Examples

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}]$

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 $[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] - λ I_{3})$

Step 1.3.2

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

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 3 - λ & 5 + 0 & 0 + 0 \\ 7 + 0 & 5 - λ & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $5$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 5 & 0 + 0 \\ 7 + 0 & 5 - λ & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 1.4.3.2

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 5 & 0 \\ 7 + 0 & 5 - λ & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 1.4.3.3

Add $7$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 5 & 0 \\ 7 & 5 - λ & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 1.4.3.4

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 5 & 0 \\ 7 & 5 - λ & 0 \\ 1 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 1.4.3.5

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 5 & 0 \\ 7 & 5 - λ & 0 \\ 1 & 1 + 0 & 0 - λ \end{array}]$

Step 1.4.3.6

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 5 & 0 \\ 7 & 5 - λ & 0 \\ 1 & 1 & 0 - λ \end{array}]$

Step 1.4.3.7

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} 3 - λ & 5 & 0 \\ 7 & 5 - λ & 0 \\ 1 & 1 & - λ \end{array}]$

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

$| \begin{array}{c} 7 & 5 - λ \\ 1 & 1 \end{array} |$

Step 1.5.1.4

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

$0 | \begin{array}{c} 7 & 5 - λ \\ 1 & 1 \end{array} |$

Step 1.5.1.5

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

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

Step 1.5.1.6

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

$0 | \begin{array}{c} 3 - λ & 5 \\ 1 & 1 \end{array} |$

Step 1.5.1.7

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

$| \begin{array}{c} 3 - λ & 5 \\ 7 & 5 - λ \end{array} |$

Step 1.5.1.8

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

$- λ | \begin{array}{c} 3 - λ & 5 \\ 7 & 5 - λ \end{array} |$

Step 1.5.1.9

Add the terms together.

$p (λ) = 0 | \begin{array}{c} 7 & 5 - λ \\ 1 & 1 \end{array} | + 0 | \begin{array}{c} 3 - λ & 5 \\ 1 & 1 \end{array} | - λ | \begin{array}{c} 3 - λ & 5 \\ 7 & 5 - λ \end{array} |$

Step 1.5.2

Multiply $0$ by $| \begin{array}{c} 7 & 5 - λ \\ 1 & 1 \end{array} |$ .

$p (λ) = 0 + 0 | \begin{array}{c} 3 - λ & 5 \\ 1 & 1 \end{array} | - λ | \begin{array}{c} 3 - λ & 5 \\ 7 & 5 - λ \end{array} |$

Step 1.5.3

Multiply $0$ by $| \begin{array}{c} 3 - λ & 5 \\ 1 & 1 \end{array} |$ .

$p (λ) = 0 + 0 - λ | \begin{array}{c} 3 - λ & 5 \\ 7 & 5 - λ \end{array} |$

Step 1.5.4

Evaluate $| \begin{array}{c} 3 - λ & 5 \\ 7 & 5 - λ \end{array} |$ .

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Step 1.5.4.1

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

$p (λ) = 0 + 0 - λ ((3 - λ) (5 - λ) - 7 \cdot 5)$

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 $(3 - λ) (5 - λ)$ using the FOIL Method.

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Step 1.5.4.2.1.1.1

Apply the distributive property.

$p (λ) = 0 + 0 - λ (3 (5 - λ) - λ (5 - λ) - 7 \cdot 5)$

Step 1.5.4.2.1.1.2

Apply the distributive property.

$p (λ) = 0 + 0 - λ (3 \cdot 5 + 3 (- λ) - λ (5 - λ) - 7 \cdot 5)$

Step 1.5.4.2.1.1.3

Apply the distributive property.

$p (λ) = 0 + 0 - λ (3 \cdot 5 + 3 (- λ) - λ \cdot 5 - λ (- λ) - 7 \cdot 5)$

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 $5$ .

