Finite Math Examples

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

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 column $1$ 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.

$| \begin{array}{c} 4 & 2 \\ 3 & 5 \end{array} |$

Step 1.1.4

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

$- 7 | \begin{array}{c} 4 & 2 \\ 3 & 5 \end{array} |$

Step 1.1.5

The minor for $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

$| \begin{array}{c} - 1 & 5 \\ 3 & 5 \end{array} |$

Step 1.1.6

Multiply element $a_{21}$ by its cofactor.

$0 | \begin{array}{c} - 1 & 5 \\ 3 & 5 \end{array} |$

Step 1.1.7

The minor for $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} - 1 & 5 \\ 4 & 2 \end{array} |$

Step 1.1.8

Multiply element $a_{31}$ by its cofactor.

$8 | \begin{array}{c} - 1 & 5 \\ 4 & 2 \end{array} |$

Step 1.1.9

Add the terms together.

$- 7 | \begin{array}{c} 4 & 2 \\ 3 & 5 \end{array} | + 0 | \begin{array}{c} - 1 & 5 \\ 3 & 5 \end{array} | + 8 | \begin{array}{c} - 1 & 5 \\ 4 & 2 \end{array} |$

Step 1.2

Multiply $0$ by $| \begin{array}{c} - 1 & 5 \\ 3 & 5 \end{array} |$ .

$- 7 | \begin{array}{c} 4 & 2 \\ 3 & 5 \end{array} | + 0 + 8 | \begin{array}{c} - 1 & 5 \\ 4 & 2 \end{array} |$

Step 1.3

Evaluate $| \begin{array}{c} 4 & 2 \\ 3 & 5 \end{array} |$ .

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Step 1.3.1

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

$- 7 (4 \cdot 5 - 3 \cdot 2) + 0 + 8 | \begin{array}{c} - 1 & 5 \\ 4 & 2 \end{array} |$

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 $4$ by $5$ .

$- 7 (20 - 3 \cdot 2) + 0 + 8 | \begin{array}{c} - 1 & 5 \\ 4 & 2 \end{array} |$

Step 1.3.2.1.2

Multiply $- 3$ by $2$ .

$- 7 (20 - 6) + 0 + 8 | \begin{array}{c} - 1 & 5 \\ 4 & 2 \end{array} |$

Step 1.3.2.2

Subtract $6$ from $20$ .

$- 7 \cdot 14 + 0 + 8 | \begin{array}{c} - 1 & 5 \\ 4 & 2 \end{array} |$

Step 1.4

Evaluate $| \begin{array}{c} - 1 & 5 \\ 4 & 2 \end{array} |$ .

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Step 1.4.1

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

$- 7 \cdot 14 + 0 + 8 (- 1 \cdot 2 - 4 \cdot 5)$

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 $- 1$ by $2$ .

$- 7 \cdot 14 + 0 + 8 (- 2 - 4 \cdot 5)$

Step 1.4.2.1.2

Multiply $- 4$ by $5$ .

$- 7 \cdot 14 + 0 + 8 (- 2 - 20)$

Step 1.4.2.2

Subtract $20$ from $- 2$ .

$- 7 \cdot 14 + 0 + 8 \cdot - 22$

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 $- 7$ by $14$ .

$- 98 + 0 + 8 \cdot - 22$

Step 1.5.1.2

Multiply $8$ by $- 22$ .

$- 98 + 0 - 176$

Step 1.5.2

Add $- 98$ and $0$ .

$- 98 - 176$

Step 1.5.3

Subtract $176$ from $- 98$ .

$- 274$

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.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{7}$ to make the entry at $1, 1$ a $1$ .

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

Step 4.1.2

Simplify $R_{1}$ .

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

Step 4.2

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{3} = R_{3} - 8 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ 0 & 4 & 2 & 0 & 1 & 0 \\ 8 - 8 \cdot 1 & 3 - 8 (\frac{1}{7}) & 5 - 8 (- \frac{5}{7}) & 0 - 8 (- \frac{1}{7}) & 0 - 8 \cdot 0 & 1 - 8 \cdot 0 \end{array}]$

Step 4.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ 0 & 4 & 2 & 0 & 1 & 0 \\ 0 & \frac{13}{7} & \frac{75}{7} & \frac{8}{7} & 0 & 1 \end{array}]$

Step 4.3

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $\frac{1}{4}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ \frac{0}{4} & \frac{4}{4} & \frac{2}{4} & \frac{0}{4} & \frac{1}{4} & \frac{0}{4} \\ 0 & \frac{13}{7} & \frac{75}{7} & \frac{8}{7} & 0 & 1 \end{array}]$

Step 4.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & \frac{1}{4} & 0 \\ 0 & \frac{13}{7} & \frac{75}{7} & \frac{8}{7} & 0 & 1 \end{array}]$

Step 4.4

Perform the row operation $R_{3} = R_{3} - \frac{13}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.4.1

Perform the row operation $R_{3} = R_{3} - \frac{13}{7} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & \frac{1}{4} & 0 \\ 0 - \frac{13}{7} \cdot 0 & \frac{13}{7} - \frac{13}{7} \cdot 1 & \frac{75}{7} - \frac{13}{7} \cdot \frac{1}{2} & \frac{8}{7} - \frac{13}{7} \cdot 0 & 0 - \frac{13}{7} \cdot \frac{1}{4} & 1 - \frac{13}{7} \cdot 0 \end{array}]$

Step 4.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{137}{14} & \frac{8}{7} & - \frac{13}{28} & 1 \end{array}]$

Step 4.5

Multiply each element of $R_{3}$ by $\frac{14}{137}$ to make the entry at $3, 3$ a $1$ .

