Linear Algebra Examples

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}]$

Step 1

Find the eigenvalues.

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Step 1.1

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

$p (λ) = determinant (A - λ I_{3})$

Step 1.2

The identity matrix or unit matrix of size $3$ is the $3 \times 3$ square matrix with ones on the main diagonal and zeros elsewhere.

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

Step 1.3

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

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Step 1.3.1

Substitute $[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] - λ I_{3})$

Step 1.3.2

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

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 1.4

Simplify.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

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

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2

Simplify each element in the matrix.

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Step 1.4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 1.4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 1.4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 1.4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - λ & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - λ \end{array}]$

Step 1.4.3

Simplify each element.

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Step 1.4.3.1

Add $0.2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.2 & 0.2 + 0 \\ 0 + 0 & 0.5 - λ & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - λ \end{array}]$

Step 1.4.3.2

Add $0.2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.2 & 0.2 \\ 0 + 0 & 0.5 - λ & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - λ \end{array}]$

Step 1.4.3.3

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.2 & 0.2 \\ 0 & 0.5 - λ & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - λ \end{array}]$

Step 1.4.3.4

Add $0.1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.2 & 0.2 \\ 0 & 0.5 - λ & 0.1 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - λ \end{array}]$

Step 1.4.3.5

Add $0.2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.2 & 0.2 \\ 0 & 0.5 - λ & 0.1 \\ 0.2 & 0.3 + 0 & 0.7 - λ \end{array}]$

Step 1.4.3.6

Add $0.3$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 0.8 - λ & 0.2 & 0.2 \\ 0 & 0.5 - λ & 0.1 \\ 0.2 & 0.3 & 0.7 - λ \end{array}]$

Step 1.5

Find the determinant.

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Step 1.5.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in 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.

$| \begin{array}{c} 0.5 - λ & 0.1 \\ 0.3 & 0.7 - λ \end{array} |$

Step 1.5.1.4

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

$(0.8 - λ) | \begin{array}{c} 0.5 - λ & 0.1 \\ 0.3 & 0.7 - λ \end{array} |$

Step 1.5.1.5

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

$| \begin{array}{c} 0.2 & 0.2 \\ 0.3 & 0.7 - λ \end{array} |$

Step 1.5.1.6

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

$0 | \begin{array}{c} 0.2 & 0.2 \\ 0.3 & 0.7 - λ \end{array} |$

Step 1.5.1.7

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

$| \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.1.8

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

$0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.1.9

Add the terms together.

$p (λ) = (0.8 - λ) | \begin{array}{c} 0.5 - λ & 0.1 \\ 0.3 & 0.7 - λ \end{array} | + 0 | \begin{array}{c} 0.2 & 0.2 \\ 0.3 & 0.7 - λ \end{array} | + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.2

Multiply $0$ by $| \begin{array}{c} 0.2 & 0.2 \\ 0.3 & 0.7 - λ \end{array} |$ .

$p (λ) = (0.8 - λ) | \begin{array}{c} 0.5 - λ & 0.1 \\ 0.3 & 0.7 - λ \end{array} | + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3

Evaluate $| \begin{array}{c} 0.5 - λ & 0.1 \\ 0.3 & 0.7 - λ \end{array} |$ .

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Step 1.5.3.1

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

$p (λ) = (0.8 - λ) ((0.5 - λ) (0.7 - λ) - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2

Simplify the determinant.

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Step 1.5.3.2.1

Simplify each term.

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Step 1.5.3.2.1.1

Expand $(0.5 - λ) (0.7 - λ)$ using the FOIL Method.

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Step 1.5.3.2.1.1.1

Apply the distributive property.

$p (λ) = (0.8 - λ) (0.5 (0.7 - λ) - λ (0.7 - λ) - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.1.2

Apply the distributive property.

$p (λ) = (0.8 - λ) (0.5 \cdot 0.7 + 0.5 (- λ) - λ (0.7 - λ) - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.1.3

Apply the distributive property.

$p (λ) = (0.8 - λ) (0.5 \cdot 0.7 + 0.5 (- λ) - λ \cdot 0.7 - λ (- λ) - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

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 $0.5$ by $0.7$ .

$p (λ) = (0.8 - λ) (0.35 + 0.5 (- λ) - λ \cdot 0.7 - λ (- λ) - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.2.1.2

Multiply $- 1$ by $0.5$ .

$p (λ) = (0.8 - λ) (0.35 - 0.5 λ - λ \cdot 0.7 - λ (- λ) - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.2.1.3

Multiply $0.7$ by $- 1$ .

