Linear Algebra Examples

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

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

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

Step 1.3.2

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

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

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.

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

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.5

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.6.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.9

Multiply $- 1$ by $1$ .

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

Step 1.4.2

Add the corresponding elements.

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

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $2$ and $0$ .

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

Step 1.4.3.2

Add $1$ and $0$ .

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

Step 1.4.3.3

Add $0$ and $0$ .

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

Step 1.4.3.4

Add $1$ and $0$ .

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

Step 1.4.3.5

Add $0$ and $0$ .

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

Step 1.4.3.6

Add $5$ and $0$ .

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

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 column $1$ by its cofactor and add.

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Step 1.5.1.1

Consider the corresponding sign chart.

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

Step 1.5.1.2

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

Step 1.5.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

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

Step 1.5.1.4

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

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

Step 1.5.1.5

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

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

Step 1.5.1.6

Multiply element $a_{21}$ by its cofactor.

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

Step 1.5.1.7

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

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

Step 1.5.1.8

Multiply element $a_{31}$ by its cofactor.

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

Step 1.5.1.9

Add the terms together.

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

Step 1.5.2

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

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

Step 1.5.3

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

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

Step 1.5.4

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

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Step 1.5.4.1

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

$p (λ) = (1 - λ) ((3 - λ) (- 1 - λ) - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2

Simplify the determinant.

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Step 1.5.4.2.1

Simplify each term.

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Step 1.5.4.2.1.1

Expand $(3 - λ) (- 1 - λ)$ using the FOIL Method.

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Step 1.5.4.2.1.1.1

Apply the distributive property.

$p (λ) = (1 - λ) (3 (- 1 - λ) - λ (- 1 - λ) - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2.1.1.2

Apply the distributive property.

$p (λ) = (1 - λ) (3 \cdot - 1 + 3 (- λ) - λ (- 1 - λ) - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2.1.1.3

Apply the distributive property.

$p (λ) = (1 - λ) (3 \cdot - 1 + 3 (- λ) - λ \cdot - 1 - λ (- λ) - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2.1.2

Simplify and combine like terms.

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Step 1.5.4.2.1.2.1

Simplify each term.

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Step 1.5.4.2.1.2.1.1

Multiply $3$ by $- 1$ .

$p (λ) = (1 - λ) (- 3 + 3 (- λ) - λ \cdot - 1 - λ (- λ) - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2.1.2.1.2

Multiply $- 1$ by $3$ .

$p (λ) = (1 - λ) (- 3 - 3 λ - λ \cdot - 1 - λ (- λ) - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2.1.2.1.3

Multiply $- λ \cdot - 1$ .

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Step 1.5.4.2.1.2.1.3.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (1 - λ) (- 3 - 3 λ + 1 λ - λ (- λ) - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2.1.2.1.3.2

Multiply $λ$ by $1$ .

$p (λ) = (1 - λ) (- 3 - 3 λ + λ - λ (- λ) - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (1 - λ) (- 3 - 3 λ + λ - 1 \cdot - 1 λ \cdot λ - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2.1.2.1.5

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

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Step 1.5.4.2.1.2.1.5.1

Move $λ$ .

$p (λ) = (1 - λ) (- 3 - 3 λ + λ - 1 \cdot - 1 (λ \cdot λ) - 5 \cdot 1) + 0 + 0$

Step 1.5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

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

Step 1.5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

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

Step 1.5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

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

Step 1.5.4.2.1.2.2

Add $- 3 λ$ and $λ$ .

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

Step 1.5.4.2.1.3

Multiply $- 5$ by $1$ .

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

Step 1.5.4.2.2

Subtract $5$ from $- 3$ .

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

Step 1.5.4.2.3

Reorder $- 2 λ$ and $λ^{2}$ .

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

Step 1.5.5

Simplify the determinant.

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Step 1.5.5.1

Combine the opposite terms in $(1 - λ) (λ^{2} - 2 λ - 8) + 0 + 0$ .

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Step 1.5.5.1.1

Add $(1 - λ) (λ^{2} - 2 λ - 8)$ and $0$ .

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

Step 1.5.5.1.2

Add $(1 - λ) (λ^{2} - 2 λ - 8)$ and $0$ .

