2}$ .

$p (λ) = 4 λ^{2} - 40 λ + 84 - (λ^{2} λ) - λ (- 10 λ) - λ \cdot 21$

Step 1.1.5.5.3.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.1.5.5.3.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 4 λ^{2} - 40 λ + 84 - (λ^{2} λ^{1}) - λ (- 10 λ) - λ \cdot 21$

Step 1.1.5.5.3.3.2.2

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

$p (λ) = 4 λ^{2} - 40 λ + 84 - λ^{2 + 1} - λ (- 10 λ) - λ \cdot 21$

Step 1.1.5.5.3.3.3

Add $2$ and $1$ .

$p (λ) = 4 λ^{2} - 40 λ + 84 - λ^{3} - λ (- 10 λ) - λ \cdot 21$

Step 1.1.5.5.3.4

Rewrite using the commutative property of multiplication.

$p (λ) = 4 λ^{2} - 40 λ + 84 - λ^{3} - 1 \cdot - 10 λ \cdot λ - λ \cdot 21$

Step 1.1.5.5.3.5

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

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Step 1.1.5.5.3.5.1

Move $λ$ .

$p (λ) = 4 λ^{2} - 40 λ + 84 - λ^{3} - 1 \cdot - 10 (λ \cdot λ) - λ \cdot 21$

Step 1.1.5.5.3.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 4 λ^{2} - 40 λ + 84 - λ^{3} - 1 \cdot - 10 λ^{2} - λ \cdot 21$

Step 1.1.5.5.3.6

Multiply $- 1$ by $- 10$ .

$p (λ) = 4 λ^{2} - 40 λ + 84 - λ^{3} + 10 λ^{2} - λ \cdot 21$

Step 1.1.5.5.3.7

Multiply $21$ by $- 1$ .

$p (λ) = 4 λ^{2} - 40 λ + 84 - λ^{3} + 10 λ^{2} - 21 λ$

Step 1.1.5.5.4

Add $4 λ^{2}$ and $10 λ^{2}$ .

$p (λ) = 14 λ^{2} - 40 λ + 84 - λ^{3} - 21 λ$

Step 1.1.5.5.5

Subtract $21 λ$ from $- 40 λ$ .

$p (λ) = 14 λ^{2} - 61 λ + 84 - λ^{3}$

Step 1.1.5.5.6

Move $84$ .

$p (λ) = 14 λ^{2} - 61 λ - λ^{3} + 84$

Step 1.1.5.5.7

Move $- 61 λ$ .

$p (λ) = 14 λ^{2} - λ^{3} - 61 λ + 84$

Step 1.1.5.5.8

Reorder $14 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 14 λ^{2} - 61 λ + 84$

Step 1.1.6

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

$- λ^{3} + 14 λ^{2} - 61 λ + 84 = 0$

Step 1.1.7

Solve for $λ$ .

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Step 1.1.7.1

Factor the left side of the equation.

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Step 1.1.7.1.1

Factor $- λ^{3} + 14 λ^{2} - 61 λ + 84$ using the rational roots test.

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Step 1.1.7.1.1.1

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.

$p = \pm 1, \pm 84, \pm 2, \pm 42, \pm 3, \pm 28, \pm 4, \pm 21, \pm 6, \pm 14, \pm 7, \pm 12$

$q = \pm 1$

Step 1.1.7.1.1.2

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

$\pm 1, \pm 84, \pm 2, \pm 42, \pm 3, \pm 28, \pm 4, \pm 21, \pm 6, \pm 14, \pm 7, \pm 12$

Step 1.1.7.1.1.3

Substitute $3$ and simplify the expression. In this case, the expression is equal to $0$ so $3$ is a root of the polynomial.

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Step 1.1.7.1.1.3.1

Substitute $3$ into the polynomial.

$- 3^{3} + 14 \cdot 3^{2} - 61 \cdot 3 + 84$

Step 1.1.7.1.1.3.2

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

$- 1 \cdot 27 + 14 \cdot 3^{2} - 61 \cdot 3 + 84$

Step 1.1.7.1.1.3.3

Multiply $- 1$ by $27$ .