$p (λ) = 0 + 0 - λ (15 + 3 (- λ) - λ \cdot 5 - λ (- λ) - 7 \cdot 5)$

Step 1.5.4.2.1.2.1.2

Multiply $- 1$ by $3$ .

$p (λ) = 0 + 0 - λ (15 - 3 λ - λ \cdot 5 - λ (- λ) - 7 \cdot 5)$

Step 1.5.4.2.1.2.1.3

Multiply $5$ by $- 1$ .

$p (λ) = 0 + 0 - λ (15 - 3 λ - 5 λ - λ (- λ) - 7 \cdot 5)$

Step 1.5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0 + 0 - λ (15 - 3 λ - 5 λ - 1 \cdot - 1 λ \cdot λ - 7 \cdot 5)$

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 $λ$ .

$p (λ) = 0 + 0 - λ (15 - 3 λ - 5 λ - 1 \cdot - 1 (λ \cdot λ) - 7 \cdot 5)$

Step 1.5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 + 0 - λ (15 - 3 λ - 5 λ - 1 \cdot - 1 λ^{2} - 7 \cdot 5)$

Step 1.5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 0 + 0 - λ (15 - 3 λ - 5 λ + 1 λ^{2} - 7 \cdot 5)$

Step 1.5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0 + 0 - λ (15 - 3 λ - 5 λ + λ^{2} - 7 \cdot 5)$

Step 1.5.4.2.1.2.2

Subtract $5 λ$ from $- 3 λ$ .

$p (λ) = 0 + 0 - λ (15 - 8 λ + λ^{2} - 7 \cdot 5)$

Step 1.5.4.2.1.3

Multiply $- 7$ by $5$ .

$p (λ) = 0 + 0 - λ (15 - 8 λ + λ^{2} - 35)$

Step 1.5.4.2.2

Subtract $35$ from $15$ .

$p (λ) = 0 + 0 - λ (- 8 λ + λ^{2} - 20)$

Step 1.5.4.2.3

Reorder $- 8 λ$ and $λ^{2}$ .

$p (λ) = 0 + 0 - λ (λ^{2} - 8 λ - 20)$

Step 1.5.5

Simplify the determinant.

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Step 1.5.5.1

Combine the opposite terms in $0 + 0 - λ (λ^{2} - 8 λ - 20)$ .

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Step 1.5.5.1.1

Add $0$ and $0$ .

$p (λ) = 0 - λ (λ^{2} - 8 λ - 20)$

Step 1.5.5.1.2

Subtract $λ (λ^{2} - 8 λ - 20)$ from $0$ .

$p (λ) = - λ (λ^{2} - 8 λ - 20)$

Step 1.5.5.2

Apply the distributive property.

$p (λ) = - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot - 20$

Step 1.5.5.3

Simplify.

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Step 1.5.5.3.1

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

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Step 1.5.5.3.1.1

Move $λ^{2}$ .

$p (λ) = - (λ^{2} λ) - λ (- 8 λ) - λ \cdot - 20$

Step 1.5.5.3.1.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.5.3.1.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - (λ^{2} λ^{1}) - λ (- 8 λ) - λ \cdot - 20$

Step 1.5.5.3.1.2.2

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

$p (λ) = - λ^{2 + 1} - λ (- 8 λ) - λ \cdot - 20$

Step 1.5.5.3.1.3

Add $2$ and $1$ .

$p (λ) = - λ^{3} - λ (- 8 λ) - λ \cdot - 20$

Step 1.5.5.3.2

Rewrite using the commutative property of multiplication.

$p (λ) = - λ^{3} - 1 \cdot - 8 λ \cdot λ - λ \cdot - 20$

Step 1.5.5.3.3

Multiply $- 20$ by $- 1$ .

$p (λ) = - λ^{3} - 1 \cdot - 8 λ \cdot λ + 20 λ$

Step 1.5.5.4

Simplify each term.

