Finite Math Examples

$[\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}]$

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

Tap for more steps...

Step 3.1

Substitute $[\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}]$ for $A$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] - λ I_{3})$

Step 3.2

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

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

Step 4

Simplify.

Tap for more steps...

Step 4.1

Simplify each term.

Tap for more steps...

Step 4.1.1

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

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

Tap for more steps...

Step 4.1.2.1

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.2.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.3.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.4.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.6.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.7.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply $- λ \cdot 0$ .

Tap for more steps...

Step 4.1.2.8.1

Multiply $0$ by $- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply $0$ by $λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply $- 1$ by $1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 1 & 0 \\ \frac{3}{2} & 0 & - 2 \\ 1 & 2 & 4 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 9 - λ & 1 + 0 & 0 + 0 \\ \frac{3}{2} + 0 & 0 - λ & - 2 + 0 \\ 1 + 0 & 2 + 0 & 4 - λ \end{array}]$

Step 4.3

Simplify each element.

Tap for more steps...

Step 4.3.1

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 1 & 0 + 0 \\ \frac{3}{2} + 0 & 0 - λ & - 2 + 0 \\ 1 + 0 & 2 + 0 & 4 - λ \end{array}]$

Step 4.3.2

Add $0$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 1 & 0 \\ \frac{3}{2} + 0 & 0 - λ & - 2 + 0 \\ 1 + 0 & 2 + 0 & 4 - λ \end{array}]$

Step 4.3.3

Add $\frac{3}{2}$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 1 & 0 \\ \frac{3}{2} & 0 - λ & - 2 + 0 \\ 1 + 0 & 2 + 0 & 4 - λ \end{array}]$

Step 4.3.4

Subtract $λ$ from $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 1 & 0 \\ \frac{3}{2} & - λ & - 2 + 0 \\ 1 + 0 & 2 + 0 & 4 - λ \end{array}]$

Step 4.3.5

Add $- 2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 1 & 0 \\ \frac{3}{2} & - λ & - 2 \\ 1 + 0 & 2 + 0 & 4 - λ \end{array}]$

Step 4.3.6

Add $1$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 1 & 0 \\ \frac{3}{2} & - λ & - 2 \\ 1 & 2 + 0 & 4 - λ \end{array}]$

Step 4.3.7

Add $2$ and $0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 1 & 0 \\ \frac{3}{2} & - λ & - 2 \\ 1 & 2 & 4 - λ \end{array}]$

Step 5

Find the determinant.

Tap for more steps...

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 row $1$ by its cofactor and add.

Tap for more steps...

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 $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

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

Step 5.1.4

Multiply element $a_{11}$ by its cofactor.

$(9 - λ) | \begin{array}{c} - λ & - 2 \\ 2 & 4 - λ \end{array} |$

Step 5.1.5

The minor for $a_{12}$ is the determinant with row $1$ and column $2$ deleted.

$| \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} |$

Step 5.1.6

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

$- 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} |$

Step 5.1.7

The minor for $a_{13}$ is the determinant with row $1$ and column $3$ deleted.

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

Step 5.1.8

Multiply element $a_{13}$ by its cofactor.

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

Step 5.1.9

Add the terms together.

$p (λ) = (9 - λ) | \begin{array}{c} - λ & - 2 \\ 2 & 4 - λ \end{array} | - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0 | \begin{array}{c} \frac{3}{2} & - λ \\ 1 & 2 \end{array} |$

Step 5.2

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

$p (λ) = (9 - λ) | \begin{array}{c} - λ & - 2 \\ 2 & 4 - λ \end{array} | - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3

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

Tap for more steps...

Step 5.3.1

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

$p (λ) = (9 - λ) (- λ (4 - λ) - 2 \cdot - 2) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3.2

Simplify the determinant.

Tap for more steps...

Step 5.3.2.1

Simplify each term.

Tap for more steps...

Step 5.3.2.1.1

Apply the distributive property.

$p (λ) = (9 - λ) (- λ \cdot 4 - λ (- λ) - 2 \cdot - 2) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3.2.1.2

Multiply $4$ by $- 1$ .

$p (λ) = (9 - λ) (- 4 λ - λ (- λ) - 2 \cdot - 2) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3.2.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (9 - λ) (- 4 λ - 1 \cdot - 1 λ \cdot λ - 2 \cdot - 2) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3.2.1.4

Simplify each term.

Tap for more steps...

Step 5.3.2.1.4.1

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

Tap for more steps...

Step 5.3.2.1.4.1.1

Move $λ$ .

$p (λ) = (9 - λ) (- 4 λ - 1 \cdot - 1 (λ \cdot λ) - 2 \cdot - 2) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3.2.1.4.1.2

Multiply $λ$ by $λ$ .