$- 27 + 14 \cdot 3^{2} - 61 \cdot 3 + 84$

Step 1.1.7.1.1.3.4

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

$- 27 + 14 \cdot 9 - 61 \cdot 3 + 84$

Step 1.1.7.1.1.3.5

Multiply $14$ by $9$ .

$- 27 + 126 - 61 \cdot 3 + 84$

Step 1.1.7.1.1.3.6

Add $- 27$ and $126$ .

$99 - 61 \cdot 3 + 84$

Step 1.1.7.1.1.3.7

Multiply $- 61$ by $3$ .

$99 - 183 + 84$

Step 1.1.7.1.1.3.8

Subtract $183$ from $99$ .

$- 84 + 84$

Step 1.1.7.1.1.3.9

Add $- 84$ and $84$ .

$0$

Step 1.1.7.1.1.4

Since $3$ is a known root, divide the polynomial by $λ - 3$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- λ^{3} + 14 λ^{2} - 61 λ + 84}{λ - 3}$

Step 1.1.7.1.1.5

Divide $- λ^{3} + 14 λ^{2} - 61 λ + 84$ by $λ - 3$ .

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Step 1.1.7.1.1.5.1

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

$λ$

$3$

$λ^{3}$

$14 λ^{2}$

$61 λ$

$84$

Step 1.1.7.1.1.5.2

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

				-	$λ^{2}$
	$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$

Step 1.1.7.1.1.5.3

Multiply the new quotient term by the divisor.

			-	$λ^{2}$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			-	$λ^{3}$	+	$3 λ^{2}$

Step 1.1.7.1.1.5.4

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

			-	$λ^{2}$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$

Step 1.1.7.1.1.5.5

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

			-	$λ^{2}$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$

Step 1.1.7.1.1.5.6

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

			-	$λ^{2}$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$

Step 1.1.7.1.1.5.7

Divide the highest order term in the dividend $11 λ^{2}$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$11 λ$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$

Step 1.1.7.1.1.5.8

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$11 λ$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$
					+	$11 λ^{2}$	-	$33 λ$

Step 1.1.7.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $11 λ^{2} - 33 λ$

			-	$λ^{2}$	+	$11 λ$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$
					-	$11 λ^{2}$	+	$33 λ$

Step 1.1.7.1.1.5.10

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

			-	$λ^{2}$	+	$11 λ$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$
					-	$11 λ^{2}$	+	$33 λ$
							-	$28 λ$

Step 1.1.7.1.1.5.11

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

			-	$λ^{2}$	+	$11 λ$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$
					-	$11 λ^{2}$	+	$33 λ$
							-	$28 λ$	+	$84$

Step 1.1.7.1.1.5.12

Divide the highest order term in the dividend $- 28 λ$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$11 λ$	-	$28$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$
					-	$11 λ^{2}$	+	$33 λ$
							-	$28 λ$	+	$84$

Step 1.1.7.1.1.5.13

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$11 λ$	-	$28$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$
					-	$11 λ^{2}$	+	$33 λ$
							-	$28 λ$	+	$84$
							-	$28 λ$	+	$84$

Step 1.1.7.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 28 λ + 84$

			-	$λ^{2}$	+	$11 λ$	-	$28$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$
					-	$11 λ^{2}$	+	$33 λ$
							-	$28 λ$	+	$84$
							+	$28 λ$	-	$84$

Step 1.1.7.1.1.5.15

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

			-	$λ^{2}$	+	$11 λ$	-	$28$
$λ$	-	$3$	-	$λ^{3}$	+	$14 λ^{2}$	-	$61 λ$	+	$84$
			+	$λ^{3}$	-	$3 λ^{2}$
					+	$11 λ^{2}$	-	$61 λ$
					-	$11 λ^{2}$	+	$33 λ$
							-	$28 λ$	+	$84$
							+	$28 λ$	-	$84$
										$0$

Step 1.1.7.1.1.5.16

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

$- λ^{2} + 11 λ - 28$

Step 1.1.7.1.1.6

Write $- λ^{3} + 14 λ^{2} - 61 λ + 84$ as a set of factors.

