Linear Algebra Examples

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

Set up the formula to find the characteristic equation $p (λ)$ .

$p (λ) = determinant (A + (- λ I_{3}))$

Substitute the known values in the formula.

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

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

Tap for more steps...

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 of the matrix $[\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}]$ .

Tap for more steps...

Simplify element $0, 0$ by multiplying $- λ \cdot 1$ .

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

Simplify element $0, 1$ by multiplying $- λ \cdot 0$ .

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

Simplify element $0, 2$ by multiplying $- λ \cdot 0$ .

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

Simplify element $1, 0$ by multiplying $- λ \cdot 0$ .

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

Simplify element $1, 1$ by multiplying $- λ \cdot 1$ .

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

Simplify element $1, 2$ by multiplying $- λ \cdot 0$ .

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

Simplify element $2, 0$ by multiplying $- λ \cdot 0$ .

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

Simplify element $2, 1$ by multiplying $- λ \cdot 0$ .

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

Simplify element $2, 2$ by multiplying $- λ \cdot 1$ .

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

Combine the similar matrices with each others.

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

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

Tap for more steps...

Combine the same size matrices $[\begin{array}{c} - 1 & 4 & 3 \\ 1 & 1 & 2 \\ - 1 & 0 & - 1 \end{array}]$ and $[\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}]$ by adding the corresponding elements of each.

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

Simplify element $0, 1$ of the matrix.

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

Simplify element $0, 2$ of the matrix.

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

Simplify element $1, 0$ of the matrix.

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

Simplify element $1, 2$ of the matrix.

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

Simplify element $2, 0$ of the matrix.

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

Simplify element $2, 1$ of the matrix.

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

The determinant of $[\begin{array}{c} - λ - 1 & 4 & 3 \\ 1 & - λ + 1 & 2 \\ - 1 & 0 & - λ - 1 \end{array}]$ is $λ (- λ - 2) (λ - 1)$ .

Tap for more steps...

Set up the determinant by breaking it into smaller components.

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

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

Tap for more steps...

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

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

Simplify the determinant.

Tap for more steps...

Simplify each term.

Tap for more steps...

Multiply $- λ - 1$ by $1$ .

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

Multiply $2$ by $1$ .

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

Simplify by multiplying through.

Tap for more steps...

Add $- 1$ and $2$ .

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

Apply the distributive property.

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

Multiply.

Tap for more steps...

Multiply $- 1$ by $- 4$ .

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

Multiply $- 4$ by $1$ .

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

The determinant of $(- λ + 1) | \begin{array}{c} - λ - 1 & 3 \\ - 1 & - λ - 1 \end{array} |$ is $- λ^{3} - λ^{2} - 2 λ + 4$ .

Tap for more steps...

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

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

Simplify the determinant.

Tap for more steps...

Simplify terms.

Tap for more steps...

Simplify each term.

Tap for more steps...

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

Tap for more steps...

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Remove parentheses.

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

Simplify and combine like terms.

Tap for more steps...

Simplify each term.

Tap for more steps...

Move $λ$ .

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

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

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

Add $1$ and $1$ .

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

Simplify $- 1 (- λ^{2})$ .

Tap for more steps...

Multiply $- 1$ by $- 1$ .

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

Multiply $λ^{2}$ by $1$ .

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

Simplify $- λ \cdot - 1$ .

Tap for more steps...

Multiply $- 1$ by $- 1$ .

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

Multiply $λ$ by $1$ .

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

Simplify $- 1 (- λ)$ .

Tap for more steps...

Multiply $- 1$ by $- 1$ .

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

Multiply $λ$ by $1$ .

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

Multiply $- 1$ by $- 1$ .

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

Add $λ$ and $λ$ .

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

Multiply $3$ by $1$ .

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

Add $1$ and $3$ .

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

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

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

Simplify terms.

Tap for more steps...

Remove unnecessary parentheses.

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

Simplify each term.

Tap for more steps...

Move $λ^{2}$ .

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

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

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

Add $2$ and $1$ .

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

Move $λ$ .