$p (λ) = (0.8 - λ) (0.35 - 0.5 λ - 0.7 λ - λ (- λ) - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = (0.8 - λ) (0.35 - 0.5 λ - 0.7 λ - 1 \cdot - 1 λ \cdot λ - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

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 $λ$ .

$p (λ) = (0.8 - λ) (0.35 - 0.5 λ - 0.7 λ - 1 \cdot - 1 (λ \cdot λ) - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = (0.8 - λ) (0.35 - 0.5 λ - 0.7 λ - 1 \cdot - 1 λ^{2} - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = (0.8 - λ) (0.35 - 0.5 λ - 0.7 λ + 1 λ^{2} - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = (0.8 - λ) (0.35 - 0.5 λ - 0.7 λ + λ^{2} - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.2.2

Subtract $0.7 λ$ from $- 0.5 λ$ .

$p (λ) = (0.8 - λ) (0.35 - 1.2 λ + λ^{2} - 0.3 \cdot 0.1) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.1.3

Multiply $- 0.3$ by $0.1$ .

$p (λ) = (0.8 - λ) (0.35 - 1.2 λ + λ^{2} - 0.03) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.2

Subtract $0.03$ from $0.35$ .

$p (λ) = (0.8 - λ) (- 1.2 λ + λ^{2} + 0.32) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.3.2.3

Reorder $- 1.2 λ$ and $λ^{2}$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 | \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$

Step 1.5.4

Evaluate $| \begin{array}{c} 0.2 & 0.2 \\ 0.5 - λ & 0.1 \end{array} |$ .

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Step 1.5.4.1

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

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.2 \cdot 0.1 - (0.5 - λ) \cdot 0.2)$

Step 1.5.4.2

Simplify the determinant.

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Step 1.5.4.2.1

Simplify each term.

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Step 1.5.4.2.1.1

Multiply $0.2$ by $0.1$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.02 - (0.5 - λ) \cdot 0.2)$

Step 1.5.4.2.1.2

Apply the distributive property.

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.02 + (- 1 \cdot 0.5 - - λ) \cdot 0.2)$

Step 1.5.4.2.1.3

Multiply $- 1$ by $0.5$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.02 + (- 0.5 - - λ) \cdot 0.2)$

Step 1.5.4.2.1.4

Multiply $- - λ$ .

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Step 1.5.4.2.1.4.1

Multiply $- 1$ by $- 1$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.02 + (- 0.5 + 1 λ) \cdot 0.2)$

Step 1.5.4.2.1.4.2

Multiply $λ$ by $1$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.02 + (- 0.5 + λ) \cdot 0.2)$

Step 1.5.4.2.1.5

Apply the distributive property.

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.02 - 0.5 \cdot 0.2 + λ \cdot 0.2)$

Step 1.5.4.2.1.6

Multiply $- 0.5$ by $0.2$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.02 - 0.1 + λ \cdot 0.2)$

Step 1.5.4.2.1.7

Move $0.2$ to the left of $λ$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.02 - 0.1 + 0.2 λ)$

Step 1.5.4.2.2

Subtract $0.1$ from $0.02$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (- 0.08 + 0.2 λ)$

Step 1.5.4.2.3

Reorder $- 0.08$ and $0.2 λ$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5

Simplify the determinant.

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Step 1.5.5.1

Add $(0.8 - λ) (λ^{2} - 1.2 λ + 0.32)$ and $0$ .

$p (λ) = (0.8 - λ) (λ^{2} - 1.2 λ + 0.32) + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2

Simplify each term.

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Step 1.5.5.2.1

Expand $(0.8 - λ) (λ^{2} - 1.2 λ + 0.32)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 0.8 λ^{2} + 0.8 (- 1.2 λ) + 0.8 \cdot 0.32 - λ \cdot λ^{2} - λ (- 1.2 λ) - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2

Simplify each term.

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Step 1.5.5.2.2.1

Multiply $- 1.2$ by $0.8$ .

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.8 \cdot 0.32 - λ \cdot λ^{2} - λ (- 1.2 λ) - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.2

Multiply $0.8$ by $0.32$ .

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - λ \cdot λ^{2} - λ (- 1.2 λ) - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.3

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

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Step 1.5.5.2.2.3.1

Move $λ^{2}$ .

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - (λ^{2} λ) - λ (- 1.2 λ) - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 1.5.5.2.2.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - (λ^{2} λ^{1}) - λ (- 1.2 λ) - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.3.2.2

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

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - λ^{2 + 1} - λ (- 1.2 λ) - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.3.3

Add $2$ and $1$ .