$p (λ) = (1 - λ) (λ^{2} - 2 λ - 8)$

Step 1.5.5.2

Expand $(1 - λ) (λ^{2} - 2 λ - 8)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 1.5.5.3

Simplify each term.

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Step 1.5.5.3.1

Multiply $λ^{2}$ by $1$ .

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

Step 1.5.5.3.2

Multiply $- 2 λ$ by $1$ .

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

Step 1.5.5.3.3

Multiply $- 8$ by $1$ .

$p (λ) = λ^{2} - 2 λ - 8 - λ \cdot λ^{2} - λ (- 2 λ) - λ \cdot - 8$

Step 1.5.5.3.4

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

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Step 1.5.5.3.4.1

Move $λ^{2}$ .

$p (λ) = λ^{2} - 2 λ - 8 - (λ^{2} λ) - λ (- 2 λ) - λ \cdot - 8$

Step 1.5.5.3.4.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.5.3.4.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = λ^{2} - 2 λ - 8 - (λ^{2} λ^{1}) - λ (- 2 λ) - λ \cdot - 8$

Step 1.5.5.3.4.2.2

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

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

Step 1.5.5.3.4.3

Add $2$ and $1$ .

$p (λ) = λ^{2} - 2 λ - 8 - λ^{3} - λ (- 2 λ) - λ \cdot - 8$

Step 1.5.5.3.5

Rewrite using the commutative property of multiplication.

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

Step 1.5.5.3.6

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

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Step 1.5.5.3.6.1

Move $λ$ .

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

Step 1.5.5.3.6.2

Multiply $λ$ by $λ$ .

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

Step 1.5.5.3.7

Multiply $- 1$ by $- 2$ .

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

Step 1.5.5.3.8

Multiply $- 8$ by $- 1$ .

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

Step 1.5.5.4

Add $λ^{2}$ and $2 λ^{2}$ .

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

Step 1.5.5.5

Add $- 2 λ$ and $8 λ$ .

$p (λ) = 3 λ^{2} + 6 λ - 8 - λ^{3}$

Step 1.5.5.6

Move $- 8$ .

$p (λ) = 3 λ^{2} + 6 λ - λ^{3} - 8$

Step 1.5.5.7

Move $6 λ$ .

$p (λ) = 3 λ^{2} - λ^{3} + 6 λ - 8$

Step 1.5.5.8

Reorder $3 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 3 λ^{2} + 6 λ - 8$

Step 1.6

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

$- λ^{3} + 3 λ^{2} + 6 λ - 8 = 0$

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

Regroup terms.

$- λ^{3} - 8 + 3 λ^{2} + 6 λ = 0$

Step 1.7.1.2

Factor $- 1$ out of $- λ^{3} - 8$ .

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Step 1.7.1.2.1

Factor $- 1$ out of $- λ^{3}$ .

$- (λ^{3}) - 8 + 3 λ^{2} + 6 λ = 0$

Step 1.7.1.2.2

Rewrite $- 8$ as $- 1 (8)$ .

$- (λ^{3}) - 1 \cdot 8 + 3 λ^{2} + 6 λ = 0$

Step 1.7.1.2.3

Factor $- 1$ out of $- (λ^{3}) - 1 (8)$ .

$- (λ^{3} + 8) + 3 λ^{2} + 6 λ = 0$

Step 1.7.1.3

Rewrite $8$ as $2^{3}$ .

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

Step 1.7.1.4

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = λ$ and $b = 2$ .

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

Step 1.7.1.5

Factor.

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Step 1.7.1.5.1

Simplify.

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Step 1.7.1.5.1.1

Multiply $2$ by $- 1$ .

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

Step 1.7.1.5.1.2

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

$- ((λ + 2) (λ^{2} - 2 λ + 4)) + 3 λ^{2} + 6 λ = 0$

Step 1.7.1.5.2

Remove unnecessary parentheses.

$- (λ + 2) (λ^{2} - 2 λ + 4) + 3 λ^{2} + 6 λ = 0$

Step 1.7.1.6

Factor $3 λ$ out of $3 λ^{2} + 6 λ$ .