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Step 1.5.5.4.1

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

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Step 1.5.5.4.1.1

Move $λ$ .

$p (λ) = - λ^{3} - 1 \cdot - 8 (λ \cdot λ) + 20 λ$

Step 1.5.5.4.1.2

Multiply $λ$ by $λ$ .

$p (λ) = - λ^{3} - 1 \cdot - 8 λ^{2} + 20 λ$

Step 1.5.5.4.2

Multiply $- 1$ by $- 8$ .

$p (λ) = - λ^{3} + 8 λ^{2} + 20 λ$

Step 1.6

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

$- λ^{3} + 8 λ^{2} + 20 λ = 0$

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 $- λ^{3} + 8 λ^{2} + 20 λ$ .

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Step 1.7.1.1.1

Factor $- λ$ out of $- λ^{3}$ .

$- λ \cdot λ^{2} + 8 λ^{2} + 20 λ = 0$

Step 1.7.1.1.2

Factor $- λ$ out of $8 λ^{2}$ .

$- λ \cdot λ^{2} - λ (- 8 λ) + 20 λ = 0$

Step 1.7.1.1.3

Factor $- λ$ out of $20 λ$ .

$- λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot - 20 = 0$

Step 1.7.1.1.4

Factor $- λ$ out of $- λ (λ^{2}) - λ (- 8 λ)$ .

$- λ (λ^{2} - 8 λ) - λ \cdot - 20 = 0$

Step 1.7.1.1.5

Factor $- λ$ out of $- λ (λ^{2} - 8 λ) - λ (- 20)$ .

$- λ (λ^{2} - 8 λ - 20) = 0$

Step 1.7.1.2

Factor.

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Step 1.7.1.2.1

Factor $λ^{2} - 8 λ - 20$ 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 $- 20$ and whose sum is $- 8$ .

$- 10, 2$

Step 1.7.1.2.1.2

Write the factored form using these integers.

$- λ ((λ - 10) (λ + 2)) = 0$

Step 1.7.1.2.2

Remove unnecessary parentheses.

$- λ (λ - 10) (λ + 2) = 0$

Step 1.7.2

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

$λ = 0$

$λ - 10 = 0$

$λ + 2 = 0$

Step 1.7.3

Set $λ$ equal to $0$ .

$λ = 0$

Step 1.7.4

Set $λ - 10$ equal to $0$ and solve for $λ$ .

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Step 1.7.4.1

Set $λ - 10$ equal to $0$ .

$λ - 10 = 0$

Step 1.7.4.2

Add $10$ to both sides of the equation.

$λ = 10$

Step 1.7.5

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

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Step 1.7.5.1

Set $λ + 2$ equal to $0$ .

$λ + 2 = 0$

Step 1.7.5.2

Subtract $2$ from both sides of the equation.

$λ = - 2$

Step 1.7.6

The final solution is all the values that make $- λ (λ - 10) (λ + 2) = 0$ true.

$λ = 0, 10, - 2$

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.

$N ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + 0 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 \cdot 1 & 0 \cdot 0 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $0$ by $1$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 & 0 \cdot 0 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $0$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $0$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 \cdot 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $0$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 \cdot 1 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.5

Multiply $0$ by $1$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \cdot 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.6

Multiply $0$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 \cdot 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.7

Multiply $0$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 \cdot 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.8

Multiply $0$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \cdot 1 \end{array}]$

Step 3.2.1.2.9

Multiply $0$ by $1$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}]$

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.

$[\begin{array}{c} 3 + 0 & 5 + 0 & 0 + 0 \\ 7 + 0 & 5 + 0 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 + 0 \end{array}]$

Step 3.2.2.2

Simplify each element.

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Step 3.2.2.2.1

Add $3$ and $0$ .

$[\begin{array}{c} 3 & 5 + 0 & 0 + 0 \\ 7 + 0 & 5 + 0 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 + 0 \end{array}]$

Step 3.2.2.2.2

Add $5$ and $0$ .