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - λ^{3} - λ (- 1.2 λ) - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - λ^{3} - 1 \cdot - 1.2 λ \cdot λ - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.5

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

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Step 1.5.5.2.2.5.1

Move $λ$ .

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - λ^{3} - 1 \cdot - 1.2 (λ \cdot λ) - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - λ^{3} - 1 \cdot - 1.2 λ^{2} - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.6

Multiply $- 1$ by $- 1.2$ .

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - λ^{3} + 1.2 λ^{2} - λ \cdot 0.32 + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.2.7

Multiply $0.32$ by $- 1$ .

$p (λ) = 0.8 λ^{2} - 0.96 λ + 0.256 - λ^{3} + 1.2 λ^{2} - 0.32 λ + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.3

Add $0.8 λ^{2}$ and $1.2 λ^{2}$ .

$p (λ) = 2 λ^{2} - 0.96 λ + 0.256 - λ^{3} - 0.32 λ + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.4

Subtract $0.32 λ$ from $- 0.96 λ$ .

$p (λ) = 2 λ^{2} - 1.28 λ + 0.256 - λ^{3} + 0.2 (0.2 λ - 0.08)$

Step 1.5.5.2.5

Apply the distributive property.

$p (λ) = 2 λ^{2} - 1.28 λ + 0.256 - λ^{3} + 0.2 (0.2 λ) + 0.2 \cdot - 0.08$

Step 1.5.5.2.6

Multiply $0.2$ by $0.2$ .

$p (λ) = 2 λ^{2} - 1.28 λ + 0.256 - λ^{3} + 0.04 λ + 0.2 \cdot - 0.08$

Step 1.5.5.2.7

Multiply $0.2$ by $- 0.08$ .

$p (λ) = 2 λ^{2} - 1.28 λ + 0.256 - λ^{3} + 0.04 λ - 0.016$

Step 1.5.5.3

Add $- 1.28 λ$ and $0.04 λ$ .

$p (λ) = 2 λ^{2} - 1.24 λ + 0.256 - λ^{3} - 0.016$

Step 1.5.5.4

Subtract $0.016$ from $0.256$ .

$p (λ) = 2 λ^{2} - 1.24 λ - λ^{3} + 0.24$

Step 1.5.5.5

Move $- 1.24 λ$ .

$p (λ) = 2 λ^{2} - λ^{3} - 1.24 λ + 0.24$

Step 1.5.5.6

Reorder $2 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 2 λ^{2} - 1.24 λ + 0.24$

Step 1.6

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

$- λ^{3} + 2 λ^{2} - 1.24 λ + 0.24 = 0$

Step 1.7

Solve for $λ$ .

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Step 1.7.1

Graph each side of the equation. The solution is the x-value of the point of intersection.

$λ = \frac{2}{5}, \frac{3}{5}, 1$

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 $λ = \frac{2}{5}$ .

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Step 3.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] - \frac{2}{5} [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $- \frac{2}{5}$ by each element of the matrix.

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} \cdot 1 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2

Simplify each element in the matrix.

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Step 3.2.1.2.1

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.2

Multiply $- \frac{2}{5} \cdot 0$ .

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Step 3.2.1.2.2.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 (\frac{2}{5}) & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.2.2

Multiply $0$ by $\frac{2}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.3

Multiply $- \frac{2}{5} \cdot 0$ .

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Step 3.2.1.2.3.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 (\frac{2}{5}) \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.3.2

Multiply $0$ by $\frac{2}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.4

Multiply $- \frac{2}{5} \cdot 0$ .

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Step 3.2.1.2.4.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 (\frac{2}{5}) & - \frac{2}{5} \cdot 1 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.4.2

Multiply $0$ by $\frac{2}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 & - \frac{2}{5} \cdot 1 & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.5

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 & - \frac{2}{5} & - \frac{2}{5} \cdot 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.6

Multiply $- \frac{2}{5} \cdot 0$ .

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Step 3.2.1.2.6.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 & - \frac{2}{5} & 0 (\frac{2}{5}) \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.6.2

Multiply $0$ by $\frac{2}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 & - \frac{2}{5} & 0 \\ - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.7

Multiply $- \frac{2}{5} \cdot 0$ .

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Step 3.2.1.2.7.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 & - \frac{2}{5} & 0 \\ 0 (\frac{2}{5}) & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.7.2

Multiply $0$ by $\frac{2}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 & - \frac{2}{5} & 0 \\ 0 & - \frac{2}{5} \cdot 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.8

Multiply $- \frac{2}{5} \cdot 0$ .

