Linear Algebra Examples

⎡ ⎢ ⎣ \begin{matrix} 9 & 8 & 7 3 & 4 & 5 2 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}]$

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

$[\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}]$ for

$A$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] - λ I_{3})$

Step 3.2

Substitute

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

$I_{3}$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] - λ [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}])$

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.

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ \cdot 1 & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2

Simplify each element in the matrix.

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Step 4.1.2.1

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & - λ \cdot 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.2.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.2.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.3.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 λ \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.3.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ - λ \cdot 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.4.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 λ & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.4.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ \cdot 1 & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.5

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & - λ \cdot 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.6.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 λ \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.6.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ - λ \cdot 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.7.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 λ & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.7.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & - λ \cdot 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.8

Multiply

$- λ \cdot 0$ .

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Step 4.1.2.8.1

Multiply

$0$ by

$- 1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 λ & - λ \cdot 1 \end{array}])$

Step 4.1.2.8.2

Multiply

$0$ by

$λ$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \cdot 1 \end{array}])$

Step 4.1.2.9

Multiply

$- 1$ by

$1$ .

$p (λ) = determinant ([\begin{array}{c} 9 & 8 & 7 \\ 3 & 4 & 5 \\ 2 & 1 & 0 \end{array}] + [\begin{array}{c} - λ & 0 & 0 \\ 0 & - λ & 0 \\ 0 & 0 & - λ \end{array}])$

Step 4.2

Add the corresponding elements.

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 + 0 & 7 + 0 \\ 3 + 0 & 4 - λ & 5 + 0 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3

Simplify each element.

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Step 4.3.1

Add

$8$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 + 0 \\ 3 + 0 & 4 - λ & 5 + 0 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.2

Add

$7$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 + 0 & 4 - λ & 5 + 0 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.3

Add

$3$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 + 0 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.4

Add

$5$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 \\ 2 + 0 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.5

Add

$2$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 \\ 2 & 1 + 0 & 0 - λ \end{array}]$

Step 4.3.6

Add

$1$ and

$0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 \\ 2 & 1 & 0 - λ \end{array}]$

Step 4.3.7

Subtract

$λ$ from

$0$ .

$p (λ) = determinant [\begin{array}{c} 9 - λ & 8 & 7 \\ 3 & 4 - λ & 5 \\ 2 & 1 & - λ \end{array}]$

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 row

$1$ 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

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} 4 - λ & 5 \\ 1 & - λ \end{array} |$

Step 5.1.4

Multiply element

$a_{11}$ by its cofactor.

$(9 - λ) | \begin{array}{c} 4 - λ & 5 \\ 1 & - λ \end{array} |$

Step 5.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

$| \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} |$

Step 5.1.6

Multiply element

$a_{12}$ by its cofactor.

$- 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} |$

Step 5.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

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

Step 5.1.8

Multiply element

$a_{13}$ by its cofactor.

$7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.1.9

Add the terms together.

$p (λ) = (9 - λ) | \begin{array}{c} 4 - λ & 5 \\ 1 & - λ \end{array} | - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2

Evaluate

$| \begin{array}{c} 4 - λ & 5 \\ 1 & - λ \end{array} |$ .

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Step 5.2.1

The determinant of a

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

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

$p (λ) = (9 - λ) ((4 - λ) (- λ) - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2

Simplify the determinant.

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Step 5.2.2.1

Simplify each term.

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Step 5.2.2.1.1

Apply the distributive property.

$p (λ) = (9 - λ) (4 (- λ) - λ (- λ) - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.2

Multiply

$- 1$ by

$4$ .

$p (λ) = (9 - λ) (- 4 λ - λ (- λ) - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.3

Rewrite using the commutative property of multiplication.

$p (λ) = (9 - λ) (- 4 λ - 1 \cdot - 1 λ \cdot λ - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.4

Simplify each term.

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Step 5.2.2.1.4.1

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 5.2.2.1.4.1.1

Move

$λ$ .

$p (λ) = (9 - λ) (- 4 λ - 1 \cdot - 1 (λ \cdot λ) - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.4.1.2

Multiply

$λ$ by

$λ$ .

$p (λ) = (9 - λ) (- 4 λ - 1 \cdot - 1 λ^{2} - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.4.2

Multiply

$- 1$ by

$- 1$ .

$p (λ) = (9 - λ) (- 4 λ + 1 λ^{2} - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.4.3

Multiply

$λ^{2}$ by

$1$ .

$p (λ) = (9 - λ) (- 4 λ + λ^{2} - 1 \cdot 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.1.5

Multiply

$- 1$ by

$5$ .

$p (λ) = (9 - λ) (- 4 λ + λ^{2} - 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.2.2.2

Reorder

$- 4 λ$ and

$λ^{2}$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 | \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} | + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.3

Evaluate

$| \begin{array}{c} 3 & 5 \\ 2 & - λ \end{array} |$ .

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Step 5.3.1

The determinant of a

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

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

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (3 (- λ) - 2 \cdot 5) + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.3.2

Simplify each term.