$p (λ) = (9 - λ) (- 4 λ - 1 \cdot - 1 λ^{2} - 2 \cdot - 2) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3.2.1.4.2

Multiply $- 1$ by $- 1$ .

$p (λ) = (9 - λ) (- 4 λ + 1 λ^{2} - 2 \cdot - 2) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3.2.1.4.3

Multiply $λ^{2}$ by $1$ .

$p (λ) = (9 - λ) (- 4 λ + λ^{2} - 2 \cdot - 2) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3.2.1.5

Multiply $- 2$ by $- 2$ .

$p (λ) = (9 - λ) (- 4 λ + λ^{2} + 4) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.3.2.2

Reorder $- 4 λ$ and $λ^{2}$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 | \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} | + 0$

Step 5.4

Evaluate $| \begin{array}{c} \frac{3}{2} & - 2 \\ 1 & 4 - λ \end{array} |$ .

Tap for more steps...

Step 5.4.1

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

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (\frac{3}{2} (4 - λ) - 1 \cdot - 2) + 0$

Step 5.4.2

Simplify the determinant.

Tap for more steps...

Step 5.4.2.1

Simplify each term.

Tap for more steps...

Step 5.4.2.1.1

Apply the distributive property.

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (\frac{3}{2} \cdot 4 + \frac{3}{2} (- λ) - 1 \cdot - 2) + 0$

Step 5.4.2.1.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 5.4.2.1.2.1

Factor $2$ out of $4$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (\frac{3}{2} \cdot (2 (2)) + \frac{3}{2} (- λ) - 1 \cdot - 2) + 0$

Step 5.4.2.1.2.2

Cancel the common factor.

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (\frac{3}{2} \cdot (2 \cdot 2) + \frac{3}{2} (- λ) - 1 \cdot - 2) + 0$

Step 5.4.2.1.2.3

Rewrite the expression.

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (3 \cdot 2 + \frac{3}{2} (- λ) - 1 \cdot - 2) + 0$

Step 5.4.2.1.3

Multiply $3$ by $2$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (6 + \frac{3}{2} (- λ) - 1 \cdot - 2) + 0$

Step 5.4.2.1.4

Combine $\frac{3}{2}$ and $λ$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (6 - \frac{3 λ}{2} - 1 \cdot - 2) + 0$

Step 5.4.2.1.5

Multiply $- 1$ by $- 2$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (6 - \frac{3 λ}{2} + 2) + 0$

Step 5.4.2.2

Add $6$ and $2$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (- \frac{3 λ}{2} + 8) + 0$

Step 5.5

Simplify the determinant.

Tap for more steps...

Step 5.5.1

Add $(9 - λ) (λ^{2} - 4 λ + 4) - 1 (- \frac{3 λ}{2} + 8)$ and $0$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ + 4) - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2

Simplify each term.

Tap for more steps...

Step 5.5.2.1

Expand $(9 - λ) (λ^{2} - 4 λ + 4)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 9 λ^{2} + 9 (- 4 λ) + 9 \cdot 4 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2

Simplify each term.

Tap for more steps...

Step 5.5.2.2.1

Multiply $- 4$ by $9$ .

$p (λ) = 9 λ^{2} - 36 λ + 9 \cdot 4 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.2

Multiply $9$ by $4$ .

$p (λ) = 9 λ^{2} - 36 λ + 36 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.3

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

Tap for more steps...

Step 5.5.2.2.3.1

Move $λ^{2}$ .

$p (λ) = 9 λ^{2} - 36 λ + 36 - (λ^{2} λ) - λ (- 4 λ) - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.3.2

Multiply $λ^{2}$ by $λ$ .

Tap for more steps...

Step 5.5.2.2.3.2.1

Raise $λ$ to the power of $1$ .

$p (λ) = 9 λ^{2} - 36 λ + 36 - (λ^{2} λ^{1}) - λ (- 4 λ) - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.3.2.2

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

$p (λ) = 9 λ^{2} - 36 λ + 36 - λ^{2 + 1} - λ (- 4 λ) - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.3.3

Add $2$ and $1$ .

$p (λ) = 9 λ^{2} - 36 λ + 36 - λ^{3} - λ (- 4 λ) - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = 9 λ^{2} - 36 λ + 36 - λ^{3} - 1 \cdot - 4 λ \cdot λ - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.5

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

Tap for more steps...

Step 5.5.2.2.5.1

Move $λ$ .

$p (λ) = 9 λ^{2} - 36 λ + 36 - λ^{3} - 1 \cdot - 4 (λ \cdot λ) - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.5.2

Multiply $λ$ by $λ$ .

$p (λ) = 9 λ^{2} - 36 λ + 36 - λ^{3} - 1 \cdot - 4 λ^{2} - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.6

Multiply $- 1$ by $- 4$ .

$p (λ) = 9 λ^{2} - 36 λ + 36 - λ^{3} + 4 λ^{2} - λ \cdot 4 - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.2.7

Multiply $4$ by $- 1$ .