$[\begin{array}{c} 3 & 5 & 0 + 0 \\ 7 + 0 & 5 + 0 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 + 0 \end{array}]$

Step 3.2.2.2.3

Add $0$ and $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 + 0 & 5 + 0 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 + 0 \end{array}]$

Step 3.2.2.2.4

Add $7$ and $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 + 0 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 + 0 \end{array}]$

Step 3.2.2.2.5

Add $5$ and $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 + 0 \end{array}]$

Step 3.2.2.2.6

Add $0$ and $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 + 0 & 1 + 0 & 0 + 0 \end{array}]$

Step 3.2.2.2.7

Add $1$ and $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 + 0 & 0 + 0 \end{array}]$

Step 3.2.2.2.8

Add $1$ and $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 + 0 \end{array}]$

Step 3.2.2.2.9

Add $0$ and $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}]$

Step 3.3

Find the null space when $λ = 0$ .

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Step 3.3.1

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

$[\begin{array}{c} 3 & 5 & 0 & 0 \\ 7 & 5 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array}]$

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 $1$ .

$[\begin{array}{c} \frac{3}{3} & \frac{5}{3} & \frac{0}{3} & \frac{0}{3} \\ 7 & 5 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 0 & 0 \\ 7 & 5 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.3.2.2.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 0 & 0 \\ 7 - 7 \cdot 1 & 5 - 7 (\frac{5}{3}) & 0 - 7 \cdot 0 & 0 - 7 \cdot 0 \\ 1 & 1 & 0 & 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 0 & 0 \\ 0 & - \frac{20}{3} & 0 & 0 \\ 1 & 1 & 0 & 0 \end{array}]$

Step 3.3.2.3

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

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Step 3.3.2.3.1

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

$[\begin{array}{c} 1 & \frac{5}{3} & 0 & 0 \\ 0 & - \frac{20}{3} & 0 & 0 \\ 1 - 1 & 1 - \frac{5}{3} & 0 - 0 & 0 - 0 \end{array}]$

Step 3.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 0 & 0 \\ 0 & - \frac{20}{3} & 0 & 0 \\ 0 & - \frac{2}{3} & 0 & 0 \end{array}]$

Step 3.3.2.4

Multiply each element of $R_{2}$ by $- \frac{3}{20}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.2.4.1

Multiply each element of $R_{2}$ by $- \frac{3}{20}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{5}{3} & 0 & 0 \\ - \frac{3}{20} \cdot 0 & - \frac{3}{20} (- \frac{20}{3}) & - \frac{3}{20} \cdot 0 & - \frac{3}{20} \cdot 0 \\ 0 & - \frac{2}{3} & 0 & 0 \end{array}]$

Step 3.3.2.4.2

Simplify $R_{2}$ .

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

Step 3.3.2.5

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

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Step 3.3.2.5.1

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

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

Step 3.3.2.5.2

Simplify $R_{3}$ .

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

Step 3.3.2.6

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

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Step 3.3.2.6.1

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

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

Step 3.3.2.6.2

Simplify $R_{1}$ .

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

Step 3.3.3

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

$x = 0$

$y = 0$

$0 = 0$

Step 3.3.4

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

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

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

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

Step 3.3.7

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

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

Step 4

Find the eigenvector using the eigenvalue $λ = 10$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] - 10 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $- 10$ by each element of the matrix.

