Linear Algebra Examples

$[\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}]$

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 $[\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] - λ I_{3})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 4 & 0 & 1 \\ 2 & 3 & 2 \\ 49 & 0 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 + 0 & 1 + 0 \\ 2 + 0 & 3 - λ & 2 + 0 \\ 49 + 0 & 0 + 0 & 4 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 1 + 0 \\ 2 + 0 & 3 - λ & 2 + 0 \\ 49 + 0 & 0 + 0 & 4 - λ \end{array}]$

Step 4.3.2

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 1 \\ 2 + 0 & 3 - λ & 2 + 0 \\ 49 + 0 & 0 + 0 & 4 - λ \end{array}]$

Step 4.3.3

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 1 \\ 2 & 3 - λ & 2 + 0 \\ 49 + 0 & 0 + 0 & 4 - λ \end{array}]$

Step 4.3.4

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 1 \\ 2 & 3 - λ & 2 \\ 49 + 0 & 0 + 0 & 4 - λ \end{array}]$

Step 4.3.5

Add $49$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 1 \\ 2 & 3 - λ & 2 \\ 49 & 0 + 0 & 4 - λ \end{array}]$

Step 4.3.6

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 4 - λ & 0 & 1 \\ 2 & 3 - λ & 2 \\ 49 & 0 & 4 - λ \end{array}]$

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.

$| \begin{array}{c} 2 & 2 \\ 49 & 4 - λ \end{array} |$

Step 5.1.4

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

$0 | \begin{array}{c} 2 & 2 \\ 49 & 4 - λ \end{array} |$

Step 5.1.5

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

$| \begin{array}{c} 4 - λ & 1 \\ 49 & 4 - λ \end{array} |$

Step 5.1.6

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

$(3 - λ) | \begin{array}{c} 4 - λ & 1 \\ 49 & 4 - λ \end{array} |$

Step 5.1.7

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

$| \begin{array}{c} 4 - λ & 1 \\ 2 & 2 \end{array} |$

Step 5.1.8

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

$0 | \begin{array}{c} 4 - λ & 1 \\ 2 & 2 \end{array} |$

Step 5.1.9

Add the terms together.

$p (λ) = 0 | \begin{array}{c} 2 & 2 \\ 49 & 4 - λ \end{array} | + (3 - λ) | \begin{array}{c} 4 - λ & 1 \\ 49 & 4 - λ \end{array} | + 0 | \begin{array}{c} 4 - λ & 1 \\ 2 & 2 \end{array} |$

Step 5.2

Multiply $0$ by $| \begin{array}{c} 2 & 2 \\ 49 & 4 - λ \end{array} |$ .

$p (λ) = 0 + (3 - λ) | \begin{array}{c} 4 - λ & 1 \\ 49 & 4 - λ \end{array} | + 0 | \begin{array}{c} 4 - λ & 1 \\ 2 & 2 \end{array} |$

Step 5.3

Multiply $0$ by $| \begin{array}{c} 4 - λ & 1 \\ 2 & 2 \end{array} |$ .

$p (λ) = 0 + (3 - λ) | \begin{array}{c} 4 - λ & 1 \\ 49 & 4 - λ \end{array} | + 0$

Step 5.4

Evaluate $| \begin{array}{c} 4 - λ & 1 \\ 49 & 4 - λ \end{array} |$ .

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Step 5.4.1

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

$p (λ) = 0 + (3 - λ) ((4 - λ) (4 - λ) - 49 \cdot 1) + 0$

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 $(4 - λ) (4 - λ)$ using the FOIL Method.

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Step 5.4.2.1.1.1

Apply the distributive property.

$p (λ) = 0 + (3 - λ) (4 (4 - λ) - λ (4 - λ) - 49 \cdot 1) + 0$

Step 5.4.2.1.1.2

Apply the distributive property.

$p (λ) = 0 + (3 - λ) (4 \cdot 4 + 4 (- λ) - λ (4 - λ) - 49 \cdot 1) + 0$

Step 5.4.2.1.1.3

Apply the distributive property.

$p (λ) = 0 + (3 - λ) (4 \cdot 4 + 4 (- λ) - λ \cdot 4 - λ (- λ) - 49 \cdot 1) + 0$

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 $4$ by $4$ .

$p (λ) = 0 + (3 - λ) (16 + 4 (- λ) - λ \cdot 4 - λ (- λ) - 49 \cdot 1) + 0$

Step 5.4.2.1.2.1.2

Multiply $- 1$ by $4$ .