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Step 1.7.1.6.1

Factor $3 λ$ out of $3 λ^{2}$ .

$- (λ + 2) (λ^{2} - 2 λ + 4) + 3 λ (λ) + 6 λ = 0$

Step 1.7.1.6.2

Factor $3 λ$ out of $6 λ$ .

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

Step 1.7.1.6.3

Factor $3 λ$ out of $3 λ (λ) + 3 λ (2)$ .

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

Step 1.7.1.7

Factor $λ + 2$ out of $- (λ + 2) (λ^{2} - 2 λ + 4) + 3 λ (λ + 2)$ .

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Step 1.7.1.7.1

Factor $λ + 2$ out of $- (λ + 2) (λ^{2} - 2 λ + 4)$ .

$(λ + 2) (- 1 (λ^{2} - 2 λ + 4)) + 3 λ (λ + 2) = 0$

Step 1.7.1.7.2

Factor $λ + 2$ out of $3 λ (λ + 2)$ .

$(λ + 2) (- 1 (λ^{2} - 2 λ + 4)) + (λ + 2) (3 λ) = 0$

Step 1.7.1.7.3

Factor $λ + 2$ out of $(λ + 2) (- 1 (λ^{2} - 2 λ + 4)) + (λ + 2) (3 λ)$ .

$(λ + 2) (- 1 (λ^{2} - 2 λ + 4) + 3 λ) = 0$

Step 1.7.1.8

Apply the distributive property.

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

Step 1.7.1.9

Simplify.

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Step 1.7.1.9.1

Rewrite $- 1 λ^{2}$ as $- λ^{2}$ .

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

Step 1.7.1.9.2

Multiply $- 2$ by $- 1$ .

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

Step 1.7.1.9.3

Multiply $- 1$ by $4$ .

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

Step 1.7.1.10

Add $2 λ$ and $3 λ$ .

$(λ + 2) (- λ^{2} + 5 λ - 4) = 0$

Step 1.7.1.11

Factor.

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Step 1.7.1.11.1

Factor by grouping.

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Step 1.7.1.11.1.1

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 $a \cdot c = - 1 \cdot - 4 = 4$ and whose sum is $b = 5$ .

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Step 1.7.1.11.1.1.1

Factor $5$ out of $5 λ$ .

$(λ + 2) (- λ^{2} + 5 (λ) - 4) = 0$

Step 1.7.1.11.1.1.2

Rewrite $5$ as $1$ plus $4$

$(λ + 2) (- λ^{2} + (1 + 4) λ - 4) = 0$

Step 1.7.1.11.1.1.3

Apply the distributive property.

$(λ + 2) (- λ^{2} + 1 λ + 4 λ - 4) = 0$

Step 1.7.1.11.1.1.4

Multiply $λ$ by $1$ .

$(λ + 2) (- λ^{2} + λ + 4 λ - 4) = 0$

Step 1.7.1.11.1.2

Factor out the greatest common factor from each group.

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Step 1.7.1.11.1.2.1

Group the first two terms and the last two terms.

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

Step 1.7.1.11.1.2.2

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

$(λ + 2) (λ (- λ + 1) - 4 (- λ + 1)) = 0$

Step 1.7.1.11.1.3

Factor the polynomial by factoring out the greatest common factor, $- λ + 1$ .

$(λ + 2) ((- λ + 1) (λ - 4)) = 0$

Step 1.7.1.11.2

Remove unnecessary parentheses.

$(λ + 2) (- λ + 1) (λ - 4) = 0$

Step 1.7.2

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

$λ + 2 = 0$

$- λ + 1 = 0$

$λ - 4 = 0$

Step 1.7.3

Set $λ + 2$ equal to $0$ and solve for $λ$ .

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Step 1.7.3.1

Set $λ + 2$ equal to $0$ .

$λ + 2 = 0$

Step 1.7.3.2

Subtract $2$ from both sides of the equation.

$λ = - 2$

Step 1.7.4

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

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Step 1.7.4.1

Set $- λ + 1$ equal to $0$ .

$- λ + 1 = 0$

Step 1.7.4.2

Solve $- λ + 1 = 0$ for $λ$ .

