Linear Algebra Examples

⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 1 2 & 3 & 2 49 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦

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

Step 1

Set up the formula to find the characteristic equation

p (λ)

$p (λ)$ .

p (λ) = determinant (A - λ I_{3})

$p (λ) = determinant (A - λ I_{3})$

Step 2

The identity matrix or unit matrix of size

3

$3$ is the

3 \times 3

$3 \times 3$ square matrix with ones on the main diagonal and zeros elsewhere.

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

Step 3

Substitute the known values into

p (λ) = determinant (A - λ I_{3})

$p (λ) = determinant (A - λ I_{3})$ .

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Step 3.1

Substitute

⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 1 2 & 3 & 2 49 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦

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

A

$A$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 1 2 & 3 & 2 49 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦ - λ I_{3} ⎞ ⎟ ⎠

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

Step 3.2

Substitute

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

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

I_{3}

$I_{3}$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 1 2 & 3 & 2 49 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦ - λ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

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

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 1 2 & 3 & 2 49 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦ - λ ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

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

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply

- λ

$- λ$ by each element of the matrix.

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 1 2 & 3 & 2 49 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - 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{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

$- 1$ by

1

$1$ .

p (λ) = determinant ⎛ ⎜ ⎝ ⎡ ⎢ ⎣ \begin{matrix} 4 & 0 & 1 2 & 3 & 2 49 & 0 & 4 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - 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{matrix} ⎤ ⎥ ⎦ ⎞ ⎟ ⎠

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

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.

$λ = 11, - 3, 3$