$p (λ) = 9 λ^{2} - 36 λ + 36 - λ^{3} + 4 λ^{2} - 4 λ - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.3

Add $9 λ^{2}$ and $4 λ^{2}$ .

$p (λ) = 13 λ^{2} - 36 λ + 36 - λ^{3} - 4 λ - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.4

Subtract $4 λ$ from $- 36 λ$ .

$p (λ) = 13 λ^{2} - 40 λ + 36 - λ^{3} - 1 (- \frac{3 λ}{2} + 8)$

Step 5.5.2.5

Apply the distributive property.

$p (λ) = 13 λ^{2} - 40 λ + 36 - λ^{3} - 1 (- \frac{3 λ}{2}) - 1 \cdot 8$

Step 5.5.2.6

Multiply $- 1 (- \frac{3 λ}{2})$ .

Tap for more steps...

Step 5.5.2.6.1

Multiply $- 1$ by $- 1$ .

$p (λ) = 13 λ^{2} - 40 λ + 36 - λ^{3} + 1 \frac{3 λ}{2} - 1 \cdot 8$

Step 5.5.2.6.2

Multiply $\frac{3 λ}{2}$ by $1$ .

$p (λ) = 13 λ^{2} - 40 λ + 36 - λ^{3} + \frac{3 λ}{2} - 1 \cdot 8$

Step 5.5.2.7

Multiply $- 1$ by $8$ .

$p (λ) = 13 λ^{2} - 40 λ + 36 - λ^{3} + \frac{3 λ}{2} - 8$

Step 5.5.3

To write $- 40 λ$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$p (λ) = 13 λ^{2} + 36 - λ^{3} - 40 λ \cdot \frac{2}{2} + \frac{3 λ}{2} - 8$

Step 5.5.4

Combine $- 40 λ$ and $\frac{2}{2}$ .

$p (λ) = 13 λ^{2} + 36 - λ^{3} + \frac{- 40 λ \cdot 2}{2} + \frac{3 λ}{2} - 8$

Step 5.5.5

Combine the numerators over the common denominator.

$p (λ) = 13 λ^{2} + 36 - λ^{3} + \frac{- 40 λ \cdot 2 + 3 λ}{2} - 8$

Step 5.5.6

Simplify each term.

Tap for more steps...

Step 5.5.6.1

Simplify the numerator.

Tap for more steps...

Step 5.5.6.1.1

Factor $λ$ out of $- 40 λ \cdot 2 + 3 λ$ .

Tap for more steps...

Step 5.5.6.1.1.1

Factor $λ$ out of $- 40 λ \cdot 2$ .

$p (λ) = 13 λ^{2} + 36 - λ^{3} + \frac{λ (- 40 \cdot 2) + 3 λ}{2} - 8$

Step 5.5.6.1.1.2

Factor $λ$ out of $3 λ$ .

$p (λ) = 13 λ^{2} + 36 - λ^{3} + \frac{λ (- 40 \cdot 2) + λ \cdot 3}{2} - 8$

Step 5.5.6.1.1.3

Factor $λ$ out of $λ (- 40 \cdot 2) + λ \cdot 3$ .

$p (λ) = 13 λ^{2} + 36 - λ^{3} + \frac{λ (- 40 \cdot 2 + 3)}{2} - 8$

Step 5.5.6.1.2

Multiply $- 40$ by $2$ .

$p (λ) = 13 λ^{2} + 36 - λ^{3} + \frac{λ (- 80 + 3)}{2} - 8$

Step 5.5.6.1.3

Add $- 80$ and $3$ .

$p (λ) = 13 λ^{2} + 36 - λ^{3} + \frac{λ \cdot - 77}{2} - 8$

Step 5.5.6.2

Move $- 77$ to the left of $λ$ .

$p (λ) = 13 λ^{2} + 36 - λ^{3} + \frac{- 77 \cdot λ}{2} - 8$

Step 5.5.6.3

Move the negative in front of the fraction.

$p (λ) = 13 λ^{2} + 36 - λ^{3} - \frac{77 λ}{2} - 8$

Step 5.5.7

Subtract $8$ from $36$ .

$p (λ) = 13 λ^{2} - λ^{3} - \frac{77 λ}{2} + 28$

Step 5.5.8

Reorder $13 λ^{2}$ and $- λ^{3}$ .

$p (λ) = - λ^{3} + 13 λ^{2} - \frac{77 λ}{2} + 28$

Step 6

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

$- λ^{3} + 13 λ^{2} - \frac{77 λ}{2} + 28 = 0$

Step 7

Solve for $λ$ .

Tap for more steps...

Step 7.1

Graph each side of the equation. The solution is the x-value of the point of intersection.

$λ \approx 1.10363103, 2.78431018, 9.11205878$