Algebra Examples

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

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

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

$A$ .

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

Step 3.2

Substitute

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

$I_{3}$ .

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

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.

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

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply

$- 1$ by

$1$ .

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

Step 4.1.2.2

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply

$0$ by

$- 1$ .

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

Step 4.1.2.2.2

Multiply

$0$ by

$λ$ .

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

Step 4.1.2.3

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply

$0$ by

$- 1$ .

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

Step 4.1.2.3.2

Multiply

$0$ by

$λ$ .

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

Step 4.1.2.4

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply

$0$ by

$- 1$ .

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

Step 4.1.2.4.2

Multiply

$0$ by

$λ$ .

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

Step 4.1.2.5

Multiply

$- 1$ by

$1$ .

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

Step 4.1.2.6

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply

$0$ by

$- 1$ .

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

Step 4.1.2.6.2

Multiply

$0$ by

$λ$ .

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

Step 4.1.2.7

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply

$0$ by

$- 1$ .

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

Step 4.1.2.7.2

Multiply

$0$ by

$λ$ .

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

Step 4.1.2.8

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply

$0$ by

$- 1$ .

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

Step 4.1.2.8.2

Multiply

$0$ by

$λ$ .

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

Step 4.1.2.9

Multiply

$- 1$ by

$1$ .

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

Step 4.2

Add the corresponding elements.

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

Step 4.3

Simplify each element.

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Step 4.3.1

Add

$4$ and

$0$ .

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

Step 4.3.2

Add

$3$ and

$0$ .

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

Step 4.3.3

Add

$1$ and

$0$ .

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

Step 4.3.4

Add

$2$ and

$0$ .

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

Step 4.3.5

Add

$- 1$ and

$0$ .

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

Step 4.3.6

Add

$0$ and

$0$ .

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

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.

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

Step 5.1.4

Multiply element

$a_{12}$ by its cofactor.

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

Step 5.1.5

The minor for

$a_{22}$ is the determinant with row

$2$ and column

$2$ deleted.

$| \begin{array}{c} - 1 - λ & 3 \\ - 1 & - 1 - λ \end{array} |$

Step 5.1.6

Multiply element

$a_{22}$ by its cofactor.

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

Step 5.1.7

The minor for

$a_{32}$ is the determinant with row

$3$ and column

$2$ deleted.

$| \begin{array}{c} - 1 - λ & 3 \\ 1 & 2 \end{array} |$

Step 5.1.8

Multiply element

$a_{32}$ by its cofactor.

$0 | \begin{array}{c} - 1 - λ & 3 \\ 1 & 2 \end{array} |$

Step 5.1.9

Add the terms together.

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

Step 5.2

Multiply

$0$ by

$| \begin{array}{c} - 1 - λ & 3 \\ 1 & 2 \end{array} |$ .

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

Step 5.3

Evaluate

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

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Step 5.3.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

Step 5.3.2

Simplify the determinant.

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Step 5.3.2.1

Simplify each term.

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Step 5.3.2.1.1

Multiply

$- 1 - λ$ by

$1$ .

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

Step 5.3.2.1.2

Multiply

$- (- 1 \cdot 2)$ .

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Step 5.3.2.1.2.1

Multiply

$- 1$ by

$2$ .

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

Step 5.3.2.1.2.2

Multiply

$- 1$ by

$- 2$ .

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

Step 5.3.2.2

Add

$- 1$ and

$2$ .

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

Step 5.4

Evaluate

$| \begin{array}{c} - 1 - λ & 3 \\ - 1 & - 1 - λ \end{array} |$ .

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Step 5.4.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

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

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

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

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Step 5.4.2.1.1.1

Apply the distributive property.

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

Step 5.4.2.1.1.2

Apply the distributive property.

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

Step 5.4.2.1.1.3

Apply the distributive property.

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

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

$- 1$ by

$- 1$ .

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

Step 5.4.2.1.2.1.2

Multiply

$- 1 (- λ)$ .

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Step 5.4.2.1.2.1.2.1

Multiply

$- 1$ by

$- 1$ .

