Finite Math Examples

$[\begin{array}{c} 2 & 3 & 8 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}]$

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.

$| \begin{array}{c} 5 & 6 \\ 8 & 9 \end{array} |$

Step 1.1.4

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

$2 | \begin{array}{c} 5 & 6 \\ 8 & 9 \end{array} |$

Step 1.1.5

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

$| \begin{array}{c} 4 & 6 \\ 7 & 9 \end{array} |$

Step 1.1.6

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

$- 3 | \begin{array}{c} 4 & 6 \\ 7 & 9 \end{array} |$

Step 1.1.7

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

$| \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

Step 1.1.8

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

$8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

Step 1.1.9

Add the terms together.

$2 | \begin{array}{c} 5 & 6 \\ 8 & 9 \end{array} | - 3 | \begin{array}{c} 4 & 6 \\ 7 & 9 \end{array} | + 8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

Step 1.2

Evaluate $| \begin{array}{c} 5 & 6 \\ 8 & 9 \end{array} |$ .

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Step 1.2.1

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

$2 (5 \cdot 9 - 8 \cdot 6) - 3 | \begin{array}{c} 4 & 6 \\ 7 & 9 \end{array} | + 8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

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 $5$ by $9$ .

$2 (45 - 8 \cdot 6) - 3 | \begin{array}{c} 4 & 6 \\ 7 & 9 \end{array} | + 8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

Step 1.2.2.1.2

Multiply $- 8$ by $6$ .

$2 (45 - 48) - 3 | \begin{array}{c} 4 & 6 \\ 7 & 9 \end{array} | + 8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

Step 1.2.2.2

Subtract $48$ from $45$ .

$2 \cdot - 3 - 3 | \begin{array}{c} 4 & 6 \\ 7 & 9 \end{array} | + 8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

Step 1.3

Evaluate $| \begin{array}{c} 4 & 6 \\ 7 & 9 \end{array} |$ .

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Step 1.3.1

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

$2 \cdot - 3 - 3 (4 \cdot 9 - 7 \cdot 6) + 8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

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 $9$ .

$2 \cdot - 3 - 3 (36 - 7 \cdot 6) + 8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

Step 1.3.2.1.2

Multiply $- 7$ by $6$ .

$2 \cdot - 3 - 3 (36 - 42) + 8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

Step 1.3.2.2

Subtract $42$ from $36$ .

$2 \cdot - 3 - 3 \cdot - 6 + 8 | \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$

Step 1.4

Evaluate $| \begin{array}{c} 4 & 5 \\ 7 & 8 \end{array} |$ .

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Step 1.4.1

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

$2 \cdot - 3 - 3 \cdot - 6 + 8 (4 \cdot 8 - 7 \cdot 5)$

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 $4$ by $8$ .

$2 \cdot - 3 - 3 \cdot - 6 + 8 (32 - 7 \cdot 5)$

Step 1.4.2.1.2

Multiply $- 7$ by $5$ .

$2 \cdot - 3 - 3 \cdot - 6 + 8 (32 - 35)$

Step 1.4.2.2

Subtract $35$ from $32$ .

$2 \cdot - 3 - 3 \cdot - 6 + 8 \cdot - 3$

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 $- 3$ .

$- 6 - 3 \cdot - 6 + 8 \cdot - 3$

Step 1.5.1.2

Multiply $- 3$ by $- 6$ .

$- 6 + 18 + 8 \cdot - 3$

Step 1.5.1.3

Multiply $8$ by $- 3$ .

$- 6 + 18 - 24$

Step 1.5.2

Add $- 6$ and $18$ .

$12 - 24$

Step 1.5.3

Subtract $24$ from $12$ .

$- 12$

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.

$[\begin{array}{c} 2 & 3 & 8 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{3}{2} & \frac{8}{2} & \frac{1}{2} & \frac{0}{2} & \frac{0}{2} \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array}]$

Step 4.2

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{2} = R_{2} - 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 4 - 4 \cdot 1 & 5 - 4 (\frac{3}{2}) & 6 - 4 \cdot 4 & 0 - 4 (\frac{1}{2}) & 1 - 4 \cdot 0 & 0 - 4 \cdot 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 & - 1 & - 10 & - 2 & 1 & 0 \\ 7 & 8 & 9 & 0 & 0 & 1 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 & - 1 & - 10 & - 2 & 1 & 0 \\ 7 - 7 \cdot 1 & 8 - 7 (\frac{3}{2}) & 9 - 7 \cdot 4 & 0 - 7 (\frac{1}{2}) & 0 - 7 \cdot 0 & 1 - 7 \cdot 0 \end{array}]$

Step 4.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 & - 1 & - 10 & - 2 & 1 & 0 \\ 0 & - \frac{5}{2} & - 19 & - \frac{7}{2} & 0 & 1 \end{array}]$