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Step 3.2.1.2.8.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 & - \frac{2}{5} & 0 \\ 0 & 0 (\frac{2}{5}) & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.8.2

Multiply $0$ by $\frac{2}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 & - \frac{2}{5} & 0 \\ 0 & 0 & - \frac{2}{5} \cdot 1 \end{array}]$

Step 3.2.1.2.9

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{2}{5} & 0 & 0 \\ 0 & - \frac{2}{5} & 0 \\ 0 & 0 & - \frac{2}{5} \end{array}]$

Step 3.2.2

Add the corresponding elements.

$[\begin{array}{c} 0.8 - \frac{2}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3

Simplify each element.

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Step 3.2.3.1

To write $0.8$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$[\begin{array}{c} 0.8 \cdot \frac{5}{5} - \frac{2}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.2

Combine $0.8$ and $\frac{5}{5}$ .

$[\begin{array}{c} \frac{0.8 \cdot 5}{5} - \frac{2}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.3

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{0.8 \cdot 5 - 2}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.4

Simplify the numerator.

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Step 3.2.3.4.1

Multiply $0.8$ by $5$ .

$[\begin{array}{c} \frac{4 - 2}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.4.2

Subtract $2$ from $4$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.5

Cancel the common factor of $2$ and $5$ .

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Step 3.2.3.5.1

Rewrite $2$ as $1 (2)$ .

$[\begin{array}{c} \frac{1 (2)}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.5.2

Cancel the common factors.

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Step 3.2.3.5.2.1

Rewrite $5$ as $1 (5)$ .

$[\begin{array}{c} \frac{1 \cdot 2}{1 \cdot 5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.5.2.2

Cancel the common factor.

$[\begin{array}{c} \frac{1 \cdot 2}{1 \cdot 5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.5.2.3

Rewrite the expression.

$[\begin{array}{c} \frac{2}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.6

Add $0.2$ and $0$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.7

Add $0.2$ and $0$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 + 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.8

Add $0$ and $0$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.5 - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.9

To write $0.5$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.5 \cdot \frac{5}{5} - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.10

Combine $0.5$ and $\frac{5}{5}$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & \frac{0.5 \cdot 5}{5} - \frac{2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.11

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & \frac{0.5 \cdot 5 - 2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.12

Simplify the numerator.

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Step 3.2.3.12.1

Multiply $0.5$ by $5$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & \frac{2.5 - 2}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.12.2

Subtract $2$ from $2.5$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & \frac{0.5}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.13

Divide $0.5$ by $5$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.14

Add $0.1$ and $0$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.15

Add $0.2$ and $0$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 \\ 0.2 & 0.3 + 0 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.16

Add $0.3$ and $0$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 \\ 0.2 & 0.3 & 0.7 - \frac{2}{5} \end{array}]$

Step 3.2.3.17

To write $0.7$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 \\ 0.2 & 0.3 & 0.7 \cdot \frac{5}{5} - \frac{2}{5} \end{array}]$

Step 3.2.3.18

Combine $0.7$ and $\frac{5}{5}$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 \\ 0.2 & 0.3 & \frac{0.7 \cdot 5}{5} - \frac{2}{5} \end{array}]$

Step 3.2.3.19

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 \\ 0.2 & 0.3 & \frac{0.7 \cdot 5 - 2}{5} \end{array}]$

Step 3.2.3.20

Simplify the numerator.

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Step 3.2.3.20.1

Multiply $0.7$ by $5$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 \\ 0.2 & 0.3 & \frac{3.5 - 2}{5} \end{array}]$

Step 3.2.3.20.2

Subtract $2$ from $3.5$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 \\ 0.2 & 0.3 & \frac{1.5}{5} \end{array}]$

Step 3.2.3.21

Divide $1.5$ by $5$ .

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 \\ 0 & 0.1 & 0.1 \\ 0.2 & 0.3 & 0.3 \end{array}]$

Step 3.3

Find the null space when $λ = \frac{2}{5}$ .