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

Step 5.4.2.1.2.1.2.2

Multiply

$λ$ by

$1$ .

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

Step 5.4.2.1.2.1.3

Multiply

$- λ \cdot - 1$ .

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Step 5.4.2.1.2.1.3.1

Multiply

$- 1$ by

$- 1$ .

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

Step 5.4.2.1.2.1.3.2

Multiply

$λ$ by

$1$ .

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

Step 5.4.2.1.2.1.4

Rewrite using the commutative property of multiplication.

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

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

$λ$ .

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

Step 5.4.2.1.2.1.5.2

Multiply

$λ$ by

$λ$ .

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

Step 5.4.2.1.2.1.6

Multiply

$- 1$ by

$- 1$ .

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

Step 5.4.2.1.2.1.7

Multiply

$λ^{2}$ by

$1$ .

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

Step 5.4.2.1.2.2

Add

$λ$ and

$λ$ .

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

Step 5.4.2.1.3

Multiply

$- (- 1 \cdot 3)$ .

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Step 5.4.2.1.3.1

Multiply

$- 1$ by

$3$ .

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

Step 5.4.2.1.3.2

Multiply

$- 1$ by

$- 3$ .

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

Step 5.4.2.2

Add

$1$ and

$3$ .

$p (λ) = - 4 (- λ + 1) + (1 - λ) (2 λ + λ^{2} + 4) + 0$

Step 5.4.2.3

Reorder

$2 λ$ and

$λ^{2}$ .

$p (λ) = - 4 (- λ + 1) + (1 - λ) (λ^{2} + 2 λ + 4) + 0$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Add

$- 4 (- λ + 1) + (1 - λ) (λ^{2} + 2 λ + 4)$ and

$0$ .

$p (λ) = - 4 (- λ + 1) + (1 - λ) (λ^{2} + 2 λ + 4)$

Step 5.5.2

Simplify each term.

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Step 5.5.2.1

Apply the distributive property.

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

Step 5.5.2.2

Multiply

$- 1$ by

$- 4$ .

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

Step 5.5.2.3

Multiply

$- 4$ by

$1$ .

$p (λ) = 4 λ - 4 + (1 - λ) (λ^{2} + 2 λ + 4)$

Step 5.5.2.4

Expand

$(1 - λ) (λ^{2} + 2 λ + 4)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 5.5.2.5

Simplify each term.

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Step 5.5.2.5.1

Multiply

$λ^{2}$ by

$1$ .

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

Step 5.5.2.5.2

Multiply

$2 λ$ by

$1$ .

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

Step 5.5.2.5.3

Multiply

$4$ by

$1$ .

$p (λ) = 4 λ - 4 + λ^{2} + 2 λ + 4 - λ \cdot λ^{2} - λ (2 λ) - λ \cdot 4$

Step 5.5.2.5.4

Multiply

$λ$ by

$λ^{2}$ by adding the exponents.

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Step 5.5.2.5.4.1

Move

$λ^{2}$ .

$p (λ) = 4 λ - 4 + λ^{2} + 2 λ + 4 - (λ^{2} λ) - λ (2 λ) - λ \cdot 4$

Step 5.5.2.5.4.2

Multiply

$λ^{2}$ by

$λ$ .

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Step 5.5.2.5.4.2.1

Raise

$λ$ to the power of

$1$ .

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

Step 5.5.2.5.4.2.2

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.5.2.5.4.3

Add

$2$ and

$1$ .

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

Step 5.5.2.5.5

Rewrite using the commutative property of multiplication.

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

Step 5.5.2.5.6

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 5.5.2.5.6.1

Move

$λ$ .

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

Step 5.5.2.5.6.2

Multiply

$λ$ by

$λ$ .

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

Step 5.5.2.5.7

Multiply

$- 1$ by

$2$ .

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

Step 5.5.2.5.8

Multiply

$4$ by

$- 1$ .

$p (λ) = 4 λ - 4 + λ^{2} + 2 λ + 4 - λ^{3} - 2 λ^{2} - 4 λ$

Step 5.5.2.6

Subtract

$2 λ^{2}$ from

$λ^{2}$ .

