Linear Algebra Examples

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

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

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

Step 1.3.2

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

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

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.

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

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.5

Multiply $- 1$ by $1$ .

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

Step 1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.6.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

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

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

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

Step 1.4.1.2.9

Multiply $- 1$ by $1$ .

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

Step 1.4.2

Add the corresponding elements.

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

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $- 1$ and $0$ .

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

Step 1.4.3.2

Add $0$ and $0$ .

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

Step 1.4.3.3

Add $- 1$ and $0$ .

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

Step 1.4.3.4

Add $- 1$ and $0$ .

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

Step 1.4.3.5

Add $0$ and $0$ .

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

Step 1.4.3.6

Add $- 1$ and $0$ .

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

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 row $1$ by its cofactor and add.

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Step 1.5.1.1

Consider the corresponding sign chart.

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

Step 1.5.1.2

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

Step 1.5.1.3

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

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

Step 1.5.1.4

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

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

Step 1.5.1.5

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

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

Step 1.5.1.6

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

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

Step 1.5.1.7

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

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

Step 1.5.1.8

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

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

Step 1.5.1.9

Add the terms together.

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

Step 1.5.2

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

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

Step 1.5.3

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

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Step 1.5.3.1

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

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

Step 1.5.3.2

Simplify the determinant.

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Step 1.5.3.2.1

Simplify each term.

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Step 1.5.3.2.1.1

Expand $(2 - λ) (1 - λ)$ using the FOIL Method.

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Step 1.5.3.2.1.1.1

Apply the distributive property.

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

Step 1.5.3.2.1.1.2

Apply the distributive property.

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

Step 1.5.3.2.1.1.3

Apply the distributive property.

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

Step 1.5.3.2.1.2

Simplify and combine like terms.

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Step 1.5.3.2.1.2.1

Simplify each term.

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Step 1.5.3.2.1.2.1.1

Multiply $2$ by $1$ .

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

Step 1.5.3.2.1.2.1.2

Multiply $- 1$ by $2$ .

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

Step 1.5.3.2.1.2.1.3

Multiply $- 1$ by $1$ .

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

Step 1.5.3.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

Step 1.5.3.2.1.2.1.5

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

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Step 1.5.3.2.1.2.1.5.1

Move $λ$ .

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

Step 1.5.3.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

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

Step 1.5.3.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

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

Step 1.5.3.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

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

Step 1.5.3.2.1.2.2

Subtract $λ$ from $- 2 λ$ .

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

Step 1.5.3.2.1.3

Multiply $- - - 1$ .

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Step 1.5.3.2.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 1.5.3.2.1.3.2

Multiply $- 1$ by $1$ .

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

Step 1.5.3.2.2

Subtract $1$ from $2$ .

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

Step 1.5.3.2.3

Reorder $- 3 λ$ and $λ^{2}$ .

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

Step 1.5.4

Evaluate $| \begin{array}{c} - 1 & - 1 \\ 0 & 1 - λ \end{array} |$ .

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Step 1.5.4.1

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

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

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

Apply the distributive property.

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

Step 1.5.4.2.1.2

Multiply $- 1$ by $1$ .

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

Step 1.5.4.2.1.3

Multiply $- - λ$ .

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Step 1.5.4.2.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 1.5.4.2.1.3.2

Multiply $λ$ by $1$ .

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

Step 1.5.4.2.1.4

Multiply $0$ by $- 1$ .

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

Step 1.5.4.2.2

Add $- 1 + λ$ and $0$ .

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

Step 1.5.4.2.3

Reorder $- 1$ and $λ$ .

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

Step 1.5.5

Simplify the determinant.

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Step 1.5.5.1

Add $(1 - λ) (λ^{2} - 3 λ + 1) + 1 (λ - 1)$ and $0$ .

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

Step 1.5.5.2

Simplify each term.

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Step 1.5.5.2.1

Expand $(1 - λ) (λ^{2} - 3 λ + 1)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 1.5.5.2.2

Simplify each term.

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Step 1.5.5.2.2.1

Multiply $λ^{2}$ by $1$ .

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

Step 1.5.5.2.2.2

Multiply $- 3 λ$ by $1$ .

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

Step 1.5.5.2.2.3

Multiply $1$ by $1$ .

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

Step 1.5.5.2.2.4

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

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Step 1.5.5.2.2.4.1

Move $λ^{2}$ .

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

Step 1.5.5.2.2.4.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.5.2.2.4.2.1

Raise $λ$ to the power of $1$ .

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

Step 1.5.5.2.2.4.2.2

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

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

Step 1.5.5.2.2.4.3

Add $2$ and $1$ .

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

Step 1.5.5.2.2.5

Rewrite using the commutative property of multiplication.

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

Step 1.5.5.2.2.6

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

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Step 1.5.5.2.2.6.1

Move $λ$ .

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

Step 1.5.5.2.2.6.2

Multiply $λ$ by $λ$ .

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

Step 1.5.5.2.2.7

Multiply $- 1$ by $- 3$ .

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

Step 1.5.5.2.2.8

Multiply $- 1$ by $1$ .

