Finite Math Examples

$[\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}]$

Step 1

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

$p (λ) = determinant (A - λ I_{3})$

Step 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 3

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

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Step 3.1

Substitute $[\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] - λ I_{3})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

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

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 0 & 0.1 & 0.9 \\ 0.6 & 0 & 0.4 \\ 0.9 & 0.1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 0 - λ & 0.1 + 0 & 0.9 + 0 \\ 0.6 + 0 & 0 - λ & 0.4 + 0 \\ 0.9 + 0 & 0.1 + 0 & 0 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 0.1 + 0 & 0.9 + 0 \\ 0.6 + 0 & 0 - λ & 0.4 + 0 \\ 0.9 + 0 & 0.1 + 0 & 0 - λ \end{array}]$

Step 4.3.2

Add $0.1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 0.1 & 0.9 + 0 \\ 0.6 + 0 & 0 - λ & 0.4 + 0 \\ 0.9 + 0 & 0.1 + 0 & 0 - λ \end{array}]$

Step 4.3.3

Add $0.9$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 0.1 & 0.9 \\ 0.6 + 0 & 0 - λ & 0.4 + 0 \\ 0.9 + 0 & 0.1 + 0 & 0 - λ \end{array}]$

Step 4.3.4

Add $0.6$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 0.1 & 0.9 \\ 0.6 & 0 - λ & 0.4 + 0 \\ 0.9 + 0 & 0.1 + 0 & 0 - λ \end{array}]$

Step 4.3.5

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 0.1 & 0.9 \\ 0.6 & - λ & 0.4 + 0 \\ 0.9 + 0 & 0.1 + 0 & 0 - λ \end{array}]$

Step 4.3.6

Add $0.4$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 0.1 & 0.9 \\ 0.6 & - λ & 0.4 \\ 0.9 + 0 & 0.1 + 0 & 0 - λ \end{array}]$

Step 4.3.7

Add $0.9$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 0.1 & 0.9 \\ 0.6 & - λ & 0.4 \\ 0.9 & 0.1 + 0 & 0 - λ \end{array}]$

Step 4.3.8

Add $0.1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 0.1 & 0.9 \\ 0.6 & - λ & 0.4 \\ 0.9 & 0.1 & 0 - λ \end{array}]$

Step 4.3.9

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} - λ & 0.1 & 0.9 \\ 0.6 & - λ & 0.4 \\ 0.9 & 0.1 & - λ \end{array}]$

Step 5

Find the determinant.

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Step 5.1

Choose the row or column with the most $0$ elements. If there are no $0$ elements choose any row or column. Multiply every element in row $1$ by its cofactor and add.

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Step 5.1.1

Consider the corresponding sign chart.

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

Step 5.1.2

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

Step 5.1.3

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

$| \begin{array}{c} - λ & 0.4 \\ 0.1 & - λ \end{array} |$

Step 5.1.4

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

$- λ | \begin{array}{c} - λ & 0.4 \\ 0.1 & - λ \end{array} |$

Step 5.1.5

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

$| \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} |$

Step 5.1.6

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

$- 0.1 | \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} |$

Step 5.1.7

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

$| \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.1.8

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

$0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.1.9

Add the terms together.

$p (λ) = - λ | \begin{array}{c} - λ & 0.4 \\ 0.1 & - λ \end{array} | - 0.1 | \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} | + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.2

Evaluate $| \begin{array}{c} - λ & 0.4 \\ 0.1 & - λ \end{array} |$ .

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Step 5.2.1

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

$p (λ) = - λ (- λ (- λ) - 0.1 \cdot 0.4) - 0.1 | \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} | + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.2.2

Simplify each term.

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Step 5.2.2.1

Rewrite using the commutative property of multiplication.

$p (λ) = - λ (- 1 \cdot - 1 λ \cdot λ - 0.1 \cdot 0.4) - 0.1 | \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} | + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.2.2.2

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

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Step 5.2.2.2.1

Move $λ$ .

$p (λ) = - λ (- 1 \cdot - 1 (λ \cdot λ) - 0.1 \cdot 0.4) - 0.1 | \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} | + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.2.2.2.2

Multiply $λ$ by $λ$ .

$p (λ) = - λ (- 1 \cdot - 1 λ^{2} - 0.1 \cdot 0.4) - 0.1 | \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} | + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.2.2.3

Multiply $- 1$ by $- 1$ .

$p (λ) = - λ (1 λ^{2} - 0.1 \cdot 0.4) - 0.1 | \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} | + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.2.2.4

Multiply $λ^{2}$ by $1$ .

$p (λ) = - λ (λ^{2} - 0.1 \cdot 0.4) - 0.1 | \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} | + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.2.2.5

Multiply $- 0.1$ by $0.4$ .