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

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

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

Add $1$ and $1$ .

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

Multiply $2$ by $- 1$ .

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

Multiply $4$ by $- 1$ .

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

Multiply $λ^{2}$ by $1$ .

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

Multiply $2 λ$ by $1$ .

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

Multiply $4$ by $1$ .

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

Simplify by adding terms.

Tap for more steps...

Add $- 2 λ^{2}$ and $λ^{2}$ .

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

Add $- 4 λ$ and $2 λ$ .

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

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

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

Subtract $2 λ$ from $4 λ$ .

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

Simplify by adding numbers.

Tap for more steps...

Add $- 4$ and $4$ .

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

Simplify by adding zeros.

Tap for more steps...

Add $- λ^{3}$ and $0$ .

$p (λ) = - λ^{3} - λ^{2} + 2 λ + 0$

Add $- λ^{3}$ and $0$ .

$p (λ) = - λ^{3} - λ^{2} + 2 λ$

Factor the characteristic polynomial.

Tap for more steps...

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

Tap for more steps...

Factor $λ$ out of $- λ^{3}$ .

Tap for more steps...

Factor $λ$ out of $λ^{3}$ .

$p (λ) = - (λ λ^{2}) - λ^{2} + 2 λ$

Move $λ$ .

$p (λ) = λ (- (λ^{2})) - λ^{2} + 2 λ$

Multiply $- 1$ by $λ^{2}$ .

$p (λ) = λ (- λ^{2}) - λ^{2} + 2 λ$

Factor $λ$ out of $- λ^{2}$ .

Tap for more steps...

Factor $λ$ out of $λ^{2}$ .

$p (λ) = λ (- λ^{2}) - (λ λ) + 2 λ$

Move $λ$ .

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

Multiply $- 1$ by $λ$ .

$p (λ) = λ (- λ^{2}) + λ (- λ) + 2 λ$

Reorder $2$ and $λ$ .

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

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

$p (λ) = λ (- λ^{2} - λ) + λ \cdot 2$

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

$p (λ) = λ (- λ^{2} - λ + 2)$

Factor.

Tap for more steps...

Factor by grouping.

Tap for more steps...

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 $b = - 1$ .

Tap for more steps...

Factor $- 1$ out of $- λ$ .

$p (λ) = λ (- λ^{2} - (λ) + 2)$

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

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

Apply the distributive property.

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

Multiply $λ$ by $1$ .

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

Remove parentheses.

$p (λ) = λ (- λ^{2} - 2 λ + λ + 2)$

Factor out the greatest common factor from each group.

Tap for more steps...

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $- λ - 2$ .

$p (λ) = λ ((- λ - 2) (λ - 1))$

Remove unnecessary parentheses.

$p (λ) = λ (- λ - 2) (λ - 1)$

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

$λ (- λ - 2) (λ - 1) = 0$

Solve the equation for $L$ .

Tap for more steps...

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

$λ = 0$

$- λ - 2 = 0$

$λ - 1 = 0$

Set the first factor equal to $0$ .

$λ = 0$

Set the next factor equal to $0$ and solve.

Tap for more steps...

Set the next factor equal to $0$ .

$- λ - 2 = 0$

Add $2$ to both sides of the equation.

$- λ = 2$

Multiply each term in $λ = - 2$ by $- 1$

Tap for more steps...

Multiply each term in $- λ = 2$ by $- 1$ .

$- λ \cdot - 1 = 2 \cdot - 1$

Simplify $- λ \cdot - 1$ .

Tap for more steps...

Multiply $- 1$ by $- 1$ .

$1 λ = 2 \cdot - 1$

Multiply $λ$ by $1$ .

$λ = 2 \cdot - 1$

Multiply $2$ by $- 1$ .

$λ = - 2$

Set the next factor equal to $0$ and solve.

Tap for more steps...

Set the next factor equal to $0$ .

$λ - 1 = 0$

Add $1$ to both sides of the equation.

$λ = 1$

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

$λ = 0, - 2, 1$

Enter YOUR Problem