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

Step 1.5.5.2.3

Add $λ^{2}$ and $3 λ^{2}$ .

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

Step 1.5.5.2.4

Subtract $λ$ from $- 3 λ$ .

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

Step 1.5.5.2.5

Multiply $λ - 1$ by $1$ .

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

Step 1.5.5.3

Combine the opposite terms in $4 λ^{2} - 4 λ + 1 - λ^{3} + λ - 1$ .

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Step 1.5.5.3.1

Subtract $1$ from $1$ .

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

Step 1.5.5.3.2

Add $4 λ^{2} - 4 λ - λ^{3} + λ$ and $0$ .

$p (λ) = 4 λ^{2} - 4 λ - λ^{3} + λ$

Step 1.5.5.4

Add $- 4 λ$ and $λ$ .

$p (λ) = 4 λ^{2} - λ^{3} - 3 λ$

Step 1.5.5.5

Reorder $4 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 4 λ^{2} - 3 λ$

Step 1.6

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

$- λ^{3} + 4 λ^{2} - 3 λ = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Factor the left side of the equation.

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Step 1.7.1.1

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

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Step 1.7.1.1.1

Factor $- λ$ out of $- λ^{3}$ .

$- λ \cdot λ^{2} + 4 λ^{2} - 3 λ = 0$

Step 1.7.1.1.2

Factor $- λ$ out of $4 λ^{2}$ .

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

Step 1.7.1.1.3

Factor $- λ$ out of $- 3 λ$ .

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

Step 1.7.1.1.4

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

$- λ (λ^{2} - 4 λ) - λ \cdot 3 = 0$

Step 1.7.1.1.5

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

$- λ (λ^{2} - 4 λ + 3) = 0$

Step 1.7.1.2

Factor.

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Step 1.7.1.2.1

Factor $λ^{2} - 4 λ + 3$ using the AC method.

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Step 1.7.1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $- 4$ .

$- 3, - 1$

Step 1.7.1.2.1.2

Write the factored form using these integers.

$- λ ((λ - 3) (λ - 1)) = 0$

Step 1.7.1.2.2

Remove unnecessary parentheses.

$- λ (λ - 3) (λ - 1) = 0$

Step 1.7.2

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

$λ = 0$

$λ - 3 = 0$

$λ - 1 = 0$

Step 1.7.3

Set $λ$ equal to $0$ .

$λ = 0$

Step 1.7.4

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

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Step 1.7.4.1

Set $λ - 3$ equal to $0$ .

$λ - 3 = 0$

Step 1.7.4.2

Add $3$ to both sides of the equation.

$λ = 3$

Step 1.7.5

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

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Step 1.7.5.1

Set $λ - 1$ equal to $0$ .

$λ - 1 = 0$

Step 1.7.5.2

Add $1$ to both sides of the equation.

$λ = 1$

Step 1.7.6

The final solution is all the values that make $- λ (λ - 3) (λ - 1) = 0$ true.

$λ = 0, 3, 1$

Step 2

The eigenvector is equal to the null space of the matrix minus the eigenvalue times the identity matrix where $N$ is the null space and $I$ is the identity matrix.

$ε_{A} = N (A - λ I_{3})$

Step 3

Find the eigenvector using the eigenvalue $λ = 0$ .

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Step 3.1

Substitute the known values into the formula.

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

Step 3.2

Simplify.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply $0$ by each element of the matrix.

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

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $0$ by $1$ .

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

Step 3.2.1.2.2

Multiply $0$ by $0$ .

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

Step 3.2.1.2.3

Multiply $0$ by $0$ .

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

Step 3.2.1.2.4

Multiply $0$ by $0$ .

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

Step 3.2.1.2.5

Multiply $0$ by $1$ .

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

Step 3.2.1.2.6

Multiply $0$ by $0$ .

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

Step 3.2.1.2.7

Multiply $0$ by $0$ .

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

Step 3.2.1.2.8

Multiply $0$ by $0$ .

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

Step 3.2.1.2.9

Multiply $0$ by $1$ .

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

Step 3.2.2

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

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Step 3.2.2.1

Add the corresponding elements.

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

Step 3.2.2.2

Simplify each element.

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Step 3.2.2.2.1

Add $1$ and $0$ .

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

Step 3.2.2.2.2

Add $- 1$ and $0$ .

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

Step 3.2.2.2.3

Add $0$ and $0$ .

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

Step 3.2.2.2.4

Add $- 1$ and $0$ .

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

Step 3.2.2.2.5

Add $2$ and $0$ .

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

Step 3.2.2.2.6

Add $- 1$ and $0$ .

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

Step 3.2.2.2.7

Add $0$ and $0$ .

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

Step 3.2.2.2.8

Add $- 1$ and $0$ .

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

Step 3.2.2.2.9

Add $1$ and $0$ .

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

Step 3.3

Find the null space when $λ = 0$ .

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Step 3.3.1

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

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

Step 3.3.2

Find the reduced row echelon form.

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Step 3.3.2.1

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

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Step 3.3.2.1.1

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

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

Step 3.3.2.1.2

Simplify $R_{2}$ .

