Linear Algebra Examples

$[\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}]$

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 $[\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] - λ I_{3})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 7 & 0 & 1 \\ - 8 & 1 & 0 \\ - 8 & 0 & 1 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 7 - λ & 0 + 0 & 1 + 0 \\ - 8 + 0 & 1 - λ & 0 + 0 \\ - 8 + 0 & 0 + 0 & 1 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 7 - λ & 0 & 1 + 0 \\ - 8 + 0 & 1 - λ & 0 + 0 \\ - 8 + 0 & 0 + 0 & 1 - λ \end{array}]$

Step 4.3.2

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 7 - λ & 0 & 1 \\ - 8 + 0 & 1 - λ & 0 + 0 \\ - 8 + 0 & 0 + 0 & 1 - λ \end{array}]$

Step 4.3.3

Add $- 8$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 7 - λ & 0 & 1 \\ - 8 & 1 - λ & 0 + 0 \\ - 8 + 0 & 0 + 0 & 1 - λ \end{array}]$

Step 4.3.4

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 7 - λ & 0 & 1 \\ - 8 & 1 - λ & 0 \\ - 8 + 0 & 0 + 0 & 1 - λ \end{array}]$

Step 4.3.5

Add $- 8$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 7 - λ & 0 & 1 \\ - 8 & 1 - λ & 0 \\ - 8 & 0 + 0 & 1 - λ \end{array}]$

Step 4.3.6

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 7 - λ & 0 & 1 \\ - 8 & 1 - λ & 0 \\ - 8 & 0 & 1 - λ \end{array}]$

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

$| \begin{array}{c} - 8 & 0 \\ - 8 & 1 - λ \end{array} |$

Step 5.1.4

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

$0 | \begin{array}{c} - 8 & 0 \\ - 8 & 1 - λ \end{array} |$

Step 5.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

$| \begin{array}{c} 7 - λ & 1 \\ - 8 & 1 - λ \end{array} |$

Step 5.1.6

Multiply element $a_{22}$ by its cofactor.

$(1 - λ) | \begin{array}{c} 7 - λ & 1 \\ - 8 & 1 - λ \end{array} |$

Step 5.1.7

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

$| \begin{array}{c} 7 - λ & 1 \\ - 8 & 0 \end{array} |$

Step 5.1.8

Multiply element $a_{32}$ by its cofactor.

$0 | \begin{array}{c} 7 - λ & 1 \\ - 8 & 0 \end{array} |$

Step 5.1.9

Add the terms together.

$p (λ) = 0 | \begin{array}{c} - 8 & 0 \\ - 8 & 1 - λ \end{array} | + (1 - λ) | \begin{array}{c} 7 - λ & 1 \\ - 8 & 1 - λ \end{array} | + 0 | \begin{array}{c} 7 - λ & 1 \\ - 8 & 0 \end{array} |$

Step 5.2

Multiply $0$ by $| \begin{array}{c} - 8 & 0 \\ - 8 & 1 - λ \end{array} |$ .

$p (λ) = 0 + (1 - λ) | \begin{array}{c} 7 - λ & 1 \\ - 8 & 1 - λ \end{array} | + 0 | \begin{array}{c} 7 - λ & 1 \\ - 8 & 0 \end{array} |$

Step 5.3

Multiply $0$ by $| \begin{array}{c} 7 - λ & 1 \\ - 8 & 0 \end{array} |$ .

$p (λ) = 0 + (1 - λ) | \begin{array}{c} 7 - λ & 1 \\ - 8 & 1 - λ \end{array} | + 0$

Step 5.4

Evaluate $| \begin{array}{c} 7 - λ & 1 \\ - 8 & 1 - λ \end{array} |$ .

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Step 5.4.1

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

$p (λ) = 0 + (1 - λ) ((7 - λ) (1 - λ) - (- 8 \cdot 1)) + 0$

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

Expand $(7 - λ) (1 - λ)$ using the FOIL Method.

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Step 5.4.2.1.1.1

Apply the distributive property.

$p (λ) = 0 + (1 - λ) (7 (1 - λ) - λ (1 - λ) - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.1.2

Apply the distributive property.

$p (λ) = 0 + (1 - λ) (7 \cdot 1 + 7 (- λ) - λ (1 - λ) - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.1.3

Apply the distributive property.

$p (λ) = 0 + (1 - λ) (7 \cdot 1 + 7 (- λ) - λ \cdot 1 - λ (- λ) - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.2

Simplify and combine like terms.

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Step 5.4.2.1.2.1

Simplify each term.

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Step 5.4.2.1.2.1.1

Multiply $7$ by $1$ .