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Step 5.3.2.1

Multiply

$- 1$ by

$3$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 2 \cdot 5) + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.3.2.2

Multiply

$- 2$ by

$5$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 | \begin{array}{c} 3 & 4 - λ \\ 2 & 1 \end{array} |$

Step 5.4

Evaluate

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

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Step 5.4.1

The determinant of a

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

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

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 \cdot 1 - 2 (4 - λ))$

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

Multiply

$3$ by

$1$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 - 2 (4 - λ))$

Step 5.4.2.1.2

Apply the distributive property.

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 - 2 \cdot 4 - 2 (- λ))$

Step 5.4.2.1.3

Multiply

$- 2$ by

$4$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 - 8 - 2 (- λ))$

Step 5.4.2.1.4

Multiply

$- 1$ by

$- 2$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (3 - 8 + 2 λ)$

Step 5.4.2.2

Subtract

$8$ from

$3$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (- 5 + 2 λ)$

Step 5.4.2.3

Reorder

$- 5$ and

$2 λ$ .

$p (λ) = (9 - λ) (λ^{2} - 4 λ - 5) - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5

Simplify the determinant.

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Step 5.5.1

Simplify each term.

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Step 5.5.1.1

Expand

$(9 - λ) (λ^{2} - 4 λ - 5)$ by multiplying each term in the first expression by each term in the second expression.

$p (λ) = 9 λ^{2} + 9 (- 4 λ) + 9 \cdot - 5 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2

Simplify each term.

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Step 5.5.1.2.1

Multiply

$- 4$ by

$9$ .

$p (λ) = 9 λ^{2} - 36 λ + 9 \cdot - 5 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.2

Multiply

$9$ by

$- 5$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ \cdot λ^{2} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.3

Multiply

$λ$ by

$λ^{2}$ by adding the exponents.

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Step 5.5.1.2.3.1

Move

$λ^{2}$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - (λ^{2} λ) - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.3.2

Multiply

$λ^{2}$ by

$λ$ .

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Step 5.5.1.2.3.2.1

Raise

$λ$ to the power of

$1$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - (λ^{2} λ^{1}) - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.3.2.2

Use the power rule

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

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{2 + 1} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.3.3

Add

$2$ and

$1$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} - λ (- 4 λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.4

Rewrite using the commutative property of multiplication.

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} - 1 \cdot - 4 λ \cdot λ - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.5

Multiply

$λ$ by

$λ$ by adding the exponents.

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Step 5.5.1.2.5.1

Move

$λ$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} - 1 \cdot - 4 (λ \cdot λ) - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.5.2

Multiply

$λ$ by

$λ$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} - 1 \cdot - 4 λ^{2} - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.6

Multiply

$- 1$ by

$- 4$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} + 4 λ^{2} - λ \cdot - 5 - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.2.7

Multiply

$- 5$ by

$- 1$ .

$p (λ) = 9 λ^{2} - 36 λ - 45 - λ^{3} + 4 λ^{2} + 5 λ - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.3

Add

$9 λ^{2}$ and

$4 λ^{2}$ .

$p (λ) = 13 λ^{2} - 36 λ - 45 - λ^{3} + 5 λ - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.4

Add

$- 36 λ$ and

$5 λ$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} - 8 (- 3 λ - 10) + 7 (2 λ - 5)$

Step 5.5.1.5

Apply the distributive property.

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} - 8 (- 3 λ) - 8 \cdot - 10 + 7 (2 λ - 5)$

Step 5.5.1.6

Multiply

$- 3$ by

$- 8$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ - 8 \cdot - 10 + 7 (2 λ - 5)$

Step 5.5.1.7

Multiply

$- 8$ by

$- 10$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ + 80 + 7 (2 λ - 5)$

Step 5.5.1.8

Apply the distributive property.

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ + 80 + 7 (2 λ) + 7 \cdot - 5$

Step 5.5.1.9

Multiply

$2$ by

$7$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ + 80 + 14 λ + 7 \cdot - 5$

Step 5.5.1.10

Multiply

$7$ by

$- 5$ .

$p (λ) = 13 λ^{2} - 31 λ - 45 - λ^{3} + 24 λ + 80 + 14 λ - 35$

Step 5.5.2

Add

$- 31 λ$ and

$24 λ$ .

$p (λ) = 13 λ^{2} - 7 λ - 45 - λ^{3} + 80 + 14 λ - 35$

Step 5.5.3

Add

$- 7 λ$ and

$14 λ$ .

$p (λ) = 13 λ^{2} + 7 λ - 45 - λ^{3} + 80 - 35$

Step 5.5.4

Add

$- 45$ and

$80$ .

$p (λ) = 13 λ^{2} + 7 λ - λ^{3} + 35 - 35$

Step 5.5.5

Combine the opposite terms in

$13 λ^{2} + 7 λ - λ^{3} + 35 - 35$ .

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Step 5.5.5.1

Subtract

$35$ from

$35$ .

$p (λ) = 13 λ^{2} + 7 λ - λ^{3} + 0$

Step 5.5.5.2

Add

$13 λ^{2} + 7 λ - λ^{3}$ and

$0$ .

$p (λ) = 13 λ^{2} + 7 λ - λ^{3}$

Step 5.5.6

Move

$7 λ$ .

$p (λ) = 13 λ^{2} - λ^{3} + 7 λ$

Step 5.5.7

Reorder

$13 λ^{2}$ and

$- λ^{3}$ .

$p (λ) = - λ^{3} + 13 λ^{2} + 7 λ$

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