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

Step 3.3.2.2

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

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Step 3.3.2.2.1

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

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

Step 3.3.2.2.2

Simplify $R_{3}$ .

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

Step 3.3.2.3

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

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Step 3.3.2.3.1

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 3.3.2.3.2

Simplify $R_{1}$ .

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

Step 3.3.3

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

$x - z = 0$

$y - z = 0$

$0 = 0$

Step 3.3.4

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

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

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

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

Step 3.3.7

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

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

Step 4

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

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Step 4.1

Substitute the known values into the formula.

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

Step 4.2

Simplify.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

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

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

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $- 3$ by $1$ .

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

Step 4.2.1.2.2

Multiply $- 3$ by $0$ .

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

Step 4.2.1.2.3

Multiply $- 3$ by $0$ .

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

Step 4.2.1.2.4

Multiply $- 3$ by $0$ .

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

Step 4.2.1.2.5

Multiply $- 3$ by $1$ .

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

Step 4.2.1.2.6

Multiply $- 3$ by $0$ .

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

Step 4.2.1.2.7

Multiply $- 3$ by $0$ .

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

Step 4.2.1.2.8

Multiply $- 3$ by $0$ .

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

Step 4.2.1.2.9

Multiply $- 3$ by $1$ .

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

Step 4.2.2

Add the corresponding elements.

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

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Subtract $3$ from $1$ .

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

Step 4.2.3.2

Add $- 1$ and $0$ .

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

Step 4.2.3.3

Add $0$ and $0$ .

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

Step 4.2.3.4

Add $- 1$ and $0$ .

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

Step 4.2.3.5

Subtract $3$ from $2$ .

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

Step 4.2.3.6

Add $- 1$ and $0$ .

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

Step 4.2.3.7

Add $0$ and $0$ .

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

Step 4.2.3.8

Add $- 1$ and $0$ .

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

Step 4.2.3.9

Subtract $3$ from $1$ .

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

Step 4.3

Find the null space when $λ = 3$ .

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Step 4.3.1

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

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

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 $1, 1$ 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 $1, 1$ a $1$ .

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

Step 4.3.2.1.2

Simplify $R_{1}$ .

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

Step 4.3.2.2

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

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Step 4.3.2.2.1

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

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

Step 4.3.2.2.2

Simplify $R_{2}$ .

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

Step 4.3.2.3

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

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Step 4.3.2.3.1

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

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

Step 4.3.2.3.2

Simplify $R_{2}$ .

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

Step 4.3.2.4

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

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Step 4.3.2.4.1

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

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

Step 4.3.2.4.2

Simplify $R_{3}$ .

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

Step 4.3.2.5

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

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Step 4.3.2.5.1

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

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

Step 4.3.2.5.2

Simplify $R_{1}$ .

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

Step 4.3.3

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

$x - z = 0$

$y + 2 z = 0$

$0 = 0$

Step 4.3.4

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

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

Step 4.3.5

Write the solution as a linear combination of vectors.

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

Step 4.3.6

Write as a solution set.

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

Step 4.3.7

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

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

Step 5

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

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Step 5.1

Substitute the known values into the formula.

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

Step 5.2

Simplify.

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Step 5.2.1

Subtract the corresponding elements.

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

Step 5.2.2

Simplify each element.

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Step 5.2.2.1

Subtract $1$ from $1$ .

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

Step 5.2.2.2

Subtract $0$ from $- 1$ .

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

Step 5.2.2.3

Subtract $0$ from $0$ .

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

Step 5.2.2.4

Subtract $0$ from $- 1$ .

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

Step 5.2.2.5

Subtract $1$ from $2$ .

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

Step 5.2.2.6

Subtract $0$ from $- 1$ .

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

Step 5.2.2.7

Subtract $0$ from $0$ .

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

Step 5.2.2.8

Subtract $0$ from $- 1$ .

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

Step 5.2.2.9

Subtract $1$ from $1$ .

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

Step 5.3

Find the null space when $λ = 1$ .

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Step 5.3.1

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

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

Step 5.3.2

Find the reduced row echelon form.

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Step 5.3.2.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

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

Step 5.3.2.2

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

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Step 5.3.2.2.1

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

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

Step 5.3.2.2.2

Simplify $R_{1}$ .

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

Step 5.3.2.3

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

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Step 5.3.2.3.1

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

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

Step 5.3.2.3.2

Simplify $R_{2}$ .

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

Step 5.3.2.4

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

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Step 5.3.2.4.1

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

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

Step 5.3.2.4.2

Simplify $R_{3}$ .

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

Step 5.3.2.5

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

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Step 5.3.2.5.1

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

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

Step 5.3.2.5.2

Simplify $R_{1}$ .

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

Step 5.3.3

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

$x + z = 0$

$y = 0$

$0 = 0$

Step 5.3.4

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

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

Step 5.3.5

Write the solution as a linear combination of vectors.

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

Step 5.3.6

Write as a solution set.

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

Step 5.3.7

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

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

Step 6

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

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