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 \cdot 1 & - 10 \cdot 0 & - 10 \cdot 0 \\ - 10 \cdot 0 & - 10 \cdot 1 & - 10 \cdot 0 \\ - 10 \cdot 0 & - 10 \cdot 0 & - 10 \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $- 10$ by $1$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 & - 10 \cdot 0 & - 10 \cdot 0 \\ - 10 \cdot 0 & - 10 \cdot 1 & - 10 \cdot 0 \\ - 10 \cdot 0 & - 10 \cdot 0 & - 10 \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $- 10$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 & 0 & - 10 \cdot 0 \\ - 10 \cdot 0 & - 10 \cdot 1 & - 10 \cdot 0 \\ - 10 \cdot 0 & - 10 \cdot 0 & - 10 \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $- 10$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 & 0 & 0 \\ - 10 \cdot 0 & - 10 \cdot 1 & - 10 \cdot 0 \\ - 10 \cdot 0 & - 10 \cdot 0 & - 10 \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $- 10$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 & 0 & 0 \\ 0 & - 10 \cdot 1 & - 10 \cdot 0 \\ - 10 \cdot 0 & - 10 \cdot 0 & - 10 \cdot 1 \end{array}]$

Step 4.2.1.2.5

Multiply $- 10$ by $1$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 & 0 & 0 \\ 0 & - 10 & - 10 \cdot 0 \\ - 10 \cdot 0 & - 10 \cdot 0 & - 10 \cdot 1 \end{array}]$

Step 4.2.1.2.6

Multiply $- 10$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 & 0 & 0 \\ 0 & - 10 & 0 \\ - 10 \cdot 0 & - 10 \cdot 0 & - 10 \cdot 1 \end{array}]$

Step 4.2.1.2.7

Multiply $- 10$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 & 0 & 0 \\ 0 & - 10 & 0 \\ 0 & - 10 \cdot 0 & - 10 \cdot 1 \end{array}]$

Step 4.2.1.2.8

Multiply $- 10$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 & 0 & 0 \\ 0 & - 10 & 0 \\ 0 & 0 & - 10 \cdot 1 \end{array}]$

Step 4.2.1.2.9

Multiply $- 10$ by $1$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} - 10 & 0 & 0 \\ 0 & - 10 & 0 \\ 0 & 0 & - 10 \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} 3 - 10 & 5 + 0 & 0 + 0 \\ 7 + 0 & 5 - 10 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - 10 \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

Subtract $10$ from $3$ .

$[\begin{array}{c} - 7 & 5 + 0 & 0 + 0 \\ 7 + 0 & 5 - 10 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - 10 \end{array}]$

Step 4.2.3.2

Add $5$ and $0$ .

$[\begin{array}{c} - 7 & 5 & 0 + 0 \\ 7 + 0 & 5 - 10 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - 10 \end{array}]$

Step 4.2.3.3

Add $0$ and $0$ .

$[\begin{array}{c} - 7 & 5 & 0 \\ 7 + 0 & 5 - 10 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - 10 \end{array}]$

Step 4.2.3.4

Add $7$ and $0$ .

$[\begin{array}{c} - 7 & 5 & 0 \\ 7 & 5 - 10 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - 10 \end{array}]$

Step 4.2.3.5

Subtract $10$ from $5$ .

$[\begin{array}{c} - 7 & 5 & 0 \\ 7 & - 5 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 - 10 \end{array}]$

Step 4.2.3.6

Add $0$ and $0$ .

$[\begin{array}{c} - 7 & 5 & 0 \\ 7 & - 5 & 0 \\ 1 + 0 & 1 + 0 & 0 - 10 \end{array}]$

Step 4.2.3.7

Add $1$ and $0$ .

$[\begin{array}{c} - 7 & 5 & 0 \\ 7 & - 5 & 0 \\ 1 & 1 + 0 & 0 - 10 \end{array}]$

Step 4.2.3.8

Add $1$ and $0$ .

$[\begin{array}{c} - 7 & 5 & 0 \\ 7 & - 5 & 0 \\ 1 & 1 & 0 - 10 \end{array}]$

Step 4.2.3.9

Subtract $10$ from $0$ .

$[\begin{array}{c} - 7 & 5 & 0 \\ 7 & - 5 & 0 \\ 1 & 1 & - 10 \end{array}]$

Step 4.3

Find the null space when $λ = 10$ .