$(λ - 3) (- λ^{2} + 11 λ - 28) = 0$

Step 1.1.7.1.2

Factor by grouping.

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Step 1.1.7.1.2.1

Factor by grouping.

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Step 1.1.7.1.2.1.1

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 $a \cdot c = - 1 \cdot - 28 = 28$ and whose sum is $b = 11$ .

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Step 1.1.7.1.2.1.1.1

Factor $11$ out of $11 λ$ .

$(λ - 3) (- λ^{2} + 11 (λ) - 28) = 0$

Step 1.1.7.1.2.1.1.2

Rewrite $11$ as $4$ plus $7$

$(λ - 3) (- λ^{2} + (4 + 7) λ - 28) = 0$

Step 1.1.7.1.2.1.1.3

Apply the distributive property.

$(λ - 3) (- λ^{2} + 4 λ + 7 λ - 28) = 0$

Step 1.1.7.1.2.1.2

Factor out the greatest common factor from each group.

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Step 1.1.7.1.2.1.2.1

Group the first two terms and the last two terms.

$(λ - 3) ((- λ^{2} + 4 λ) + 7 λ - 28) = 0$

Step 1.1.7.1.2.1.2.2

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

$(λ - 3) (λ (- λ + 4) - 7 (- λ + 4)) = 0$

Step 1.1.7.1.2.1.3

Factor the polynomial by factoring out the greatest common factor, $- λ + 4$ .

$(λ - 3) ((- λ + 4) (λ - 7)) = 0$

Step 1.1.7.1.2.2

Remove unnecessary parentheses.

$(λ - 3) (- λ + 4) (λ - 7) = 0$

Step 1.1.7.2

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

$λ - 3 = 0$

$- λ + 4 = 0$

$λ - 7 = 0$

Step 1.1.7.3

Set $λ - 3$ equal to $0$ and solve for $λ$ .

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Step 1.1.7.3.1

Set $λ - 3$ equal to $0$ .

$λ - 3 = 0$

Step 1.1.7.3.2

Add $3$ to both sides of the equation.

$λ = 3$

Step 1.1.7.4

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

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Step 1.1.7.4.1

Set $- λ + 4$ equal to $0$ .

$- λ + 4 = 0$

Step 1.1.7.4.2

Solve $- λ + 4 = 0$ for $λ$ .

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Step 1.1.7.4.2.1

Subtract $4$ from both sides of the equation.

$- λ = - 4$

Step 1.1.7.4.2.2

Divide each term in $- λ = - 4$ by $- 1$ and simplify.

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Step 1.1.7.4.2.2.1

Divide each term in $- λ = - 4$ by $- 1$ .

$\frac{- λ}{- 1} = \frac{- 4}{- 1}$

Step 1.1.7.4.2.2.2

Simplify the left side.

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Step 1.1.7.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{λ}{1} = \frac{- 4}{- 1}$

Step 1.1.7.4.2.2.2.2

Divide $λ$ by $1$ .

$λ = \frac{- 4}{- 1}$

Step 1.1.7.4.2.2.3

Simplify the right side.

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Step 1.1.7.4.2.2.3.1

Divide $- 4$ by $- 1$ .

$λ = 4$

Step 1.1.7.5

Set $λ - 7$ equal to $0$ and solve for $λ$ .

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Step 1.1.7.5.1

Set $λ - 7$ equal to $0$ .

$λ - 7 = 0$

Step 1.1.7.5.2

Add $7$ to both sides of the equation.

$λ = 7$

Step 1.1.7.6

The final solution is all the values that make $(λ - 3) (- λ + 4) (λ - 7) = 0$ true.

$λ = 3, 4, 7$

Step 1.2

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 1.3

Find the eigenvector using the eigenvalue $λ = 3$ .

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Step 1.3.1

Substitute the known values into the formula.

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

Step 1.3.2

Simplify.

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Step 1.3.2.1

Simplify each term.

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Step 1.3.2.1.1

Multiply $- 3$ by each element of the matrix.

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

Step 1.3.2.1.2

Simplify each element in the matrix.

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Step 1.3.2.1.2.1

Multiply $- 3$ by $1$ .

