Precalculus Examples

$[\begin{array}{c} 5 & 8 & 1 \\ 6 & 1 & 9 \\ 7 & 5 & 7 \end{array}]$

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.

$| \begin{array}{c} 1 & 9 \\ 5 & 7 \end{array} |$

Step 1.1.4

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

$5 | \begin{array}{c} 1 & 9 \\ 5 & 7 \end{array} |$

Step 1.1.5

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

$| \begin{array}{c} 6 & 9 \\ 7 & 7 \end{array} |$

Step 1.1.6

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

$- 8 | \begin{array}{c} 6 & 9 \\ 7 & 7 \end{array} |$

Step 1.1.7

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

$| \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.1.8

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

$1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.1.9

Add the terms together.

$5 | \begin{array}{c} 1 & 9 \\ 5 & 7 \end{array} | - 8 | \begin{array}{c} 6 & 9 \\ 7 & 7 \end{array} | + 1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.2

Evaluate $| \begin{array}{c} 1 & 9 \\ 5 & 7 \end{array} |$ .

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Step 1.2.1

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

$5 (1 \cdot 7 - 5 \cdot 9) - 8 | \begin{array}{c} 6 & 9 \\ 7 & 7 \end{array} | + 1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.2.2

Simplify the determinant.

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Step 1.2.2.1

Simplify each term.

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Step 1.2.2.1.1

Multiply $7$ by $1$ .

$5 (7 - 5 \cdot 9) - 8 | \begin{array}{c} 6 & 9 \\ 7 & 7 \end{array} | + 1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.2.2.1.2

Multiply $- 5$ by $9$ .

$5 (7 - 45) - 8 | \begin{array}{c} 6 & 9 \\ 7 & 7 \end{array} | + 1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.2.2.2

Subtract $45$ from $7$ .

$5 \cdot - 38 - 8 | \begin{array}{c} 6 & 9 \\ 7 & 7 \end{array} | + 1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.3

Evaluate $| \begin{array}{c} 6 & 9 \\ 7 & 7 \end{array} |$ .

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Step 1.3.1

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

$5 \cdot - 38 - 8 (6 \cdot 7 - 7 \cdot 9) + 1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

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 $6$ by $7$ .

$5 \cdot - 38 - 8 (42 - 7 \cdot 9) + 1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.3.2.1.2

Multiply $- 7$ by $9$ .

$5 \cdot - 38 - 8 (42 - 63) + 1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.3.2.2

Subtract $63$ from $42$ .

$5 \cdot - 38 - 8 \cdot - 21 + 1 | \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$

Step 1.4

Evaluate $| \begin{array}{c} 6 & 1 \\ 7 & 5 \end{array} |$ .

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Step 1.4.1

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

$5 \cdot - 38 - 8 \cdot - 21 + 1 (6 \cdot 5 - 7 \cdot 1)$

Step 1.4.2

Simplify the determinant.

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Step 1.4.2.1

Simplify each term.

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Step 1.4.2.1.1

Multiply $6$ by $5$ .

$5 \cdot - 38 - 8 \cdot - 21 + 1 (30 - 7 \cdot 1)$

Step 1.4.2.1.2

Multiply $- 7$ by $1$ .

$5 \cdot - 38 - 8 \cdot - 21 + 1 (30 - 7)$

Step 1.4.2.2

Subtract $7$ from $30$ .

$5 \cdot - 38 - 8 \cdot - 21 + 1 \cdot 23$

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 $5$ by $- 38$ .

$- 190 - 8 \cdot - 21 + 1 \cdot 23$

Step 1.5.1.2

Multiply $- 8$ by $- 21$ .

$- 190 + 168 + 1 \cdot 23$

Step 1.5.1.3

Multiply $23$ by $1$ .

$- 190 + 168 + 23$

Step 1.5.2

Add $- 190$ and $168$ .

$- 22 + 23$

Step 1.5.3

Add $- 22$ and $23$ .

$1$

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.

$[\begin{array}{c} 5 & 8 & 1 & 1 & 0 & 0 \\ 6 & 1 & 9 & 0 & 1 & 0 \\ 7 & 5 & 7 & 0 & 0 & 1 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $\frac{1}{5}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{5}{5} & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & \frac{0}{5} & \frac{0}{5} \\ 6 & 1 & 9 & 0 & 1 & 0 \\ 7 & 5 & 7 & 0 & 0 & 1 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 6 & 1 & 9 & 0 & 1 & 0 \\ 7 & 5 & 7 & 0 & 0 & 1 \end{array}]$

Step 4.2

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 6 - 6 \cdot 1 & 1 - 6 (\frac{8}{5}) & 9 - 6 (\frac{1}{5}) & 0 - 6 (\frac{1}{5}) & 1 - 6 \cdot 0 & 0 - 6 \cdot 0 \\ 7 & 5 & 7 & 0 & 0 & 1 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & - \frac{43}{5} & \frac{39}{5} & - \frac{6}{5} & 1 & 0 \\ 7 & 5 & 7 & 0 & 0 & 1 \end{array}]$

Step 4.3

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.3.1

Perform the row operation $R_{3} = R_{3} - 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & - \frac{43}{5} & \frac{39}{5} & - \frac{6}{5} & 1 & 0 \\ 7 - 7 \cdot 1 & 5 - 7 (\frac{8}{5}) & 7 - 7 (\frac{1}{5}) & 0 - 7 (\frac{1}{5}) & 0 - 7 \cdot 0 & 1 - 7 \cdot 0 \end{array}]$

Step 4.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & - \frac{43}{5} & \frac{39}{5} & - \frac{6}{5} & 1 & 0 \\ 0 & - \frac{31}{5} & \frac{28}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

Step 4.4

Multiply each element of $R_{2}$ by $- \frac{5}{43}$ to make the entry at $2, 2$ a $1$ .