$p (λ) = 0 + (1 - λ) (7 + 7 (- λ) - λ \cdot 1 - λ (- λ) - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.2.1.2

Multiply $- 1$ by $7$ .

$p (λ) = 0 + (1 - λ) (7 - 7 λ - λ \cdot 1 - λ (- λ) - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.2.1.3

Multiply $- 1$ by $1$ .

$p (λ) = 0 + (1 - λ) (7 - 7 λ - λ - λ (- λ) - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0 + (1 - λ) (7 - 7 λ - λ - 1 \cdot - 1 λ \cdot λ - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.2.1.5

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

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Step 5.4.2.1.2.1.5.1

Move $λ$ .

$p (λ) = 0 + (1 - λ) (7 - 7 λ - λ - 1 \cdot - 1 (λ \cdot λ) - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 + (1 - λ) (7 - 7 λ - λ - 1 \cdot - 1 λ^{2} - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 0 + (1 - λ) (7 - 7 λ - λ + 1 λ^{2} - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0 + (1 - λ) (7 - 7 λ - λ + λ^{2} - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.2.2

Subtract $λ$ from $- 7 λ$ .

$p (λ) = 0 + (1 - λ) (7 - 8 λ + λ^{2} - (- 8 \cdot 1)) + 0$

Step 5.4.2.1.3

Multiply $- (- 8 \cdot 1)$ .

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Step 5.4.2.1.3.1

Multiply $- 8$ by $1$ .

$p (λ) = 0 + (1 - λ) (7 - 8 λ + λ^{2} - - 8) + 0$

Step 5.4.2.1.3.2

Multiply $- 1$ by $- 8$ .

$p (λ) = 0 + (1 - λ) (7 - 8 λ + λ^{2} + 8) + 0$

Step 5.4.2.2

Add $7$ and $8$ .

$p (λ) = 0 + (1 - λ) (- 8 λ + λ^{2} + 15) + 0$

Step 5.4.2.3

Reorder $- 8 λ$ and $λ^{2}$ .

$p (λ) = 0 + (1 - λ) (λ^{2} - 8 λ + 15) + 0$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Combine the opposite terms in $0 + (1 - λ) (λ^{2} - 8 λ + 15) + 0$ .

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Step 5.5.1.1

Add $0$ and $(1 - λ) (λ^{2} - 8 λ + 15)$ .

$p (λ) = (1 - λ) (λ^{2} - 8 λ + 15) + 0$

Step 5.5.1.2

Add $(1 - λ) (λ^{2} - 8 λ + 15)$ and $0$ .

$p (λ) = (1 - λ) (λ^{2} - 8 λ + 15)$

Step 5.5.2

Expand $(1 - λ) (λ^{2} - 8 λ + 15)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 1 λ^{2} + 1 (- 8 λ) + 1 \cdot 15 - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot 15$

Step 5.5.3

Simplify each term.

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Step 5.5.3.1

Multiply $λ^{2}$ by $1$ .

$p (λ) = λ^{2} + 1 (- 8 λ) + 1 \cdot 15 - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot 15$

Step 5.5.3.2

Multiply $- 8 λ$ by $1$ .

$p (λ) = λ^{2} - 8 λ + 1 \cdot 15 - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot 15$

Step 5.5.3.3

Multiply $15$ by $1$ .

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

Step 5.5.3.4

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

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Step 5.5.3.4.1

Move $λ^{2}$ .

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

Step 5.5.3.4.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.5.3.4.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = λ^{2} - 8 λ + 15 - (λ^{2} λ^{1}) - λ (- 8 λ) - λ \cdot 15$

Step 5.5.3.4.2.2

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

$p (λ) = λ^{2} - 8 λ + 15 - λ^{2 + 1} - λ (- 8 λ) - λ \cdot 15$

Step 5.5.3.4.3

Add $2$ and $1$ .

$p (λ) = λ^{2} - 8 λ + 15 - λ^{3} - λ (- 8 λ) - λ \cdot 15$

Step 5.5.3.5

Rewrite using the commutative property of multiplication.

$p (λ) = λ^{2} - 8 λ + 15 - λ^{3} - 1 \cdot - 8 λ \cdot λ - λ \cdot 15$

Step 5.5.3.6

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

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Step 5.5.3.6.1

Move $λ$ .

$p (λ) = λ^{2} - 8 λ + 15 - λ^{3} - 1 \cdot - 8 (λ \cdot λ) - λ \cdot 15$

Step 5.5.3.6.2

Multiply $λ$ by $λ$ .

$p (λ) = λ^{2} - 8 λ + 15 - λ^{3} - 1 \cdot - 8 λ^{2} - λ \cdot 15$

Step 5.5.3.7

Multiply $- 1$ by $- 8$ .