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Step 4.5.1

Multiply each element of $R_{3}$ by $\frac{14}{137}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & \frac{1}{4} & 0 \\ \frac{14}{137} \cdot 0 & \frac{14}{137} \cdot 0 & \frac{14}{137} \cdot \frac{137}{14} & \frac{14}{137} \cdot \frac{8}{7} & \frac{14}{137} (- \frac{13}{28}) & \frac{14}{137} \cdot 1 \end{array}]$

Step 4.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ 0 & 1 & \frac{1}{2} & 0 & \frac{1}{4} & 0 \\ 0 & 0 & 1 & \frac{16}{137} & - \frac{13}{274} & \frac{14}{137} \end{array}]$

Step 4.6

Perform the row operation $R_{2} = R_{2} - \frac{1}{2} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 4.6.1

Perform the row operation $R_{2} = R_{2} - \frac{1}{2} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ 0 - \frac{1}{2} \cdot 0 & 1 - \frac{1}{2} \cdot 0 & \frac{1}{2} - \frac{1}{2} \cdot 1 & 0 - \frac{1}{2} \cdot \frac{16}{137} & \frac{1}{4} - \frac{1}{2} (- \frac{13}{274}) & 0 - \frac{1}{2} \cdot \frac{14}{137} \\ 0 & 0 & 1 & \frac{16}{137} & - \frac{13}{274} & \frac{14}{137} \end{array}]$

Step 4.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & - \frac{5}{7} & - \frac{1}{7} & 0 & 0 \\ 0 & 1 & 0 & - \frac{8}{137} & \frac{75}{274} & - \frac{7}{137} \\ 0 & 0 & 1 & \frac{16}{137} & - \frac{13}{274} & \frac{14}{137} \end{array}]$

Step 4.7

Perform the row operation $R_{1} = R_{1} + \frac{5}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 4.7.1

Perform the row operation $R_{1} = R_{1} + \frac{5}{7} R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 + \frac{5}{7} \cdot 0 & \frac{1}{7} + \frac{5}{7} \cdot 0 & - \frac{5}{7} + \frac{5}{7} \cdot 1 & - \frac{1}{7} + \frac{5}{7} \cdot \frac{16}{137} & 0 + \frac{5}{7} (- \frac{13}{274}) & 0 + \frac{5}{7} \cdot \frac{14}{137} \\ 0 & 1 & 0 & - \frac{8}{137} & \frac{75}{274} & - \frac{7}{137} \\ 0 & 0 & 1 & \frac{16}{137} & - \frac{13}{274} & \frac{14}{137} \end{array}]$

Step 4.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{7} & 0 & - \frac{57}{959} & - \frac{65}{1918} & \frac{10}{137} \\ 0 & 1 & 0 & - \frac{8}{137} & \frac{75}{274} & - \frac{7}{137} \\ 0 & 0 & 1 & \frac{16}{137} & - \frac{13}{274} & \frac{14}{137} \end{array}]$

Step 4.8

Perform the row operation $R_{1} = R_{1} - \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 4.8.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{7} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{7} \cdot 0 & \frac{1}{7} - \frac{1}{7} \cdot 1 & 0 - \frac{1}{7} \cdot 0 & - \frac{57}{959} - \frac{1}{7} (- \frac{8}{137}) & - \frac{65}{1918} - \frac{1}{7} \cdot \frac{75}{274} & \frac{10}{137} - \frac{1}{7} (- \frac{7}{137}) \\ 0 & 1 & 0 & - \frac{8}{137} & \frac{75}{274} & - \frac{7}{137} \\ 0 & 0 & 1 & \frac{16}{137} & - \frac{13}{274} & \frac{14}{137} \end{array}]$

Step 4.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{7}{137} & - \frac{10}{137} & \frac{11}{137} \\ 0 & 1 & 0 & - \frac{8}{137} & \frac{75}{274} & - \frac{7}{137} \\ 0 & 0 & 1 & \frac{16}{137} & - \frac{13}{274} & \frac{14}{137} \end{array}]$

Step 5

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

$[\begin{array}{c} - \frac{7}{137} & - \frac{10}{137} & \frac{11}{137} \\ - \frac{8}{137} & \frac{75}{274} & - \frac{7}{137} \\ \frac{16}{137} & - \frac{13}{274} & \frac{14}{137} \end{array}]$