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Step 4.3.1

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

$[\begin{array}{c} - 7 & 5 & 0 & 0 \\ 7 & - 5 & 0 & 0 \\ 1 & 1 & - 10 & 0 \end{array}]$

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 $- \frac{1}{7}$ 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 $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{7} \cdot - 7 & - \frac{1}{7} \cdot 5 & - \frac{1}{7} \cdot 0 & - \frac{1}{7} \cdot 0 \\ 7 & - 5 & 0 & 0 \\ 1 & 1 & - 10 & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{5}{7} & 0 & 0 \\ 7 & - 5 & 0 & 0 \\ 1 & 1 & - 10 & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.3.2.2.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{5}{7} & 0 & 0 \\ 7 - 7 \cdot 1 & - 5 - 7 (- \frac{5}{7}) & 0 - 7 \cdot 0 & 0 - 7 \cdot 0 \\ 1 & 1 & - 10 & 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{7} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & - 10 & 0 \end{array}]$

Step 4.3.2.3

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

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Step 4.3.2.3.1

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

$[\begin{array}{c} 1 & - \frac{5}{7} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 - 1 & 1 + \frac{5}{7} & - 10 - 0 & 0 - 0 \end{array}]$

Step 4.3.2.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{5}{7} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & \frac{12}{7} & - 10 & 0 \end{array}]$

Step 4.3.2.4

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

$[\begin{array}{c} 1 & - \frac{5}{7} & 0 & 0 \\ 0 & \frac{12}{7} & - 10 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.5

Multiply each element of $R_{2}$ by $\frac{7}{12}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.2.5.1

Multiply each element of $R_{2}$ by $\frac{7}{12}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{5}{7} & 0 & 0 \\ \frac{7}{12} \cdot 0 & \frac{7}{12} \cdot \frac{12}{7} & \frac{7}{12} \cdot - 10 & \frac{7}{12} \cdot 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{5}{7} & 0 & 0 \\ 0 & 1 & - \frac{35}{6} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.6

Perform the row operation $R_{1} = R_{1} + \frac{5}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 4.3.2.6.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{5}{7} \cdot 0 & - \frac{5}{7} + \frac{5}{7} \cdot 1 & 0 + \frac{5}{7} (- \frac{35}{6}) & 0 + \frac{5}{7} \cdot 0 \\ 0 & 1 & - \frac{35}{6} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.2.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{25}{6} & 0 \\ 0 & 1 & - \frac{35}{6} & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 4.3.3

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

$x - \frac{25}{6} z = 0$

$y - \frac{35}{6} z = 0$

$0 = 0$

Step 4.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} \frac{25 z}{6} \\ \frac{35 z}{6} \\ z \end{array}]$

Step 4.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} \frac{25}{6} \\ \frac{35}{6} \\ 1 \end{array}]$

Step 4.3.6

Write as a solution set.

${z [\begin{array}{c} \frac{25}{6} \\ \frac{35}{6} \\ 1 \end{array}] | z \in R}$

Step 4.3.7

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

${[\begin{array}{c} \frac{25}{6} \\ \frac{35}{6} \\ 1 \end{array}]}$

Step 5

Find the eigenvector using the eigenvalue $λ = - 2$ .

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Step 5.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + 2 [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $2$ by each element of the matrix.

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 \cdot 1 & 2 \cdot 0 & 2 \cdot 0 \\ 2 \cdot 0 & 2 \cdot 1 & 2 \cdot 0 \\ 2 \cdot 0 & 2 \cdot 0 & 2 \cdot 1 \end{array}]$

Step 5.2.1.2

Simplify each element in the matrix.