$p (λ) = - λ (λ^{2} - 0.04) - 0.1 | \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} | + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.3

Evaluate $| \begin{array}{c} 0.6 & 0.4 \\ 0.9 & - λ \end{array} |$ .

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Step 5.3.1

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

$p (λ) = - λ (λ^{2} - 0.04) - 0.1 (0.6 (- λ) - 0.9 \cdot 0.4) + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.3.2

Simplify each term.

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Step 5.3.2.1

Multiply $- 1$ by $0.6$ .

$p (λ) = - λ (λ^{2} - 0.04) - 0.1 (- 0.6 λ - 0.9 \cdot 0.4) + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.3.2.2

Multiply $- 0.9$ by $0.4$ .

$p (λ) = - λ (λ^{2} - 0.04) - 0.1 (- 0.6 λ - 0.36) + 0.9 | \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$

Step 5.4

Evaluate $| \begin{array}{c} 0.6 & - λ \\ 0.9 & 0.1 \end{array} |$ .

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Step 5.4.1

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

$p (λ) = - λ (λ^{2} - 0.04) - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.6 \cdot 0.1 - 0.9 (- λ))$

Step 5.4.2

Simplify the determinant.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Multiply $0.6$ by $0.1$ .

$p (λ) = - λ (λ^{2} - 0.04) - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.06 - 0.9 (- λ))$

Step 5.4.2.1.2

Multiply $- 1$ by $- 0.9$ .

$p (λ) = - λ (λ^{2} - 0.04) - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.06 + 0.9 λ)$

Step 5.4.2.2

Reorder $0.06$ and $0.9 λ$ .

$p (λ) = - λ (λ^{2} - 0.04) - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.9 λ + 0.06)$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Simplify each term.

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Step 5.5.1.1

Apply the distributive property.

$p (λ) = - λ \cdot λ^{2} - λ \cdot - 0.04 - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.9 λ + 0.06)$

Step 5.5.1.2

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

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Step 5.5.1.2.1

Move $λ^{2}$ .

$p (λ) = - (λ^{2} λ) - λ \cdot - 0.04 - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.9 λ + 0.06)$

Step 5.5.1.2.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.5.1.2.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = - (λ^{2} λ^{1}) - λ \cdot - 0.04 - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.9 λ + 0.06)$

Step 5.5.1.2.2.2

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

$p (λ) = - λ^{2 + 1} - λ \cdot - 0.04 - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.9 λ + 0.06)$

Step 5.5.1.2.3

Add $2$ and $1$ .

$p (λ) = - λ^{3} - λ \cdot - 0.04 - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.9 λ + 0.06)$

Step 5.5.1.3

Multiply $- 0.04$ by $- 1$ .

$p (λ) = - λ^{3} + 0.04 λ - 0.1 (- 0.6 λ - 0.36) + 0.9 (0.9 λ + 0.06)$

Step 5.5.1.4

Apply the distributive property.

$p (λ) = - λ^{3} + 0.04 λ - 0.1 (- 0.6 λ) - 0.1 \cdot - 0.36 + 0.9 (0.9 λ + 0.06)$

Step 5.5.1.5

Multiply $- 0.6$ by $- 0.1$ .

$p (λ) = - λ^{3} + 0.04 λ + 0.06 λ - 0.1 \cdot - 0.36 + 0.9 (0.9 λ + 0.06)$

Step 5.5.1.6

Multiply $- 0.1$ by $- 0.36$ .

$p (λ) = - λ^{3} + 0.04 λ + 0.06 λ + 0.036 + 0.9 (0.9 λ + 0.06)$

Step 5.5.1.7

Apply the distributive property.

$p (λ) = - λ^{3} + 0.04 λ + 0.06 λ + 0.036 + 0.9 (0.9 λ) + 0.9 \cdot 0.06$

Step 5.5.1.8

Multiply $0.9$ by $0.9$ .

$p (λ) = - λ^{3} + 0.04 λ + 0.06 λ + 0.036 + 0.81 λ + 0.9 \cdot 0.06$

Step 5.5.1.9

Multiply $0.9$ by $0.06$ .

$p (λ) = - λ^{3} + 0.04 λ + 0.06 λ + 0.036 + 0.81 λ + 0.054$

Step 5.5.2

Add $0.04 λ$ and $0.06 λ$ .

$p (λ) = - λ^{3} + 0.1 λ + 0.036 + 0.81 λ + 0.054$

Step 5.5.3

Add $0.1 λ$ and $0.81 λ$ .

$p (λ) = - λ^{3} + 0.91 λ + 0.036 + 0.054$

Step 5.5.4

Add $0.036$ and $0.054$ .

$p (λ) = - λ^{3} + 0.91 λ + 0.09$

Step 6

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

$- λ^{3} + 0.91 λ + 0.09 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

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

$λ \approx - 0.9, - 0.1, 1$