$p (λ) = 0 + (3 - λ) (16 - 4 λ - λ \cdot 4 - λ (- λ) - 49 \cdot 1) + 0$

Step 5.4.2.1.2.1.3

Multiply $4$ by $- 1$ .

$p (λ) = 0 + (3 - λ) (16 - 4 λ - 4 λ - λ (- λ) - 49 \cdot 1) + 0$

Step 5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

$p (λ) = 0 + (3 - λ) (16 - 4 λ - 4 λ - 1 \cdot - 1 λ \cdot λ - 49 \cdot 1) + 0$

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 $λ$ .

$p (λ) = 0 + (3 - λ) (16 - 4 λ - 4 λ - 1 \cdot - 1 (λ \cdot λ) - 49 \cdot 1) + 0$

Step 5.4.2.1.2.1.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 0 + (3 - λ) (16 - 4 λ - 4 λ - 1 \cdot - 1 λ^{2} - 49 \cdot 1) + 0$

Step 5.4.2.1.2.1.6

Multiply $- 1$ by $- 1$ .

$p (λ) = 0 + (3 - λ) (16 - 4 λ - 4 λ + 1 λ^{2} - 49 \cdot 1) + 0$

Step 5.4.2.1.2.1.7

Multiply $λ^{2}$ by $1$ .

$p (λ) = 0 + (3 - λ) (16 - 4 λ - 4 λ + λ^{2} - 49 \cdot 1) + 0$

Step 5.4.2.1.2.2

Subtract $4 λ$ from $- 4 λ$ .

$p (λ) = 0 + (3 - λ) (16 - 8 λ + λ^{2} - 49 \cdot 1) + 0$

Step 5.4.2.1.3

Multiply $- 49$ by $1$ .

$p (λ) = 0 + (3 - λ) (16 - 8 λ + λ^{2} - 49) + 0$

Step 5.4.2.2

Subtract $49$ from $16$ .

$p (λ) = 0 + (3 - λ) (- 8 λ + λ^{2} - 33) + 0$

Step 5.4.2.3

Reorder $- 8 λ$ and $λ^{2}$ .

$p (λ) = 0 + (3 - λ) (λ^{2} - 8 λ - 33) + 0$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Combine the opposite terms in $0 + (3 - λ) (λ^{2} - 8 λ - 33) + 0$ .

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Step 5.5.1.1

Add $0$ and $(3 - λ) (λ^{2} - 8 λ - 33)$ .

$p (λ) = (3 - λ) (λ^{2} - 8 λ - 33) + 0$

Step 5.5.1.2

Add $(3 - λ) (λ^{2} - 8 λ - 33)$ and $0$ .

$p (λ) = (3 - λ) (λ^{2} - 8 λ - 33)$

Step 5.5.2

Expand $(3 - λ) (λ^{2} - 8 λ - 33)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 3 λ^{2} + 3 (- 8 λ) + 3 \cdot - 33 - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot - 33$

Step 5.5.3

Simplify each term.

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Step 5.5.3.1

Multiply $- 8$ by $3$ .

$p (λ) = 3 λ^{2} - 24 λ + 3 \cdot - 33 - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot - 33$

Step 5.5.3.2

Multiply $3$ by $- 33$ .

$p (λ) = 3 λ^{2} - 24 λ - 99 - λ \cdot λ^{2} - λ (- 8 λ) - λ \cdot - 33$

Step 5.5.3.3

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

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Step 5.5.3.3.1

Move $λ^{2}$ .

$p (λ) = 3 λ^{2} - 24 λ - 99 - (λ^{2} λ) - λ (- 8 λ) - λ \cdot - 33$

Step 5.5.3.3.2

Multiply $λ^{2}$ by $λ$ .

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Step 5.5.3.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 3 λ^{2} - 24 λ - 99 - (λ^{2} λ^{1}) - λ (- 8 λ) - λ \cdot - 33$

Step 5.5.3.3.2.2

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

$p (λ) = 3 λ^{2} - 24 λ - 99 - λ^{2 + 1} - λ (- 8 λ) - λ \cdot - 33$

Step 5.5.3.3.3

Add $2$ and $1$ .

$p (λ) = 3 λ^{2} - 24 λ - 99 - λ^{3} - λ (- 8 λ) - λ \cdot - 33$

Step 5.5.3.4

Rewrite using the commutative property of multiplication.