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

Step 1.3.2.1.2.2

Multiply $- 3$ by $0$ .

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

Step 1.3.2.1.2.3

Multiply $- 3$ by $0$ .

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

Step 1.3.2.1.2.4

Multiply $- 3$ by $0$ .

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

Step 1.3.2.1.2.5

Multiply $- 3$ by $1$ .

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

Step 1.3.2.1.2.6

Multiply $- 3$ by $0$ .

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

Step 1.3.2.1.2.7

Multiply $- 3$ by $0$ .

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

Step 1.3.2.1.2.8

Multiply $- 3$ by $0$ .

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

Step 1.3.2.1.2.9

Multiply $- 3$ by $1$ .

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

Step 1.3.2.2

Add the corresponding elements.

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

Step 1.3.2.3

Simplify each element.

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Step 1.3.2.3.1

Subtract $3$ from $5$ .

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

Step 1.3.2.3.2

Add $2$ and $0$ .

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

Step 1.3.2.3.3

Add $0$ and $0$ .

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

Step 1.3.2.3.4

Add $2$ and $0$ .

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

Step 1.3.2.3.5

Subtract $3$ from $5$ .

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

Step 1.3.2.3.6

Add $0$ and $0$ .

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

Step 1.3.2.3.7

Add $4$ and $0$ .

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

Step 1.3.2.3.8

Add $- 1$ and $0$ .

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

Step 1.3.2.3.9

Subtract $3$ from $4$ .

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

Step 1.3.3

Find the null space when $λ = 3$ .

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Step 1.3.3.1

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

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

Step 1.3.3.2

Find the reduced row echelon form.

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Step 1.3.3.2.1

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

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Step 1.3.3.2.1.1

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

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

Step 1.3.3.2.1.2

Simplify $R_{1}$ .

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

Step 1.3.3.2.2

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 1.3.3.2.2.1

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

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

Step 1.3.3.2.2.2

Simplify $R_{2}$ .

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

Step 1.3.3.2.3

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

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Step 1.3.3.2.3.1

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

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

Step 1.3.3.2.3.2

Simplify $R_{3}$ .

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

Step 1.3.3.2.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

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

Step 1.3.3.2.5

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

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Step 1.3.3.2.5.1

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

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

Step 1.3.3.2.5.2

Simplify $R_{2}$ .

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

Step 1.3.3.2.6

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

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Step 1.3.3.2.6.1

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

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

Step 1.3.3.2.6.2

Simplify $R_{1}$ .

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

Step 1.3.3.3

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

$x + \frac{1}{5} z = 0$

$y - \frac{1}{5} z = 0$

$0 = 0$

Step 1.3.3.4

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

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

Step 1.3.3.5

Write the solution as a linear combination of vectors.

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

Step 1.3.3.6

Write as a solution set.

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

Step 1.3.3.7

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

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

Step 1.4

Find the eigenvector using the eigenvalue $λ = 4$ .

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Step 1.4.1

Substitute the known values into the formula.

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

Step 1.4.2

Simplify.

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Step 1.4.2.1

Simplify each term.

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Step 1.4.2.1.1

Multiply $- 4$ by each element of the matrix.

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

Step 1.4.2.1.2

Simplify each element in the matrix.

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Step 1.4.2.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.4.2.1.2.2

Multiply $- 4$ by $0$ .

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

Step 1.4.2.1.2.3

Multiply $- 4$ by $0$ .

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

Step 1.4.2.1.2.4

Multiply $- 4$ by $0$ .

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

Step 1.4.2.1.2.5

Multiply $- 4$ by $1$ .

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

Step 1.4.2.1.2.6

Multiply $- 4$ by $0$ .

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

Step 1.4.2.1.2.7

Multiply $- 4$ by $0$ .

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

Step 1.4.2.1.2.8

Multiply $- 4$ by $0$ .

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

Step 1.4.2.1.2.9

Multiply $- 4$ by $1$ .

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

Step 1.4.2.2

Add the corresponding elements.

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

Step 1.4.2.3

Simplify each element.

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Step 1.4.2.3.1

Subtract $4$ from $5$ .