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Step 3.3.1

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

$[\begin{array}{c} \frac{2}{5} & 0.2 & 0.2 & 0 \\ 0 & 0.1 & 0.1 & 0 \\ 0.2 & 0.3 & 0.3 & 0 \end{array}]$

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 $\frac{5}{2}$ 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 $\frac{5}{2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{2} \cdot \frac{2}{5} & \frac{5}{2} \cdot 0.2 & \frac{5}{2} \cdot 0.2 & \frac{5}{2} \cdot 0 \\ 0 & 0.1 & 0.1 & 0 \\ 0.2 & 0.3 & 0.3 & 0 \end{array}]$

Step 3.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0.1 & 0.1 & 0 \\ 0.2 & 0.3 & 0.3 & 0 \end{array}]$

Step 3.3.2.2

Perform the row operation $R_{3} = R_{3} - 0.2 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.2.2.1

Perform the row operation $R_{3} = R_{3} - 0.2 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0.1 & 0.1 & 0 \\ 0.2 - 0.2 \cdot 1 & 0.3 - 0.2 (\frac{1}{2}) & 0.3 - 0.2 (\frac{1}{2}) & 0 - 0.2 \cdot 0 \end{array}]$

Step 3.3.2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0.1 & 0.1 & 0 \\ 0 & \frac{1}{5} & \frac{1}{5} & 0 \end{array}]$

Step 3.3.2.3

Multiply each element of $R_{2}$ by $\frac{1}{0.1}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.2.3.1

Multiply each element of $R_{2}$ by $\frac{1}{0.1}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{2} & \frac{1}{2} & 0 \\ \frac{0}{0.1} & \frac{0.1}{0.1} & \frac{0.1}{0.1} & \frac{0}{0.1} \\ 0 & \frac{1}{5} & \frac{1}{5} & 0 \end{array}]$

Step 3.3.2.3.2

Simplify $R_{2}$ .

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

Step 3.3.2.4

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

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Step 3.3.2.4.1

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

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

Step 3.3.2.4.2

Simplify $R_{3}$ .

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

Step 3.3.2.5

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 3.3.2.5.1

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

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

Step 3.3.2.5.2

Simplify $R_{1}$ .

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

Step 3.3.3

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

$x = 0$

$y + z = 0$

$0 = 0$

Step 3.3.4

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

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

Step 3.3.5

Write the solution as a linear combination of vectors.

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

Step 3.3.6

Write as a solution set.

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

Step 3.3.7

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

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

Step 4

Find the eigenvector using the eigenvalue $λ = \frac{3}{5}$ .

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Step 4.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] - \frac{3}{5} [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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 $- \frac{3}{5}$ by each element of the matrix.

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} \cdot 1 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2

Simplify each element in the matrix.

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Step 4.2.1.2.1

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.2

Multiply $- \frac{3}{5} \cdot 0$ .

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Step 4.2.1.2.2.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 (\frac{3}{5}) & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.2.2

Multiply $0$ by $\frac{3}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.3

Multiply $- \frac{3}{5} \cdot 0$ .

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Step 4.2.1.2.3.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 (\frac{3}{5}) \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.3.2

Multiply $0$ by $\frac{3}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.4

Multiply $- \frac{3}{5} \cdot 0$ .

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Step 4.2.1.2.4.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 (\frac{3}{5}) & - \frac{3}{5} \cdot 1 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.4.2

Multiply $0$ by $\frac{3}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 & - \frac{3}{5} \cdot 1 & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.5

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 & - \frac{3}{5} & - \frac{3}{5} \cdot 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.6

Multiply $- \frac{3}{5} \cdot 0$ .

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Step 4.2.1.2.6.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 & - \frac{3}{5} & 0 (\frac{3}{5}) \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.6.2

Multiply $0$ by $\frac{3}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 & - \frac{3}{5} & 0 \\ - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.7

Multiply $- \frac{3}{5} \cdot 0$ .

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Step 4.2.1.2.7.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 & - \frac{3}{5} & 0 \\ 0 (\frac{3}{5}) & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.7.2

Multiply $0$ by $\frac{3}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 & - \frac{3}{5} & 0 \\ 0 & - \frac{3}{5} \cdot 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.8

Multiply $- \frac{3}{5} \cdot 0$ .

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Step 4.2.1.2.8.1

Multiply $0$ by $- 1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 & - \frac{3}{5} & 0 \\ 0 & 0 (\frac{3}{5}) & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.8.2

Multiply $0$ by $\frac{3}{5}$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 & - \frac{3}{5} & 0 \\ 0 & 0 & - \frac{3}{5} \cdot 1 \end{array}]$

Step 4.2.1.2.9

Multiply $- 1$ by $1$ .

$[\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] + [\begin{array}{c} - \frac{3}{5} & 0 & 0 \\ 0 & - \frac{3}{5} & 0 \\ 0 & 0 & - \frac{3}{5} \end{array}]$

Step 4.2.2

Add the corresponding elements.