$p (λ) = 3 λ^{2} - 24 λ - 99 - λ^{3} - 1 \cdot - 8 λ \cdot λ - λ \cdot - 33$

Step 5.5.3.5

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

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Step 5.5.3.5.1

Move $λ$ .

$p (λ) = 3 λ^{2} - 24 λ - 99 - λ^{3} - 1 \cdot - 8 (λ \cdot λ) - λ \cdot - 33$

Step 5.5.3.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 3 λ^{2} - 24 λ - 99 - λ^{3} - 1 \cdot - 8 λ^{2} - λ \cdot - 33$

Step 5.5.3.6

Multiply $- 1$ by $- 8$ .

$p (λ) = 3 λ^{2} - 24 λ - 99 - λ^{3} + 8 λ^{2} - λ \cdot - 33$

Step 5.5.3.7

Multiply $- 33$ by $- 1$ .

$p (λ) = 3 λ^{2} - 24 λ - 99 - λ^{3} + 8 λ^{2} + 33 λ$

Step 5.5.4

Add $3 λ^{2}$ and $8 λ^{2}$ .

$p (λ) = 11 λ^{2} - 24 λ - 99 - λ^{3} + 33 λ$

Step 5.5.5

Add $- 24 λ$ and $33 λ$ .

$p (λ) = 11 λ^{2} + 9 λ - 99 - λ^{3}$

Step 5.5.6

Move $- 99$ .

$p (λ) = 11 λ^{2} + 9 λ - λ^{3} - 99$

Step 5.5.7

Move $9 λ$ .

$p (λ) = 11 λ^{2} - λ^{3} + 9 λ - 99$

Step 5.5.8

Reorder $11 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 11 λ^{2} + 9 λ - 99$

Step 6

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

$- λ^{3} + 11 λ^{2} + 9 λ - 99 = 0$

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 out the greatest common factor from each group.

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Step 7.1.1.1

Group the first two terms and the last two terms.

$(- λ^{3} + 11 λ^{2}) + 9 λ - 99 = 0$

Step 7.1.1.2

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

$λ^{2} (- λ + 11) - 9 (- λ + 11) = 0$

Step 7.1.2

Factor the polynomial by factoring out the greatest common factor, $- λ + 11$ .

$(- λ + 11) (λ^{2} - 9) = 0$

Step 7.1.3

Rewrite $9$ as $3^{2}$ .

$(- λ + 11) (λ^{2} - 3^{2}) = 0$

Step 7.1.4

Factor.

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Step 7.1.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = λ$ and $b = 3$ .

$(- λ + 11) ((λ + 3) (λ - 3)) = 0$

Step 7.1.4.2

Remove unnecessary parentheses.

$(- λ + 11) (λ + 3) (λ - 3) = 0$

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 $0$ .

$- λ + 11 = 0$

$λ + 3 = 0$

$λ - 3 = 0$

Step 7.3

Set $- λ + 11$ equal to $0$ and solve for $λ$ .

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Step 7.3.1

Set $- λ + 11$ equal to $0$ .

$- λ + 11 = 0$

Step 7.3.2

Solve $- λ + 11 = 0$ for $λ$ .

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Step 7.3.2.1

Subtract $11$ from both sides of the equation.

$- λ = - 11$

Step 7.3.2.2

Divide each term in $- λ = - 11$ by $- 1$ and simplify.

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Step 7.3.2.2.1

Divide each term in $- λ = - 11$ by $- 1$ .

$\frac{- λ}{- 1} = \frac{- 11}{- 1}$

Step 7.3.2.2.2

Simplify the left side.

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Step 7.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{λ}{1} = \frac{- 11}{- 1}$

Step 7.3.2.2.2.2

Divide $λ$ by $1$ .

$λ = \frac{- 11}{- 1}$

Step 7.3.2.2.3

Simplify the right side.

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Step 7.3.2.2.3.1

Divide $- 11$ by $- 1$ .

$λ = 11$

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

Subtract $3$ from both sides of the equation.

$λ = - 3$

Step 7.5

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

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Step 7.5.1

Set $λ - 3$ equal to $0$ .

$λ - 3 = 0$

Step 7.5.2

Add $3$ to both sides of the equation.

$λ = 3$

Step 7.6

The final solution is all the values that make $(- λ + 11) (λ + 3) (λ - 3) = 0$ true.

$λ = 11, - 3, 3$