$p (λ) = 4 λ - 4 - λ^{2} + 2 λ + 4 - λ^{3} - 4 λ$

Step 5.5.2.7

Subtract

$4 λ$ from

$2 λ$ .

$p (λ) = 4 λ - 4 - λ^{2} - 2 λ + 4 - λ^{3}$

Step 5.5.3

Combine the opposite terms in

$4 λ - 4 - λ^{2} - 2 λ + 4 - λ^{3}$ .

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Step 5.5.3.1

Add

$- 4$ and

$4$ .

$p (λ) = 4 λ - λ^{2} - 2 λ + 0 - λ^{3}$

Step 5.5.3.2

Add

$4 λ - λ^{2} - 2 λ$ and

$0$ .

$p (λ) = 4 λ - λ^{2} - 2 λ - λ^{3}$

Step 5.5.4

Subtract

$2 λ$ from

$4 λ$ .

$p (λ) = - λ^{2} + 2 λ - λ^{3}$

Step 5.5.5

Move

$2 λ$ .

$p (λ) = - λ^{2} - λ^{3} + 2 λ$

Step 5.5.6

Reorder

$- λ^{2}$ and

$- λ^{3}$ .

$p (λ) = - λ^{3} - λ^{2} + 2 λ$

Step 6

Set the characteristic polynomial equal to

$0$ to find the eigenvalues

$λ$ .

$- λ^{3} - λ^{2} + 2 λ = 0$

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

$- λ$ out of

$- λ^{3} - λ^{2} + 2 λ$ .

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Step 7.1.1.1

Factor

$- λ$ out of

$- λ^{3}$ .

$- λ \cdot λ^{2} - λ^{2} + 2 λ = 0$

Step 7.1.1.2

Factor

$- λ$ out of

$- λ^{2}$ .

$- λ \cdot λ^{2} - λ \cdot λ + 2 λ = 0$

Step 7.1.1.3

Factor

$- λ$ out of

$2 λ$ .

$- λ \cdot λ^{2} - λ \cdot λ - λ \cdot - 2 = 0$

Step 7.1.1.4

Factor

$- λ$ out of

$- λ (λ^{2}) - λ (λ)$ .

$- λ (λ^{2} + λ) - λ \cdot - 2 = 0$

Step 7.1.1.5

Factor

$- λ$ out of

$- λ (λ^{2} + λ) - λ (- 2)$ .

$- λ (λ^{2} + λ - 2) = 0$

Step 7.1.2

Factor.

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Step 7.1.2.1

Factor

$λ^{2} + λ - 2$ using the AC method.

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Step 7.1.2.1.1

Consider the form

$x^{2} + b x + c$ . Find a pair of integers whose product is

$c$ and whose sum is

$b$ . In this case, whose product is

$- 2$ and whose sum is

$1$ .

$- 1, 2$

Step 7.1.2.1.2

Write the factored form using these integers.

$- λ ((λ - 1) (λ + 2)) = 0$

Step 7.1.2.2

Remove unnecessary parentheses.

$- λ (λ - 1) (λ + 2) = 0$

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

$0$ .

$λ = 0$

$λ - 1 = 0$

$λ + 2 = 0$

Step 7.3

Set

$λ$ equal to

$0$ .

$λ = 0$

Step 7.4

Set

$λ - 1$ equal to

$0$ and solve for

$λ$ .

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Step 7.4.1

Set

$λ - 1$ equal to

$0$ .

$λ - 1 = 0$

Step 7.4.2

Add

$1$ to both sides of the equation.

$λ = 1$

Step 7.5

Set

$λ + 2$ equal to

$0$ and solve for

$λ$ .

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Step 7.5.1

Set

$λ + 2$ equal to

$0$ .

$λ + 2 = 0$

Step 7.5.2

Subtract

$2$ from both sides of the equation.

$λ = - 2$

Step 7.6

The final solution is all the values that make

$- λ (λ - 1) (λ + 2) = 0$ true.

$λ = 0, 1, - 2$

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