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Step 5.2.1.2.1

Multiply $2$ by $1$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 & 2 \cdot 0 & 2 \cdot 0 \\ 2 \cdot 0 & 2 \cdot 1 & 2 \cdot 0 \\ 2 \cdot 0 & 2 \cdot 0 & 2 \cdot 1 \end{array}]$

Step 5.2.1.2.2

Multiply $2$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 & 0 & 2 \cdot 0 \\ 2 \cdot 0 & 2 \cdot 1 & 2 \cdot 0 \\ 2 \cdot 0 & 2 \cdot 0 & 2 \cdot 1 \end{array}]$

Step 5.2.1.2.3

Multiply $2$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 & 0 & 0 \\ 2 \cdot 0 & 2 \cdot 1 & 2 \cdot 0 \\ 2 \cdot 0 & 2 \cdot 0 & 2 \cdot 1 \end{array}]$

Step 5.2.1.2.4

Multiply $2$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 & 0 & 0 \\ 0 & 2 \cdot 1 & 2 \cdot 0 \\ 2 \cdot 0 & 2 \cdot 0 & 2 \cdot 1 \end{array}]$

Step 5.2.1.2.5

Multiply $2$ by $1$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 & 0 & 0 \\ 0 & 2 & 2 \cdot 0 \\ 2 \cdot 0 & 2 \cdot 0 & 2 \cdot 1 \end{array}]$

Step 5.2.1.2.6

Multiply $2$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 \cdot 0 & 2 \cdot 0 & 2 \cdot 1 \end{array}]$

Step 5.2.1.2.7

Multiply $2$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 2 \cdot 0 & 2 \cdot 1 \end{array}]$

Step 5.2.1.2.8

Multiply $2$ by $0$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \cdot 1 \end{array}]$

Step 5.2.1.2.9

Multiply $2$ by $1$ .

$[\begin{array}{c} 3 & 5 & 0 \\ 7 & 5 & 0 \\ 1 & 1 & 0 \end{array}] + [\begin{array}{c} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}]$

Step 5.2.2

Add the corresponding elements.

$[\begin{array}{c} 3 + 2 & 5 + 0 & 0 + 0 \\ 7 + 0 & 5 + 2 & 0 + 0 \\ 1 + 0 & 1 + 0 & 0 + 2 \end{array}]$

Step 5.2.3

Simplify each element.

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Step 5.2.3.1

Add $3$ and $2$ .

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

Step 5.2.3.2

Add $5$ and $0$ .

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

Step 5.2.3.3

Add $0$ and $0$ .

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

Step 5.2.3.4

Add $7$ and $0$ .

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

Step 5.2.3.5

Add $5$ and $2$ .

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

Step 5.2.3.6

Add $0$ and $0$ .

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

Step 5.2.3.7

Add $1$ and $0$ .

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

Step 5.2.3.8

Add $1$ and $0$ .

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

Step 5.2.3.9

Add $0$ and $2$ .

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

Step 5.3

Find the null space when $λ = - 2$ .

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Step 5.3.1

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

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

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 $\frac{1}{5}$ 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 $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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

Step 5.3.2.1.2

Simplify $R_{1}$ .

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

Step 5.3.2.2

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 5.3.2.2.1

Perform the row operation $R_{2} = R_{2} - 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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

Step 5.3.2.2.2

Simplify $R_{2}$ .

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

Step 5.3.2.3

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

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Step 5.3.2.3.1

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

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

Step 5.3.2.3.2

Simplify $R_{3}$ .

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

Step 5.3.2.4

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

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

Step 5.3.2.5

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

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Step 5.3.2.5.1

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

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

Step 5.3.2.5.2

Simplify $R_{2}$ .

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

Step 5.3.3

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

$x + y = 0$

$z = 0$

$0 = 0$

Step 5.3.4

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

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

Step 5.3.5

Write the solution as a linear combination of vectors.

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

Step 5.3.6

Write as a solution set.

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

Step 5.3.7

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

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

Step 6

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

${[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}], [\begin{array}{c} \frac{25}{6} \\ \frac{35}{6} \\ 1 \end{array}], [\begin{array}{c} - 1 \\ 1 \\ 0 \end{array}]}$

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