$[\begin{array}{c} 0.8 - \frac{3}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3

Simplify each element.

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Step 4.2.3.1

To write $0.8$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$[\begin{array}{c} 0.8 \cdot \frac{5}{5} - \frac{3}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.2

Combine $0.8$ and $\frac{5}{5}$ .

$[\begin{array}{c} \frac{0.8 \cdot 5}{5} - \frac{3}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.3

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{0.8 \cdot 5 - 3}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.4

Simplify the numerator.

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Step 4.2.3.4.1

Multiply $0.8$ by $5$ .

$[\begin{array}{c} \frac{4 - 3}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.4.2

Subtract $3$ from $4$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.5

Cancel the common factor of $1$ and $5$ .

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Step 4.2.3.5.1

Rewrite $1$ as $1 (1)$ .

$[\begin{array}{c} \frac{1 (1)}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.5.2

Cancel the common factors.

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Step 4.2.3.5.2.1

Rewrite $5$ as $1 (5)$ .

$[\begin{array}{c} \frac{1 \cdot 1}{1 \cdot 5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.5.2.2

Cancel the common factor.

$[\begin{array}{c} \frac{1 \cdot 1}{1 \cdot 5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.5.2.3

Rewrite the expression.

$[\begin{array}{c} \frac{1}{5} & 0.2 + 0 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.6

Add $0.2$ and $0$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 + 0 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.7

Add $0.2$ and $0$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 + 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.8

Add $0$ and $0$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & 0.5 - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.9

To write $0.5$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & 0.5 \cdot \frac{5}{5} - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.10

Combine $0.5$ and $\frac{5}{5}$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & \frac{0.5 \cdot 5}{5} - \frac{3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.11

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & \frac{0.5 \cdot 5 - 3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.12

Simplify the numerator.

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Step 4.2.3.12.1

Multiply $0.5$ by $5$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & \frac{2.5 - 3}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.12.2

Subtract $3$ from $2.5$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & \frac{- 0.5}{5} & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.13

Divide $- 0.5$ by $5$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 + 0 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.14

Add $0.1$ and $0$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 \\ 0.2 + 0 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.15

Add $0.2$ and $0$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 \\ 0.2 & 0.3 + 0 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.16

Add $0.3$ and $0$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 \\ 0.2 & 0.3 & 0.7 - \frac{3}{5} \end{array}]$

Step 4.2.3.17

To write $0.7$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 \\ 0.2 & 0.3 & 0.7 \cdot \frac{5}{5} - \frac{3}{5} \end{array}]$

Step 4.2.3.18

Combine $0.7$ and $\frac{5}{5}$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 \\ 0.2 & 0.3 & \frac{0.7 \cdot 5}{5} - \frac{3}{5} \end{array}]$

Step 4.2.3.19

Combine the numerators over the common denominator.

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 \\ 0.2 & 0.3 & \frac{0.7 \cdot 5 - 3}{5} \end{array}]$

Step 4.2.3.20

Simplify the numerator.

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Step 4.2.3.20.1

Multiply $0.7$ by $5$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 \\ 0.2 & 0.3 & \frac{3.5 - 3}{5} \end{array}]$

Step 4.2.3.20.2

Subtract $3$ from $3.5$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 \\ 0.2 & 0.3 & \frac{0.5}{5} \end{array}]$

Step 4.2.3.21

Divide $0.5$ by $5$ .

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 \\ 0 & - 0.1 & 0.1 \\ 0.2 & 0.3 & 0.1 \end{array}]$

Step 4.3

Find the null space when $λ = \frac{3}{5}$ .

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Step 4.3.1

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

$[\begin{array}{c} \frac{1}{5} & 0.2 & 0.2 & 0 \\ 0 & - 0.1 & 0.1 & 0 \\ 0.2 & 0.3 & 0.1 & 0 \end{array}]$

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 $5$ 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 $5$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} 5 (\frac{1}{5}) & 5 \cdot 0.2 & 5 \cdot 0.2 & 5 \cdot 0 \\ 0 & - 0.1 & 0.1 & 0 \\ 0.2 & 0.3 & 0.1 & 0 \end{array}]$

Step 4.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 1 & 1 & 0 \\ 0 & - 0.1 & 0.1 & 0 \\ 0.2 & 0.3 & 0.1 & 0 \end{array}]$

Step 4.3.2.2

Perform the row operation $R_{3} = R_{3} - 0.2 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.3.2.2.1