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Step 4.4.1

Multiply each element of $R_{2}$ by $- \frac{5}{43}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ - \frac{5}{43} \cdot 0 & - \frac{5}{43} (- \frac{43}{5}) & - \frac{5}{43} \cdot \frac{39}{5} & - \frac{5}{43} (- \frac{6}{5}) & - \frac{5}{43} \cdot 1 & - \frac{5}{43} \cdot 0 \\ 0 & - \frac{31}{5} & \frac{28}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

Step 4.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{39}{43} & \frac{6}{43} & - \frac{5}{43} & 0 \\ 0 & - \frac{31}{5} & \frac{28}{5} & - \frac{7}{5} & 0 & 1 \end{array}]$

Step 4.5

Perform the row operation $R_{3} = R_{3} + \frac{31}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.5.1

Perform the row operation $R_{3} = R_{3} + \frac{31}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{39}{43} & \frac{6}{43} & - \frac{5}{43} & 0 \\ 0 + \frac{31}{5} \cdot 0 & - \frac{31}{5} + \frac{31}{5} \cdot 1 & \frac{28}{5} + \frac{31}{5} (- \frac{39}{43}) & - \frac{7}{5} + \frac{31}{5} \cdot \frac{6}{43} & 0 + \frac{31}{5} (- \frac{5}{43}) & 1 + \frac{31}{5} \cdot 0 \end{array}]$

Step 4.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{39}{43} & \frac{6}{43} & - \frac{5}{43} & 0 \\ 0 & 0 & - \frac{1}{43} & - \frac{23}{43} & - \frac{31}{43} & 1 \end{array}]$

Step 4.6

Multiply each element of $R_{3}$ by $- 43$ to make the entry at $3, 3$ a $1$ .

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Step 4.6.1

Multiply each element of $R_{3}$ by $- 43$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{39}{43} & \frac{6}{43} & - \frac{5}{43} & 0 \\ - 43 \cdot 0 & - 43 \cdot 0 & - 43 (- \frac{1}{43}) & - 43 (- \frac{23}{43}) & - 43 (- \frac{31}{43}) & - 43 \cdot 1 \end{array}]$

Step 4.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & - \frac{39}{43} & \frac{6}{43} & - \frac{5}{43} & 0 \\ 0 & 0 & 1 & 23 & 31 & - 43 \end{array}]$

Step 4.7

Perform the row operation $R_{2} = R_{2} + \frac{39}{43} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 4.7.1

Perform the row operation $R_{2} = R_{2} + \frac{39}{43} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 + \frac{39}{43} \cdot 0 & 1 + \frac{39}{43} \cdot 0 & - \frac{39}{43} + \frac{39}{43} \cdot 1 & \frac{6}{43} + \frac{39}{43} \cdot 23 & - \frac{5}{43} + \frac{39}{43} \cdot 31 & 0 + \frac{39}{43} \cdot - 43 \\ 0 & 0 & 1 & 23 & 31 & - 43 \end{array}]$

Step 4.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{8}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 \\ 0 & 1 & 0 & 21 & 28 & - 39 \\ 0 & 0 & 1 & 23 & 31 & - 43 \end{array}]$

Step 4.8

Perform the row operation $R_{1} = R_{1} - \frac{1}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 4.8.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{5} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{5} \cdot 0 & \frac{8}{5} - \frac{1}{5} \cdot 0 & \frac{1}{5} - \frac{1}{5} \cdot 1 & \frac{1}{5} - \frac{1}{5} \cdot 23 & 0 - \frac{1}{5} \cdot 31 & 0 - \frac{1}{5} \cdot - 43 \\ 0 & 1 & 0 & 21 & 28 & - 39 \\ 0 & 0 & 1 & 23 & 31 & - 43 \end{array}]$

Step 4.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{8}{5} & 0 & - \frac{22}{5} & - \frac{31}{5} & \frac{43}{5} \\ 0 & 1 & 0 & 21 & 28 & - 39 \\ 0 & 0 & 1 & 23 & 31 & - 43 \end{array}]$

Step 4.9

Perform the row operation $R_{1} = R_{1} - \frac{8}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 4.9.1

Perform the row operation $R_{1} = R_{1} - \frac{8}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{8}{5} \cdot 0 & \frac{8}{5} - \frac{8}{5} \cdot 1 & 0 - \frac{8}{5} \cdot 0 & - \frac{22}{5} - \frac{8}{5} \cdot 21 & - \frac{31}{5} - \frac{8}{5} \cdot 28 & \frac{43}{5} - \frac{8}{5} \cdot - 39 \\ 0 & 1 & 0 & 21 & 28 & - 39 \\ 0 & 0 & 1 & 23 & 31 & - 43 \end{array}]$

Step 4.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - 38 & - 51 & 71 \\ 0 & 1 & 0 & 21 & 28 & - 39 \\ 0 & 0 & 1 & 23 & 31 & - 43 \end{array}]$

Step 5

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

$[\begin{array}{c} - 38 & - 51 & 71 \\ 21 & 28 & - 39 \\ 23 & 31 & - 43 \end{array}]$