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

Step 1.4.2.3.2

Add $2$ and $0$ .

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

Step 1.4.2.3.3

Add $0$ and $0$ .

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

Step 1.4.2.3.4

Add $2$ and $0$ .

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

Step 1.4.2.3.5

Subtract $4$ from $5$ .

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

Step 1.4.2.3.6

Add $0$ and $0$ .

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

Step 1.4.2.3.7

Add $4$ and $0$ .

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

Step 1.4.2.3.8

Add $- 1$ and $0$ .

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

Step 1.4.2.3.9

Subtract $4$ from $4$ .

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

Step 1.4.3

Find the null space when $λ = 4$ .

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Step 1.4.3.1

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

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

Step 1.4.3.2

Find the reduced row echelon form.

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Step 1.4.3.2.1

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 1.4.3.2.1.1

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

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

Step 1.4.3.2.1.2

Simplify $R_{2}$ .

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

Step 1.4.3.2.2

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

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Step 1.4.3.2.2.1

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

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

Step 1.4.3.2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & - 9 & 0 & 0 \end{array}]$

Step 1.4.3.2.3

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

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Step 1.4.3.2.3.1

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

$[\begin{array}{c} 1 & 2 & 0 & 0 \\ - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 0 & - \frac{1}{3} \cdot 0 \\ 0 & - 9 & 0 & 0 \end{array}]$

Step 1.4.3.2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & - 9 & 0 & 0 \end{array}]$

Step 1.4.3.2.4

Perform the row operation $R_{3} = R_{3} + 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 1.4.3.2.4.1

Perform the row operation $R_{3} = R_{3} + 9 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 + 9 \cdot 0 & - 9 + 9 \cdot 1 & 0 + 9 \cdot 0 & 0 + 9 \cdot 0 \end{array}]$

Step 1.4.3.2.4.2

Simplify $R_{3}$ .

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

Step 1.4.3.2.5

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

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Step 1.4.3.2.5.1

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

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

Step 1.4.3.2.5.2

Simplify $R_{1}$ .

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

Step 1.4.3.3

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

$x = 0$

$y = 0$

$0 = 0$

Step 1.4.3.4

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

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

Step 1.4.3.5

Write the solution as a linear combination of vectors.

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

Step 1.4.3.6

Write as a solution set.

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

Step 1.4.3.7

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

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

Step 1.5

Find the eigenvector using the eigenvalue $λ = 7$ .

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Step 1.5.1

Substitute the known values into the formula.

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

Step 1.5.2

Simplify.

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Step 1.5.2.1

Simplify each term.

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Step 1.5.2.1.1

Multiply $- 7$ by each element of the matrix.

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

Step 1.5.2.1.2

Simplify each element in the matrix.

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Step 1.5.2.1.2.1

Multiply $- 7$ by $1$ .

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

Step 1.5.2.1.2.2

Multiply $- 7$ by $0$ .

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

Step 1.5.2.1.2.3

Multiply $- 7$ by $0$ .

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

Step 1.5.2.1.2.4

Multiply $- 7$ by $0$ .

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

Step 1.5.2.1.2.5

Multiply $- 7$ by $1$ .

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

Step 1.5.2.1.2.6

Multiply $- 7$ by $0$ .

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

Step 1.5.2.1.2.7

Multiply $- 7$ by $0$ .

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

Step 1.5.2.1.2.8

Multiply $- 7$ by $0$ .

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

Step 1.5.2.1.2.9

Multiply $- 7$ by $1$ .

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

Step 1.5.2.2

Add the corresponding elements.

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

Step 1.5.2.3

Simplify each element.

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Step 1.5.2.3.1

Subtract $7$ from $5$ .

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

Step 1.5.2.3.2

Add $2$ and $0$ .

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

Step 1.5.2.3.3

Add $0$ and $0$ .

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

Step 1.5.2.3.4

Add $2$ and $0$ .

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

Step 1.5.2.3.5

Subtract $7$ from $5$ .

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

Step 1.5.2.3.6

Add $0$ and $0$ .

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

Step 1.5.2.3.7

Add $4$ and $0$ .