Perform the row operation $R_{3} = R_{3} - 0.2 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 0 \\ 0 & - 0.1 & 0.1 & 0 \\ 0.2 - 0.2 \cdot 1 & 0.3 - 0.2 \cdot 1 & 0.1 - 0.2 \cdot 1 & 0 - 0.2 \cdot 0 \end{array}]$

Step 4.3.2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 & 0 \\ 0 & - 0.1 & 0.1 & 0 \\ 0 & 0.1 & - 0.1 & 0 \end{array}]$

Step 4.3.2.3

Multiply each element of $R_{2}$ by $\frac{1}{- 0.1}$ 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 $\frac{1}{- 0.1}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 & 0 \\ \frac{0}{- 0.1} & \frac{- 0.1}{- 0.1} & \frac{0.1}{- 0.1} & \frac{0}{- 0.1} \\ 0 & 0.1 & - 0.1 & 0 \end{array}]$

Step 4.3.2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 & 0 \\ 0 & 1 & - 1 & 0 \\ 0 & 0.1 & - 0.1 & 0 \end{array}]$

Step 4.3.2.4

Perform the row operation $R_{3} = R_{3} - 0.1 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.3.2.4.1

Perform the row operation $R_{3} = R_{3} - 0.1 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 & 0 \\ 0 & 1 & - 1 & 0 \\ 0 - 0.1 \cdot 0 & 0.1 - 0.1 \cdot 1 & - 0.1 - 0.1 \cdot - 1 & 0 - 0.1 \cdot 0 \end{array}]$

Step 4.3.2.4.2

Simplify $R_{3}$ .

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

Step 4.3.2.5

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

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Step 4.3.2.5.1

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

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

Step 4.3.2.5.2

Simplify $R_{1}$ .

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

Step 4.3.3

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

$x + 2 z = 0$

$y - z = 0$

$0 = 0$

Step 4.3.4

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

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

Step 4.3.5

Write the solution as a linear combination of vectors.

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

Step 4.3.6

Write as a solution set.

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

Step 4.3.7

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

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

Step 5

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

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Step 5.1

Substitute the known values into the formula.

$N ([\begin{array}{c} 0.8 & 0.2 & 0.2 \\ 0 & 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 \end{array}] - [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 5.2

Simplify.

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Step 5.2.1

Subtract the corresponding elements.

$[\begin{array}{c} 0.8 - 1 & 0.2 - 0 & 0.2 - 0 \\ 0 - 0 & 0.5 - 1 & 0.1 - 0 \\ 0.2 - 0 & 0.3 - 0 & 0.7 - 1 \end{array}]$

Step 5.2.2

Simplify each element.

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Step 5.2.2.1

Subtract $1$ from $0.8$ .

$[\begin{array}{c} - 0.2 & 0.2 - 0 & 0.2 - 0 \\ 0 - 0 & 0.5 - 1 & 0.1 - 0 \\ 0.2 - 0 & 0.3 - 0 & 0.7 - 1 \end{array}]$

Step 5.2.2.2

Subtract $0$ from $0.2$ .

$[\begin{array}{c} - 0.2 & 0.2 & 0.2 - 0 \\ 0 - 0 & 0.5 - 1 & 0.1 - 0 \\ 0.2 - 0 & 0.3 - 0 & 0.7 - 1 \end{array}]$

Step 5.2.2.3

Subtract $0$ from $0.2$ .

$[\begin{array}{c} - 0.2 & 0.2 & 0.2 \\ 0 - 0 & 0.5 - 1 & 0.1 - 0 \\ 0.2 - 0 & 0.3 - 0 & 0.7 - 1 \end{array}]$

Step 5.2.2.4

Subtract $0$ from $0$ .

$[\begin{array}{c} - 0.2 & 0.2 & 0.2 \\ 0 & 0.5 - 1 & 0.1 - 0 \\ 0.2 - 0 & 0.3 - 0 & 0.7 - 1 \end{array}]$

Step 5.2.2.5

Subtract $1$ from $0.5$ .

$[\begin{array}{c} - 0.2 & 0.2 & 0.2 \\ 0 & - 0.5 & 0.1 - 0 \\ 0.2 - 0 & 0.3 - 0 & 0.7 - 1 \end{array}]$

Step 5.2.2.6

Subtract $0$ from $0.1$ .

$[\begin{array}{c} - 0.2 & 0.2 & 0.2 \\ 0 & - 0.5 & 0.1 \\ 0.2 - 0 & 0.3 - 0 & 0.7 - 1 \end{array}]$

Step 5.2.2.7

Subtract $0$ from $0.2$ .

