Linear 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

Substitute the known values in the formula.

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

Step 3

Subtract the eigenvalue times the identity matrix from the original matrix.

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

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

Simplify each element in the matrix.

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Rearrange $- λ \cdot 1$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 1$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 0$ .

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

Rearrange $- λ \cdot 1$ .

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

Add the corresponding elements.

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

Simplify each element of the matrix $[\begin{array}{c} - 1 - λ & 4 + 0 & 3 + 0 \\ 1 + 0 & 1 - λ & 2 + 0 \\ - 1 + 0 & 0 + 0 & - 1 - λ \end{array}]$ .

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Simplify $4 + 0$ .

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

Simplify $3 + 0$ .

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

Simplify $1 + 0$ .

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

Simplify $2 + 0$ .

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

Simplify $- 1 + 0$ .

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

Simplify $0 + 0$ .

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

Step 4

Find the determinant of $[\begin{array}{c} - 1 - λ & 4 & 3 \\ 1 & 1 - λ & 2 \\ - 1 & 0 & - 1 - λ \end{array}]$ .

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Set up the determinant by breaking it into smaller components.

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

Find the determinant of $- 4 | \begin{array}{c} 1 & 2 \\ - 1 & - 1 - λ \end{array} |$ .

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

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

Simplify the determinant.

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Simplify each term.

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Multiply $- 1 - λ$ by $1$ .

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

Multiply $2$ by $1$ .

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

Simplify by multiplying through.

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Add $- 1$ and $2$ .

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

Apply the distributive property.

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

Multiply.

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Multiply $- 1$ by $- 4$ .

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

Multiply $- 4$ by $1$ .

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

Find the determinant of $(1 - λ) | \begin{array}{c} - 1 - λ & 3 \\ - 1 & - 1 - λ \end{array} |$ .

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

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

Simplify the determinant.

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Simplify terms.

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Simplify each term.

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

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Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

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Simplify each term.

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Multiply $- 1$ by $- 1$ .

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

Multiply $- 1 (- λ)$ .

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Multiply $- 1$ by $- 1$ .

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

Multiply $λ$ by $1$ .

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

Multiply $- λ \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

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

Multiply $λ$ by $1$ .

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

Rewrite using the commutative property of multiplication.

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

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

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Move $λ$ .

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

Multiply $λ$ by $λ$ .

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

Multiply $- 1$ by $- 1$ .

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

Multiply $λ^{2}$ by $1$ .

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

Add $λ$ and $λ$ .

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

Multiply $3$ by $1$ .

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

Add $1$ and $3$ .

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

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

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

Simplify terms.

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Simplify each term.

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Multiply $2 λ$ by $1$ .

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

Multiply $λ^{2}$ by $1$ .

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

Multiply $4$ by $1$ .

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

Rewrite using the commutative property of multiplication.

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

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

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Move $λ$ .

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

Multiply $λ$ by $λ$ .

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

Multiply $- 1$ by $2$ .

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

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

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Move $λ^{2}$ .

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

Multiply $λ^{2}$ by $λ$ .

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Raise $λ$ to the power of $1$ .

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

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

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

Add $2$ and $1$ .

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

Multiply $4$ by $- 1$ .

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

Simplify by adding terms.

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Subtract $4 λ$ from $2 λ$ .

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

Subtract $2 λ^{2}$ from $λ^{2}$ .

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

Since the matrix is multiplied by $0$ , the determinant is $0$ .

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

Combine the opposite terms in $4 λ - 4 - λ^{2} + 4 - λ^{3} - 2 λ + 0$ .

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Add $4 λ - 4 - λ^{2} + 4 - λ^{3} - 2 λ$ and $0$ .

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

Add $- 4$ and $4$ .

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

Add $4 λ - λ^{2}$ and $0$ .

$p (λ) = 4 λ - λ^{2} - λ^{3} - 2 λ$

Subtract $2 λ$ from $4 λ$ .

$p (λ) = - λ^{2} - λ^{3} + 2 λ$

Factor the characteristic polynomial.

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Factor $λ$ out of $- λ^{2} - λ^{3} + 2 λ$ .

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Factor $λ$ out of $- λ^{2}$ .

$p (λ) = λ (- λ) - λ^{3} + 2 λ$

Factor $λ$ out of $- λ^{3}$ .

$p (λ) = λ (- λ) + λ (- λ^{2}) + 2 λ$

Factor $λ$ out of $2 λ$ .

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

Factor $λ$ out of $λ (- λ) + λ (- λ^{2})$ .

$p (λ) = λ (- λ - λ^{2}) + λ \cdot 2$

Factor $λ$ out of $λ (- λ - λ^{2}) + λ \cdot 2$ .

$p (λ) = λ (- λ - λ^{2} + 2)$

Let $u = λ$ . Substitute $u$ for all occurrences of $λ$ .

$p (λ) = λ (- u - u^{2} + 2)$

Factor by grouping.

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Reorder terms.

$p (λ) = λ (- u^{2} - u + 2)$

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 $a \cdot c = - 1 \cdot 2 = - 2$ and whose sum is $b = - 1$ .

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Factor $- 1$ out of $- u$ .

$p (λ) = λ (- u^{2} - (u) + 2)$

Rewrite $- 1$ as $1$ plus $- 2$

$p (λ) = λ (- u^{2} + (1 - 2) u + 2)$

Apply the distributive property.

$p (λ) = λ (- u^{2} + 1 u - 2 u + 2)$

Multiply $u$ by $1$ .

$p (λ) = λ (- u^{2} + u - 2 u + 2)$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$p (λ) = λ ((- u^{2} + u) - 2 u + 2)$

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

$p (λ) = λ (u (- u + 1) + 2 (- u + 1))$

Factor the polynomial by factoring out the greatest common factor, $- u + 1$ .

$p (λ) = λ ((- u + 1) (u + 2))$

Factor.

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Replace all occurrences of $u$ with $λ$ .

$p (λ) = λ ((- λ + 1) (λ + 2))$

Remove unnecessary parentheses.

$p (λ) = λ (- λ + 1) (λ + 2)$

Step 5

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

$λ (- λ + 1) (λ + 2) = 0$

Step 6

Find the roots of $λ (- λ + 1) (λ + 2) = 0$ by solving for $λ$ .

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

$λ = 0$

$- λ + 1 = 0$

$λ + 2 = 0$

Set $λ$ equal to $0$ .

$λ = 0$

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

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Set $- λ + 1$ equal to $0$ .

$- λ + 1 = 0$

Solve $- λ + 1 = 0$ for $λ$ .

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Subtract $1$ from both sides of the equation.

$- λ = - 1$

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

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

$\frac{- λ}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

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Dividing two negative values results in a positive value.

$\frac{λ}{1} = \frac{- 1}{- 1}$

Divide $λ$ by $1$ .

$λ = \frac{- 1}{- 1}$

Simplify the right side.

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Divide $- 1$ by $- 1$ .

$λ = 1$

Set $λ + 2$ equal to $0$ and solve for $λ$ .

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Set $λ + 2$ equal to $0$ .

$λ + 2 = 0$

Subtract $2$ from both sides of the equation.

$λ = - 2$

The final solution is all the values that make $λ (- λ + 1) (λ + 2) = 0$ true.

$λ = 0, 1, - 2$

Step 7

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