$p (λ) = λ^{2} - 8 λ + 15 - λ^{3} + 8 λ^{2} - λ \cdot 15$

Step 5.5.3.8

Multiply $15$ by $- 1$ .

$p (λ) = λ^{2} - 8 λ + 15 - λ^{3} + 8 λ^{2} - 15 λ$

Step 5.5.4

Add $λ^{2}$ and $8 λ^{2}$ .

$p (λ) = 9 λ^{2} - 8 λ + 15 - λ^{3} - 15 λ$

Step 5.5.5

Subtract $15 λ$ from $- 8 λ$ .

$p (λ) = 9 λ^{2} - 23 λ + 15 - λ^{3}$

Step 5.5.6

Move $15$ .

$p (λ) = 9 λ^{2} - 23 λ - λ^{3} + 15$

Step 5.5.7

Move $- 23 λ$ .

$p (λ) = 9 λ^{2} - λ^{3} - 23 λ + 15$

Step 5.5.8

Reorder $9 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 9 λ^{2} - 23 λ + 15$

Step 6

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

$- λ^{3} + 9 λ^{2} - 23 λ + 15 = 0$

Step 7

Solve for $λ$ .

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Step 7.1

Factor the left side of the equation.

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Step 7.1.1

Factor $- λ^{3} + 9 λ^{2} - 23 λ + 15$ using the rational roots test.

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Step 7.1.1.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 15, \pm 3, \pm 5$

$q = \pm 1$

Step 7.1.1.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 15, \pm 3, \pm 5$

Step 7.1.1.3

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

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Step 7.1.1.3.1

Substitute $1$ into the polynomial.

$- 1^{3} + 9 \cdot 1^{2} - 23 \cdot 1 + 15$

Step 7.1.1.3.2

Raise $1$ to the power of $3$ .

$- 1 \cdot 1 + 9 \cdot 1^{2} - 23 \cdot 1 + 15$

Step 7.1.1.3.3

Multiply $- 1$ by $1$ .

$- 1 + 9 \cdot 1^{2} - 23 \cdot 1 + 15$

Step 7.1.1.3.4

Raise $1$ to the power of $2$ .

$- 1 + 9 \cdot 1 - 23 \cdot 1 + 15$

Step 7.1.1.3.5

Multiply $9$ by $1$ .

$- 1 + 9 - 23 \cdot 1 + 15$

Step 7.1.1.3.6

Add $- 1$ and $9$ .

$8 - 23 \cdot 1 + 15$

Step 7.1.1.3.7

Multiply $- 23$ by $1$ .

$8 - 23 + 15$

Step 7.1.1.3.8

Subtract $23$ from $8$ .

$- 15 + 15$

Step 7.1.1.3.9

Add $- 15$ and $15$ .

$0$

Step 7.1.1.4

Since $1$ is a known root, divide the polynomial by $λ - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{- λ^{3} + 9 λ^{2} - 23 λ + 15}{λ - 1}$

Step 7.1.1.5

Divide $- λ^{3} + 9 λ^{2} - 23 λ + 15$ by $λ - 1$ .

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Step 7.1.1.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$λ$

$1$

$λ^{3}$

$9 λ^{2}$

$23 λ$

$15$

Step 7.1.1.5.2

Divide the highest order term in the dividend $- λ^{3}$ by the highest order term in divisor $λ$ .

				-	$λ^{2}$
	$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$

Step 7.1.1.5.3

Multiply the new quotient term by the divisor.

			-	$λ^{2}$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			-	$λ^{3}$	+	$λ^{2}$

Step 7.1.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $- λ^{3} + λ^{2}$

			-	$λ^{2}$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$

Step 7.1.1.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$λ^{2}$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$

Step 7.1.1.5.6

Pull the next terms from the original dividend down into the current dividend.

			-	$λ^{2}$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$

Step 7.1.1.5.7

Divide the highest order term in the dividend $8 λ^{2}$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$8 λ$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$

Step 7.1.1.5.8

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$8 λ$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$
					+	$8 λ^{2}$	-	$8 λ$

Step 7.1.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $8 λ^{2} - 8 λ$

			-	$λ^{2}$	+	$8 λ$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$
					-	$8 λ^{2}$	+	$8 λ$

Step 7.1.1.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$λ^{2}$	+	$8 λ$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$
					-	$8 λ^{2}$	+	$8 λ$
							-	$15 λ$

Step 7.1.1.5.11

Pull the next terms from the original dividend down into the current dividend.