$[\begin{array}{c} - 0.2 & 0.2 & 0.2 \\ 0 & - 0.5 & 0.1 \\ 0.2 & 0.3 - 0 & 0.7 - 1 \end{array}]$

Step 5.2.2.8

Subtract $0$ from $0.3$ .

$[\begin{array}{c} - 0.2 & 0.2 & 0.2 \\ 0 & - 0.5 & 0.1 \\ 0.2 & 0.3 & 0.7 - 1 \end{array}]$

Step 5.2.2.9

Subtract $1$ from $0.7$ .

$[\begin{array}{c} - 0.2 & 0.2 & 0.2 \\ 0 & - 0.5 & 0.1 \\ 0.2 & 0.3 & - 0.3 \end{array}]$

Step 5.3

Find the null space when $λ = 1$ .

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Step 5.3.1

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

$[\begin{array}{c} - 0.2 & 0.2 & 0.2 & 0 \\ 0 & - 0.5 & 0.1 & 0 \\ 0.2 & 0.3 & - 0.3 & 0 \end{array}]$

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 $\frac{1}{- 0.2}$ 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 $\frac{1}{- 0.2}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{- 0.2}{- 0.2} & \frac{0.2}{- 0.2} & \frac{0.2}{- 0.2} & \frac{0}{- 0.2} \\ 0 & - 0.5 & 0.1 & 0 \\ 0.2 & 0.3 & - 0.3 & 0 \end{array}]$

Step 5.3.2.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & - 0.5 & 0.1 & 0 \\ 0.2 & 0.3 & - 0.3 & 0 \end{array}]$

Step 5.3.2.2

Perform the row operation $R_{3} = R_{3} - 0.2 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 5.3.2.2.1

Perform the row operation $R_{3} = R_{3} - 0.2 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & - 0.5 & 0.1 & 0 \\ 0.2 - 0.2 \cdot 1 & 0.3 - 0.2 \cdot - 1 & - 0.3 - 0.2 \cdot - 1 & 0 - 0.2 \cdot 0 \end{array}]$

Step 5.3.2.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & - 0.5 & 0.1 & 0 \\ 0 & 0.5 & - 0.1 & 0 \end{array}]$

Step 5.3.2.3

Multiply each element of $R_{2}$ by $\frac{1}{- 0.5}$ 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 $\frac{1}{- 0.5}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ \frac{0}{- 0.5} & \frac{- 0.5}{- 0.5} & \frac{0.1}{- 0.5} & \frac{0}{- 0.5} \\ 0 & 0.5 & - 0.1 & 0 \end{array}]$

Step 5.3.2.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & 1 & - 0.2 & 0 \\ 0 & 0.5 & - 0.1 & 0 \end{array}]$

Step 5.3.2.4

Perform the row operation $R_{3} = R_{3} - 0.5 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 5.3.2.4.1

Perform the row operation $R_{3} = R_{3} - 0.5 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & 1 & - 0.2 & 0 \\ 0 - 0.5 \cdot 0 & 0.5 - 0.5 \cdot 1 & - 0.1 - 0.5 \cdot - 0.2 & 0 - 0.5 \cdot 0 \end{array}]$

Step 5.3.2.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & 0 \\ 0 & 1 & - 0.2 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & - 1 - 0.2 & 0 + 0 \\ 0 & 1 & - 0.2 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.2.5.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 1.2 & 0 \\ 0 & 1 & - 0.2 & 0 \\ 0 & 0 & 0 & 0 \end{array}]$

Step 5.3.3

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

$x - 1.2 z = 0$

$y - 0.2 z = 0$

$0 = 0$

Step 5.3.4

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

$[\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 1.2 z \\ 0.2 z \\ z \end{array}]$

Step 5.3.5

Write the solution as a linear combination of vectors.

$[\begin{array}{c} x \\ y \\ z \end{array}] = z [\begin{array}{c} 1.2 \\ 0.2 \\ 1 \end{array}]$

Step 5.3.6

Write as a solution set.

${z [\begin{array}{c} 1.2 \\ 0.2 \\ 1 \end{array}] | z \in R}$

Step 5.3.7

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

${[\begin{array}{c} 1.2 \\ 0.2 \\ 1 \end{array}]}$

Step 6

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

${[\begin{array}{c} 0 \\ - 1 \\ 1 \end{array}], [\begin{array}{c} - 2 \\ 1 \\ 1 \end{array}], [\begin{array}{c} 1.2 \\ 0.2 \\ 1 \end{array}]}$