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

Step 1.5.2.3.8

Add $- 1$ and $0$ .

$[\begin{array}{c} - 2 & 2 & 0 \\ 2 & - 2 & 0 \\ 4 & - 1 & 4 - 7 \end{array}]$

Step 1.5.2.3.9

Subtract $7$ from $4$ .

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

Step 1.5.3

Find the null space when $λ = 7$ .

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Step 1.5.3.1

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

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

Step 1.5.3.2

Find the reduced row echelon form.

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Step 1.5.3.2.1

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

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Step 1.5.3.2.1.1

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

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

Step 1.5.3.2.1.2

Simplify $R_{1}$ .

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

Step 1.5.3.2.2

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 1.5.3.2.2.1

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

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

Step 1.5.3.2.2.2

Simplify $R_{2}$ .

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

Step 1.5.3.2.3

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

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Step 1.5.3.2.3.1

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

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

Step 1.5.3.2.3.2

Simplify $R_{3}$ .

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

Step 1.5.3.2.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

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

Step 1.5.3.2.5

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

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Step 1.5.3.2.5.1

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

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

Step 1.5.3.2.5.2

Simplify $R_{2}$ .

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

Step 1.5.3.2.6

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

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Step 1.5.3.2.6.1

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

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

Step 1.5.3.2.6.2

Simplify $R_{1}$ .

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

Step 1.5.3.3

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

$x - z = 0$

$y - z = 0$

$0 = 0$

Step 1.5.3.4

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

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

Step 1.5.3.5

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 1.5.3.6

Write as a solution set.

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

Step 1.5.3.7

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

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

Step 1.6

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

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

Step 2

Define $P$ as a matrix of the eigenvectors.

$P = [\begin{array}{c} - \frac{1}{5} & 0 & 1 \\ \frac{1}{5} & 0 & 1 \\ 1 & 1 & 1 \end{array}]$

Step 3

Find the inverse of $P$ .

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Step 3.1

Find the determinant.

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Step 3.1.1

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 column $2$ by its cofactor and add.

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Step 3.1.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 3.1.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 3.1.1.3

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

$| \begin{array}{c} \frac{1}{5} & 1 \\ 1 & 1 \end{array} |$

Step 3.1.1.4

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

$0 | \begin{array}{c} \frac{1}{5} & 1 \\ 1 & 1 \end{array} |$

Step 3.1.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

$| \begin{array}{c} - \frac{1}{5} & 1 \\ 1 & 1 \end{array} |$

Step 3.1.1.6

Multiply element $a_{22}$ by its cofactor.

$0 | \begin{array}{c} - \frac{1}{5} & 1 \\ 1 & 1 \end{array} |$

Step 3.1.1.7

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

$| \begin{array}{c} - \frac{1}{5} & 1 \\ \frac{1}{5} & 1 \end{array} |$

Step 3.1.1.8

Multiply element $a_{32}$ by its cofactor.

$- 1 | \begin{array}{c} - \frac{1}{5} & 1 \\ \frac{1}{5} & 1 \end{array} |$

Step 3.1.1.9

Add the terms together.

$0 | \begin{array}{c} \frac{1}{5} & 1 \\ 1 & 1 \end{array} | + 0 | \begin{array}{c} - \frac{1}{5} & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} - \frac{1}{5} & 1 \\ \frac{1}{5} & 1 \end{array} |$

Step 3.1.2

Multiply $0$ by $| \begin{array}{c} \frac{1}{5} & 1 \\ 1 & 1 \end{array} |$ .

$0 + 0 | \begin{array}{c} - \frac{1}{5} & 1 \\ 1 & 1 \end{array} | - 1 | \begin{array}{c} - \frac{1}{5} & 1 \\ \frac{1}{5} & 1 \end{array} |$

Step 3.1.3

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

$0 + 0 - 1 | \begin{array}{c} - \frac{1}{5} & 1 \\ \frac{1}{5} & 1 \end{array} |$

Step 3.1.4

Evaluate $| \begin{array}{c} - \frac{1}{5} & 1 \\ \frac{1}{5} & 1 \end{array} |$ .