Step 4.4

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

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Step 4.4.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ - 0 & - - 1 & - - 10 & - - 2 & - 1 \cdot 1 & - 0 \\ 0 & - \frac{5}{2} & - 19 & - \frac{7}{2} & 0 & 1 \end{array}]$

Step 4.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 10 & 2 & - 1 & 0 \\ 0 & - \frac{5}{2} & - 19 & - \frac{7}{2} & 0 & 1 \end{array}]$

Step 4.5

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

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Step 4.5.1

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

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 10 & 2 & - 1 & 0 \\ 0 + \frac{5}{2} \cdot 0 & - \frac{5}{2} + \frac{5}{2} \cdot 1 & - 19 + \frac{5}{2} \cdot 10 & - \frac{7}{2} + \frac{5}{2} \cdot 2 & 0 + \frac{5}{2} \cdot - 1 & 1 + \frac{5}{2} \cdot 0 \end{array}]$

Step 4.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 10 & 2 & - 1 & 0 \\ 0 & 0 & 6 & \frac{3}{2} & - \frac{5}{2} & 1 \end{array}]$

Step 4.6

Multiply each element of $R_{3}$ by $\frac{1}{6}$ to make the entry at $3, 3$ a $1$ .

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Step 4.6.1

Multiply each element of $R_{3}$ by $\frac{1}{6}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 10 & 2 & - 1 & 0 \\ \frac{0}{6} & \frac{0}{6} & \frac{6}{6} & \frac{\frac{3}{2}}{6} & \frac{- \frac{5}{2}}{6} & \frac{1}{6} \end{array}]$

Step 4.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 10 & 2 & - 1 & 0 \\ 0 & 0 & 1 & \frac{1}{4} & - \frac{5}{12} & \frac{1}{6} \end{array}]$

Step 4.7

Perform the row operation $R_{2} = R_{2} - 10 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 4.7.1

Perform the row operation $R_{2} = R_{2} - 10 R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 - 10 \cdot 0 & 1 - 10 \cdot 0 & 10 - 10 \cdot 1 & 2 - 10 (\frac{1}{4}) & - 1 - 10 (- \frac{5}{12}) & 0 - 10 (\frac{1}{6}) \\ 0 & 0 & 1 & \frac{1}{4} & - \frac{5}{12} & \frac{1}{6} \end{array}]$

Step 4.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 4 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 & - \frac{1}{2} & \frac{19}{6} & - \frac{5}{3} \\ 0 & 0 & 1 & \frac{1}{4} & - \frac{5}{12} & \frac{1}{6} \end{array}]$

Step 4.8

Perform the row operation $R_{1} = R_{1} - 4 R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 4.8.1

Perform the row operation $R_{1} = R_{1} - 4 R_{3}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 4 \cdot 0 & \frac{3}{2} - 4 \cdot 0 & 4 - 4 \cdot 1 & \frac{1}{2} - 4 (\frac{1}{4}) & 0 - 4 (- \frac{5}{12}) & 0 - 4 (\frac{1}{6}) \\ 0 & 1 & 0 & - \frac{1}{2} & \frac{19}{6} & - \frac{5}{3} \\ 0 & 0 & 1 & \frac{1}{4} & - \frac{5}{12} & \frac{1}{6} \end{array}]$

Step 4.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{3}{2} & 0 & - \frac{1}{2} & \frac{5}{3} & - \frac{2}{3} \\ 0 & 1 & 0 & - \frac{1}{2} & \frac{19}{6} & - \frac{5}{3} \\ 0 & 0 & 1 & \frac{1}{4} & - \frac{5}{12} & \frac{1}{6} \end{array}]$

Step 4.9

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

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Step 4.9.1

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

$[\begin{array}{c} 1 - \frac{3}{2} \cdot 0 & \frac{3}{2} - \frac{3}{2} \cdot 1 & 0 - \frac{3}{2} \cdot 0 & - \frac{1}{2} - \frac{3}{2} (- \frac{1}{2}) & \frac{5}{3} - \frac{3}{2} \cdot \frac{19}{6} & - \frac{2}{3} - \frac{3}{2} (- \frac{5}{3}) \\ 0 & 1 & 0 & - \frac{1}{2} & \frac{19}{6} & - \frac{5}{3} \\ 0 & 0 & 1 & \frac{1}{4} & - \frac{5}{12} & \frac{1}{6} \end{array}]$

Step 4.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{1}{4} & - \frac{37}{12} & \frac{11}{6} \\ 0 & 1 & 0 & - \frac{1}{2} & \frac{19}{6} & - \frac{5}{3} \\ 0 & 0 & 1 & \frac{1}{4} & - \frac{5}{12} & \frac{1}{6} \end{array}]$

Step 5

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

$[\begin{array}{c} \frac{1}{4} & - \frac{37}{12} & \frac{11}{6} \\ - \frac{1}{2} & \frac{19}{6} & - \frac{5}{3} \\ \frac{1}{4} & - \frac{5}{12} & \frac{1}{6} \end{array}]$