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Step 1.7.4.2.1

Subtract $1$ from both sides of the equation.

$- λ = - 1$

Step 1.7.4.2.2

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

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Step 1.7.4.2.2.1

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

$\frac{- λ}{- 1} = \frac{- 1}{- 1}$

Step 1.7.4.2.2.2

Simplify the left side.

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Step 1.7.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{λ}{1} = \frac{- 1}{- 1}$

Step 1.7.4.2.2.2.2

Divide $λ$ by $1$ .

$λ = \frac{- 1}{- 1}$

Step 1.7.4.2.2.3

Simplify the right side.

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Step 1.7.4.2.2.3.1

Divide $- 1$ by $- 1$ .

$λ = 1$

Step 1.7.5

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

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Step 1.7.5.1

Set $λ - 4$ equal to $0$ .

$λ - 4 = 0$

Step 1.7.5.2

Add $4$ to both sides of the equation.

$λ = 4$

Step 1.7.6

The final solution is all the values that make $(λ + 2) (- λ + 1) (λ - 4) = 0$ true.

$λ = - 2, 1, 4$

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 $λ = - 2$ .

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Step 3.1

Substitute the known values into the formula.

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

Step 3.2

Simplify.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply $2$ by each element of the matrix.

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

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $2$ by $1$ .

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

Step 3.2.1.2.2

Multiply $2$ by $0$ .

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

Step 3.2.1.2.3

Multiply $2$ by $0$ .

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

Step 3.2.1.2.4

Multiply $2$ by $0$ .

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

Step 3.2.1.2.5

Multiply $2$ by $1$ .

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

Step 3.2.1.2.6

Multiply $2$ by $0$ .

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

Step 3.2.1.2.7

Multiply $2$ by $0$ .

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

Step 3.2.1.2.8

Multiply $2$ by $0$ .

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

Step 3.2.1.2.9

Multiply $2$ by $1$ .

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

Step 3.2.2

Add the corresponding elements.

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

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

Add $1$ and $2$ .

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

Step 3.2.3.2

Add $2$ and $0$ .

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

Step 3.2.3.3

Add $1$ and $0$ .

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

Step 3.2.3.4

Add $0$ and $0$ .

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

Step 3.2.3.5

Add $3$ and $2$ .

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

Step 3.2.3.6

Add $1$ and $0$ .

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

Step 3.2.3.7

Add $0$ and $0$ .

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

Step 3.2.3.8

Add $5$ and $0$ .

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

Step 3.2.3.9

Add $- 1$ and $2$ .

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

Step 3.3

Find the null space when $λ = - 2$ .

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Step 3.3.1

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

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

Step 3.3.2

Find the reduced row echelon form.

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Step 3.3.2.1

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

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Step 3.3.2.1.1

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

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

Step 3.3.2.1.2

Simplify $R_{1}$ .

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

Step 3.3.2.2

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

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Step 3.3.2.2.1

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

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

Step 3.3.2.2.2

Simplify $R_{2}$ .

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

Step 3.3.2.3

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.3.2.3.1

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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

Step 3.3.2.3.2

Simplify $R_{3}$ .

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

Step 3.3.2.4

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

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Step 3.3.2.4.1

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

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

Step 3.3.2.4.2

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 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 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 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 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 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 4

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

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Step 4.1

Substitute the known values into the formula.

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

Step 4.2

Simplify.

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Step 4.2.1

Subtract the corresponding elements.

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

Step 4.2.2

Simplify each element.

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Step 4.2.2.1

Subtract $1$ from $1$ .

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

Step 4.2.2.2

Subtract $0$ from $2$ .

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

Step 4.2.2.3

Subtract $0$ from $1$ .

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

Step 4.2.2.4

Subtract $0$ from $0$ .

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

Step 4.2.2.5

Subtract $1$ from $3$ .

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

Step 4.2.2.6

Subtract $0$ from $1$ .

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

Step 4.2.2.7

Subtract $0$ from $0$ .

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

Step 4.2.2.8

Subtract $0$ from $5$ .

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

Step 4.2.2.9

Subtract $1$ from $- 1$ .

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

Step 4.3

Find the null space when $λ = 1$ .

