Precalculus Examples

⎡ ⎢ ⎣ \begin{matrix} 0 & 2 & 2 2 & 0 & 1 1 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦

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

Step 1

Find the determinant.

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Step 1.1

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 1.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 1.1.2

The cofactor is the minor with the sign changed if the indices match a

$-$ position on the sign chart.

Step 1.1.3

The minor for

$a_{11}$ is the determinant with row

$1$ and column

$1$ deleted.

$| \begin{array}{c} 0 & 1 \\ 1 & 0 \end{array} |$

Step 1.1.4

Multiply element

$a_{11}$ by its cofactor.

$0 | \begin{array}{c} 0 & 1 \\ 1 & 0 \end{array} |$

Step 1.1.5

The minor for

$a_{12}$ is the determinant with row

$1$ and column

$2$ deleted.

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

Step 1.1.6

Multiply element

$a_{12}$ by its cofactor.

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

Step 1.1.7

The minor for

$a_{13}$ is the determinant with row

$1$ and column

$3$ deleted.

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

Step 1.1.8

Multiply element

$a_{13}$ by its cofactor.

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

Step 1.1.9

Add the terms together.

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

Step 1.2

Multiply

$0$ by

$| \begin{array}{c} 0 & 1 \\ 1 & 0 \end{array} |$ .

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

Step 1.3

Evaluate

$| \begin{array}{c} 2 & 1 \\ 1 & 0 \end{array} |$ .

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Step 1.3.1

The determinant of a

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

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

$0 - 2 (2 \cdot 0 - 1 \cdot 1) + 2 | \begin{array}{c} 2 & 0 \\ 1 & 1 \end{array} |$

Step 1.3.2

Simplify the determinant.

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Step 1.3.2.1

Simplify each term.

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Step 1.3.2.1.1

Multiply

$2$ by

$0$ .

$0 - 2 (0 - 1 \cdot 1) + 2 | \begin{array}{c} 2 & 0 \\ 1 & 1 \end{array} |$

Step 1.3.2.1.2

Multiply

$- 1$ by

$1$ .

$0 - 2 (0 - 1) + 2 | \begin{array}{c} 2 & 0 \\ 1 & 1 \end{array} |$

Step 1.3.2.2

Subtract

$1$ from

$0$ .

$0 - 2 \cdot - 1 + 2 | \begin{array}{c} 2 & 0 \\ 1 & 1 \end{array} |$

Step 1.4

Evaluate

$| \begin{array}{c} 2 & 0 \\ 1 & 1 \end{array} |$ .

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Step 1.4.1

The determinant of a

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

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

$0 - 2 \cdot - 1 + 2 (2 \cdot 1 - 1 \cdot 0)$

Step 1.4.2

Simplify the determinant.

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Step 1.4.2.1

Multiply

$2$ by

$1$ .

$0 - 2 \cdot - 1 + 2 (2 - 1 \cdot 0)$

Step 1.4.2.2

Subtract

$0$ from

$2$ .

$0 - 2 \cdot - 1 + 2 \cdot 2$

Step 1.5

Simplify the determinant.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

Multiply

$- 2$ by

$- 1$ .

$0 + 2 + 2 \cdot 2$

Step 1.5.1.2

Multiply

$2$ by

$2$ .

$0 + 2 + 4$

Step 1.5.2

Add

$0$ and

$2$ .

$2 + 4$

Step 1.5.3

Add

$2$ and

$4$ .

$6$

Step 2

Since the determinant is non-zero, the inverse exists.

Step 3

Set up a

$3 \times 6$ matrix where the left half is the original matrix and the right half is its identity matrix.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

Swap

$R_{2}$ with

$R_{1}$ to put a nonzero entry at

$1, 1$ .

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

Step 4.2

Multiply each element of

$R_{1}$ by

$\frac{1}{2}$ to make the entry at

$1, 1$ a

$1$ .