			-	$λ^{2}$	+	$8 λ$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$
					-	$8 λ^{2}$	+	$8 λ$
							-	$15 λ$	+	$15$

Step 7.1.1.5.12

Divide the highest order term in the dividend $- 15 λ$ by the highest order term in divisor $λ$ .

			-	$λ^{2}$	+	$8 λ$	-	$15$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$
					-	$8 λ^{2}$	+	$8 λ$
							-	$15 λ$	+	$15$

Step 7.1.1.5.13

Multiply the new quotient term by the divisor.

			-	$λ^{2}$	+	$8 λ$	-	$15$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$
					-	$8 λ^{2}$	+	$8 λ$
							-	$15 λ$	+	$15$
							-	$15 λ$	+	$15$

Step 7.1.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- 15 λ + 15$

			-	$λ^{2}$	+	$8 λ$	-	$15$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$
					-	$8 λ^{2}$	+	$8 λ$
							-	$15 λ$	+	$15$
							+	$15 λ$	-	$15$

Step 7.1.1.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

			-	$λ^{2}$	+	$8 λ$	-	$15$
$λ$	-	$1$	-	$λ^{3}$	+	$9 λ^{2}$	-	$23 λ$	+	$15$
			+	$λ^{3}$	-	$λ^{2}$
					+	$8 λ^{2}$	-	$23 λ$
					-	$8 λ^{2}$	+	$8 λ$
							-	$15 λ$	+	$15$
							+	$15 λ$	-	$15$
										$0$

Step 7.1.1.5.16

Since the remander is $0$ , the final answer is the quotient.

$- λ^{2} + 8 λ - 15$

Step 7.1.1.6

Write $- λ^{3} + 9 λ^{2} - 23 λ + 15$ as a set of factors.

$(λ - 1) (- λ^{2} + 8 λ - 15) = 0$

Step 7.1.2

Factor by grouping.

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Step 7.1.2.1

Factor by grouping.

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Step 7.1.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 15 = 15$ and whose sum is $b = 8$ .

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Step 7.1.2.1.1.1

Factor $8$ out of $8 λ$ .

$(λ - 1) (- λ^{2} + 8 (λ) - 15) = 0$

Step 7.1.2.1.1.2

Rewrite $8$ as $3$ plus $5$

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

Step 7.1.2.1.1.3

Apply the distributive property.

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

Step 7.1.2.1.2

Factor out the greatest common factor from each group.

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Step 7.1.2.1.2.1

Group the first two terms and the last two terms.

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

Step 7.1.2.1.2.2

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

$(λ - 1) (λ (- λ + 3) - 5 (- λ + 3)) = 0$

Step 7.1.2.1.3

Factor the polynomial by factoring out the greatest common factor, $- λ + 3$ .

$(λ - 1) ((- λ + 3) (λ - 5)) = 0$

Step 7.1.2.2

Remove unnecessary parentheses.

$(λ - 1) (- λ + 3) (λ - 5) = 0$

Step 7.2

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

$λ - 1 = 0$

$- λ + 3 = 0$

$λ - 5 = 0$

Step 7.3

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

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Step 7.3.1

Set $λ - 1$ equal to $0$ .

$λ - 1 = 0$

Step 7.3.2

Add $1$ to both sides of the equation.

$λ = 1$

Step 7.4

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

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Step 7.4.1

Set $- λ + 3$ equal to $0$ .

$- λ + 3 = 0$

Step 7.4.2

Solve $- λ + 3 = 0$ for $λ$ .

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Step 7.4.2.1

Subtract $3$ from both sides of the equation.

$- λ = - 3$

Step 7.4.2.2

Divide each term in $- λ = - 3$ by $- 1$ and simplify.

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Step 7.4.2.2.1

Divide each term in $- λ = - 3$ by $- 1$ .

$\frac{- λ}{- 1} = \frac{- 3}{- 1}$

Step 7.4.2.2.2

Simplify the left side.

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Step 7.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{λ}{1} = \frac{- 3}{- 1}$

Step 7.4.2.2.2.2

Divide $λ$ by $1$ .

$λ = \frac{- 3}{- 1}$

Step 7.4.2.2.3

Simplify the right side.

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Step 7.4.2.2.3.1

Divide $- 3$ by $- 1$ .

$λ = 3$

Step 7.5

Set $λ - 5$ equal to $0$ and solve for $λ$ .

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Step 7.5.1

Set $λ - 5$ equal to $0$ .

$λ - 5 = 0$

Step 7.5.2

Add $5$ to both sides of the equation.

$λ = 5$

Step 7.6

The final solution is all the values that make $(λ - 1) (- λ + 3) (λ - 5) = 0$ true.

$λ = 1, 3, 5$