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Step 3.1.4.1

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

$0 + 0 - 1 (- \frac{1}{5} \cdot 1 - \frac{1}{5} \cdot 1)$

Step 3.1.4.2

Simplify the determinant.

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Step 3.1.4.2.1

Simplify each term.

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Step 3.1.4.2.1.1

Multiply $- 1$ by $1$ .

$0 + 0 - 1 (- \frac{1}{5} - \frac{1}{5} \cdot 1)$

Step 3.1.4.2.1.2

Multiply $- 1$ by $1$ .

$0 + 0 - 1 (- \frac{1}{5} - \frac{1}{5})$

Step 3.1.4.2.2

Combine the numerators over the common denominator.

$0 + 0 - 1 \frac{- 1 - 1}{5}$

Step 3.1.4.2.3

Subtract $1$ from $- 1$ .

$0 + 0 - 1 (\frac{- 2}{5})$

Step 3.1.4.2.4

Move the negative in front of the fraction.

$0 + 0 - 1 (- \frac{2}{5})$

Step 3.1.5

Simplify the determinant.

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Step 3.1.5.1

Multiply $- 1 (- \frac{2}{5})$ .

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Step 3.1.5.1.1

Multiply $- 1$ by $- 1$ .

$0 + 0 + 1 (\frac{2}{5})$

Step 3.1.5.1.2

Multiply $\frac{2}{5}$ by $1$ .

$0 + 0 + \frac{2}{5}$

Step 3.1.5.2

Add $0$ and $0$ .

$0 + \frac{2}{5}$

Step 3.1.5.3

Add $0$ and $\frac{2}{5}$ .

$\frac{2}{5}$

Step 3.2

Since the determinant is non-zero, the inverse exists.

Step 3.3

Set up a $3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

$P^{- 1} = [\begin{array}{c} - \frac{1}{5} & 0 & 1 & 1 & 0 & 0 \\ \frac{1}{5} & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Step 3.4

Find the reduced row echelon form.

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Step 3.4.1

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

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Step 3.4.1.1

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

$P^{- 1} = [\begin{array}{c} - 5 (- \frac{1}{5}) & - 5 \cdot 0 & - 5 \cdot 1 & - 5 \cdot 1 & - 5 \cdot 0 & - 5 \cdot 0 \\ \frac{1}{5} & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Step 3.4.1.2

Simplify $R_{1}$ .

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ \frac{1}{5} & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Step 3.4.2

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

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Step 3.4.2.1

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

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ \frac{1}{5} - \frac{1}{5} \cdot 1 & 0 - \frac{1}{5} \cdot 0 & 1 - \frac{1}{5} \cdot - 5 & 0 - \frac{1}{5} \cdot - 5 & 1 - \frac{1}{5} \cdot 0 & 0 - \frac{1}{5} \cdot 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Step 3.4.2.2

Simplify $R_{2}$ .

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ 0 & 0 & 2 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}]$

Step 3.4.3

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

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Step 3.4.3.1

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

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ 0 & 0 & 2 & 1 & 1 & 0 \\ 1 - 1 & 1 - 0 & 1 + 5 & 0 + 5 & 0 - 0 & 1 - 0 \end{array}]$

Step 3.4.3.2

Simplify $R_{3}$ .

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ 0 & 0 & 2 & 1 & 1 & 0 \\ 0 & 1 & 6 & 5 & 0 & 1 \end{array}]$

Step 3.4.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ 0 & 1 & 6 & 5 & 0 & 1 \\ 0 & 0 & 2 & 1 & 1 & 0 \end{array}]$

Step 3.4.5

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

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Step 3.4.5.1

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

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ 0 & 1 & 6 & 5 & 0 & 1 \\ \frac{0}{2} & \frac{0}{2} & \frac{2}{2} & \frac{1}{2} & \frac{1}{2} & \frac{0}{2} \end{array}]$

Step 3.4.5.2

Simplify $R_{3}$ .

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ 0 & 1 & 6 & 5 & 0 & 1 \\ 0 & 0 & 1 & \frac{1}{2} & \frac{1}{2} & 0 \end{array}]$

Step 3.4.6

Perform the row operation $R_{2} = R_{2} - 6 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 3.4.6.1

Perform the row operation $R_{2} = R_{2} - 6 R_{3}$ to make the entry at $2, 3$ a $0$ .