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Step 4.2.1

Multiply each element of

$R_{1}$ by

$\frac{1}{2}$ to make the entry at

$1, 1$ a

$1$ .

$[\begin{array}{c} \frac{2}{2} & \frac{0}{2} & \frac{1}{2} & \frac{0}{2} & \frac{1}{2} & \frac{0}{2} \\ 0 & 2 & 2 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.2.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & 2 & 2 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \end{array}]$

Step 4.3

Perform the row operation

$R_{3} = R_{3} - R_{1}$ to make the entry at

$3, 1$ a

$0$ .

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Step 4.3.1

Perform the row operation

$R_{3} = R_{3} - R_{1}$ to make the entry at

$3, 1$ a

$0$ .

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

Step 4.3.2

Simplify

$R_{3}$ .

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

Step 4.4

Multiply each element of

$R_{2}$ by

$\frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

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Step 4.4.1

Multiply each element of

$R_{2}$ by

$\frac{1}{2}$ to make the entry at

$2, 2$ a

$1$ .

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

Step 4.4.2

Simplify

$R_{2}$ .

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

Step 4.5

Perform the row operation

$R_{3} = R_{3} - R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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Step 4.5.1

Perform the row operation

$R_{3} = R_{3} - R_{2}$ to make the entry at

$3, 2$ a

$0$ .

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

Step 4.5.2

Simplify

$R_{3}$ .

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

Step 4.6

Multiply each element of

$R_{3}$ by

$- \frac{2}{3}$ to make the entry at

$3, 3$ a

$1$ .

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Step 4.6.1

Multiply each element of

$R_{3}$ by

$- \frac{2}{3}$ to make the entry at

$3, 3$ a

$1$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & 1 & 1 & \frac{1}{2} & 0 & 0 \\ - \frac{2}{3} \cdot 0 & - \frac{2}{3} \cdot 0 & - \frac{2}{3} (- \frac{3}{2}) & - \frac{2}{3} (- \frac{1}{2}) & - \frac{2}{3} (- \frac{1}{2}) & - \frac{2}{3} \cdot 1 \end{array}]$

Step 4.6.2

Simplify

$R_{3}$ .

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

Step 4.7

Perform the row operation

$R_{2} = R_{2} - R_{3}$ to make the entry at

$2, 3$ a

$0$ .

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Step 4.7.1

Perform the row operation

$R_{2} = R_{2} - R_{3}$ to make the entry at

$2, 3$ a

$0$ .

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

Step 4.7.2

Simplify

$R_{2}$ .

$[\begin{array}{c} 1 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{2}{3} \\ 0 & 0 & 1 & \frac{1}{3} & \frac{1}{3} & - \frac{2}{3} \end{array}]$

Step 4.8

Perform the row operation

$R_{1} = R_{1} - \frac{1}{2} R_{3}$ to make the entry at

$1, 3$ a

$0$ .

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Step 4.8.1

Perform the row operation

$R_{1} = R_{1} - \frac{1}{2} R_{3}$ to make the entry at

$1, 3$ a

$0$ .

$[\begin{array}{c} 1 - \frac{1}{2} \cdot 0 & 0 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot \frac{1}{3} & \frac{1}{2} - \frac{1}{2} \cdot \frac{1}{3} & 0 - \frac{1}{2} (- \frac{2}{3}) \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{2}{3} \\ 0 & 0 & 1 & \frac{1}{3} & \frac{1}{3} & - \frac{2}{3} \end{array}]$

Step 4.8.2

Simplify

$R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{1}{6} & \frac{1}{3} & \frac{1}{3} \\ 0 & 1 & 0 & \frac{1}{6} & - \frac{1}{3} & \frac{2}{3} \\ 0 & 0 & 1 & \frac{1}{3} & \frac{1}{3} & - \frac{2}{3} \end{array}]$

Step 5

The right half of the reduced row echelon form is the inverse.

$[\begin{array}{c} - \frac{1}{6} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} & - \frac{2}{3} \end{array}]$

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