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Step 4.3.1

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

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

Step 4.3.2

Find the reduced row echelon form.

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Step 4.3.2.1

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

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Step 4.3.2.1.1

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

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

Step 4.3.2.1.2

Simplify $R_{1}$ .

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

Step 4.3.2.2

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

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Step 4.3.2.2.1

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

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

Step 4.3.2.2.2

Simplify $R_{2}$ .

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

Step 4.3.2.3

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 2$ a $0$ .

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Step 4.3.2.3.1

Perform the row operation $R_{3} = R_{3} - 5 R_{1}$ to make the entry at $3, 2$ a $0$ .

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

Step 4.3.2.3.2

Simplify $R_{3}$ .

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

Step 4.3.2.4

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

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

Step 4.3.2.5

Multiply each element of $R_{2}$ by $- \frac{2}{9}$ to make the entry at $2, 3$ a $1$ .

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Step 4.3.2.5.1

Multiply each element of $R_{2}$ by $- \frac{2}{9}$ to make the entry at $2, 3$ a $1$ .

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

Step 4.3.2.5.2

Simplify $R_{2}$ .

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

Step 4.3.2.6

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

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

Step 4.3.2.6.2

Simplify $R_{1}$ .

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

Step 4.3.3

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

$y = 0$

$z = 0$

$0 = 0$

Step 4.3.4

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

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

Step 4.3.5

Write the solution as a linear combination of vectors.

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

Step 4.3.6

Write as a solution set.

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

Step 4.3.7

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

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

Step 5

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

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Step 5.1

Substitute the known values into the formula.

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

Step 5.2

Simplify.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

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

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

Step 5.2.1.2

Simplify each element in the matrix.

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Step 5.2.1.2.1

Multiply $- 4$ by $1$ .

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

Step 5.2.1.2.2

Multiply $- 4$ by $0$ .

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

Step 5.2.1.2.3

Multiply $- 4$ by $0$ .

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

Step 5.2.1.2.4

Multiply $- 4$ by $0$ .

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

Step 5.2.1.2.5

Multiply $- 4$ by $1$ .

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

Step 5.2.1.2.6

Multiply $- 4$ by $0$ .

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

Step 5.2.1.2.7

Multiply $- 4$ by $0$ .

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

Step 5.2.1.2.8

Multiply $- 4$ by $0$ .

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

Step 5.2.1.2.9

Multiply $- 4$ by $1$ .

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

Step 5.2.2

Add the corresponding elements.

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

Step 5.2.3

Simplify each element.

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Step 5.2.3.1

Subtract $4$ from $1$ .

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

Step 5.2.3.2

Add $2$ and $0$ .

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

Step 5.2.3.3

Add $1$ and $0$ .

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

Step 5.2.3.4

Add $0$ and $0$ .

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

Step 5.2.3.5

Subtract $4$ from $3$ .

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

Step 5.2.3.6

Add $1$ and $0$ .

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

Step 5.2.3.7

Add $0$ and $0$ .

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

Step 5.2.3.8

Add $5$ and $0$ .

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

Step 5.2.3.9

Subtract $4$ from $- 1$ .

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

Step 5.3

Find the null space when $λ = 4$ .

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Step 5.3.1

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

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

Step 5.3.2

Find the reduced row echelon form.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

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

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

Step 5.3.2.1.2

Simplify $R_{1}$ .

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

Step 5.3.2.2

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

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Step 5.3.2.2.1

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

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

Step 5.3.2.2.2

Simplify $R_{2}$ .

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

Step 5.3.2.3

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 5.3.2.3.1

Perform the row operation $R_{3} = R_{3} - 5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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

Step 5.3.2.3.2

Simplify $R_{3}$ .

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

Step 5.3.2.4

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

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Step 5.3.2.4.1

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

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

Step 5.3.2.4.2

Simplify $R_{1}$ .

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

Step 5.3.3

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

$x - z = 0$

$y - z = 0$

$0 = 0$

Step 5.3.4

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

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

Step 5.3.5

Write the solution as a linear combination of vectors.

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

Step 5.3.6

Write as a solution set.

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

Step 5.3.7

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

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

Step 6

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

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