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ 0 - 6 \cdot 0 & 1 - 6 \cdot 0 & 6 - 6 \cdot 1 & 5 - 6 (\frac{1}{2}) & 0 - 6 (\frac{1}{2}) & 1 - 6 \cdot 0 \\ 0 & 0 & 1 & \frac{1}{2} & \frac{1}{2} & 0 \end{array}]$

Step 3.4.6.2

Simplify $R_{2}$ .

$P^{- 1} = [\begin{array}{c} 1 & 0 & - 5 & - 5 & 0 & 0 \\ 0 & 1 & 0 & 2 & - 3 & 1 \\ 0 & 0 & 1 & \frac{1}{2} & \frac{1}{2} & 0 \end{array}]$

Step 3.4.7

Perform the row operation $R_{1} = R_{1} + 5 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 3.4.7.1

Perform the row operation $R_{1} = R_{1} + 5 R_{3}$ to make the entry at $1, 3$ a $0$ .

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

Step 3.4.7.2

Simplify $R_{1}$ .

$P^{- 1} = [\begin{array}{c} 1 & 0 & 0 & - \frac{5}{2} & \frac{5}{2} & 0 \\ 0 & 1 & 0 & 2 & - 3 & 1 \\ 0 & 0 & 1 & \frac{1}{2} & \frac{1}{2} & 0 \end{array}]$

Step 3.5

The right half of the reduced row echelon form is the inverse.

$P^{- 1} = [\begin{array}{c} - \frac{5}{2} & \frac{5}{2} & 0 \\ 2 & - 3 & 1 \\ \frac{1}{2} & \frac{1}{2} & 0 \end{array}]$

Step 4

Use the similarity transformation to find the diagonal matrix $D$ .

$D = P^{- 1} A P$

Step 5

Substitute the matrices.

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

Step 6

Simplify.

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Step 6.1

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

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Step 6.1.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $3 \times 3$ and the second matrix is $3 \times 3$ .

Step 6.1.2

Multiply each row in the first matrix by each column in the second matrix.

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

Step 6.1.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} - \frac{15}{2} & \frac{15}{2} & 0 \\ 8 & - 12 & 4 \\ \frac{7}{2} & \frac{7}{2} & 0 \end{array}] [\begin{array}{c} - \frac{1}{5} & 0 & 1 \\ \frac{1}{5} & 0 & 1 \\ 1 & 1 & 1 \end{array}]$

Step 6.2

Multiply $[\begin{array}{c} - \frac{15}{2} & \frac{15}{2} & 0 \\ 8 & - 12 & 4 \\ \frac{7}{2} & \frac{7}{2} & 0 \end{array}] [\begin{array}{c} - \frac{1}{5} & 0 & 1 \\ \frac{1}{5} & 0 & 1 \\ 1 & 1 & 1 \end{array}]$ .

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Step 6.2.1

Step 6.2.2

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} - \frac{15}{2} (- \frac{1}{5}) + \frac{15}{2} \cdot \frac{1}{5} + 0 \cdot 1 & - \frac{15}{2} \cdot 0 + \frac{15}{2} \cdot 0 + 0 \cdot 1 & - \frac{15}{2} \cdot 1 + \frac{15}{2} \cdot 1 + 0 \cdot 1 \\ 8 (- \frac{1}{5}) - 12 (\frac{1}{5}) + 4 \cdot 1 & 8 \cdot 0 - 12 \cdot 0 + 4 \cdot 1 & 8 \cdot 1 - 12 \cdot 1 + 4 \cdot 1 \\ \frac{7}{2} (- \frac{1}{5}) + \frac{7}{2} \cdot \frac{1}{5} + 0 \cdot 1 & \frac{7}{2} \cdot 0 + \frac{7}{2} \cdot 0 + 0 \cdot 1 & \frac{7}{2} \cdot 1 + \frac{7}{2} \cdot 1 + 0 \cdot 1 \end{array}]$

Step 6.2.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 7 \end{array}]$

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