Finite Math Examples

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

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.

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

Step 1.1.4

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

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

Step 1.1.5

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

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

Step 1.1.6

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

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

Step 1.1.7

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

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

Step 1.1.8

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

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

Step 1.1.9

Add the terms together.

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

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 $- 1$ .

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

Step 1.3.2.1.2

Multiply $- (- 1 \cdot 1)$ .

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Step 1.3.2.1.2.1

Multiply $- 1$ by $1$ .

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

Step 1.3.2.1.2.2

Multiply $- 1$ by $- 1$ .

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

Step 1.3.2.2

Add $2$ and $1$ .

$- 1 \cdot 3 - 2 | \begin{array}{c} - 2 & 1 \\ 0 & - 1 \end{array} | + 0$

Step 1.4

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

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Step 1.4.1

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

$- 1 \cdot 3 - 2 (- 2 \cdot - 1 + 0 \cdot 1) + 0$

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 $- 2$ by $- 1$ .

$- 1 \cdot 3 - 2 (2 + 0 \cdot 1) + 0$

Step 1.4.2.1.2

Multiply $0$ by $1$ .

$- 1 \cdot 3 - 2 (2 + 0) + 0$

Step 1.4.2.2

Add $2$ and $0$ .

$- 1 \cdot 3 - 2 \cdot 2 + 0$

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 $- 1$ by $3$ .

$- 3 - 2 \cdot 2 + 0$

Step 1.5.1.2

Multiply $- 2$ by $2$ .

$- 3 - 4 + 0$

Step 1.5.2

Subtract $4$ from $- 3$ .

$- 7 + 0$

Step 1.5.3

Add $- 7$ and $0$ .

$- 7$

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.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify $R_{1}$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

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

Step 4.3.2

Simplify $R_{2}$ .

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

Step 4.4

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

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Step 4.4.1

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

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

Step 4.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & 0 & - 1 & 0 & 0 \\ 0 & 1 & - \frac{1}{6} & \frac{1}{3} & - \frac{1}{6} & 0 \\ 0 & 0 & - \frac{7}{6} & \frac{1}{3} & - \frac{1}{6} & 1 \end{array}]$

Step 4.5

Multiply each element of $R_{3}$ by $- \frac{6}{7}$ to make the entry at $3, 3$ a $1$ .

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Step 4.5.1

Multiply each element of $R_{3}$ by $- \frac{6}{7}$ to make the entry at $3, 3$ a $1$ .

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

Step 4.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 2 & 0 & - 1 & 0 & 0 \\ 0 & 1 & - \frac{1}{6} & \frac{1}{3} & - \frac{1}{6} & 0 \\ 0 & 0 & 1 & - \frac{2}{7} & \frac{1}{7} & - \frac{6}{7} \end{array}]$

Step 4.6

Perform the row operation $R_{2} = R_{2} + \frac{1}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 4.6.1

Perform the row operation $R_{2} = R_{2} + \frac{1}{6} R_{3}$ to make the entry at $2, 3$ a $0$ .

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

Step 4.6.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 2 & 0 & - 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{2}{7} & - \frac{1}{7} & - \frac{1}{7} \\ 0 & 0 & 1 & - \frac{2}{7} & \frac{1}{7} & - \frac{6}{7} \end{array}]$

Step 4.7

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

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Step 4.7.1

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

$[\begin{array}{c} 1 + 2 \cdot 0 & - 2 + 2 \cdot 1 & 0 + 2 \cdot 0 & - 1 + 2 (\frac{2}{7}) & 0 + 2 (- \frac{1}{7}) & 0 + 2 (- \frac{1}{7}) \\ 0 & 1 & 0 & \frac{2}{7} & - \frac{1}{7} & - \frac{1}{7} \\ 0 & 0 & 1 & - \frac{2}{7} & \frac{1}{7} & - \frac{6}{7} \end{array}]$

Step 4.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{3}{7} & - \frac{2}{7} & - \frac{2}{7} \\ 0 & 1 & 0 & \frac{2}{7} & - \frac{1}{7} & - \frac{1}{7} \\ 0 & 0 & 1 & - \frac{2}{7} & \frac{1}{7} & - \frac{6}{7} \end{array}]$

Step 5

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

$[\begin{array}{c} - \frac{3}{7} & - \frac{2}{7} & - \frac{2}{7} \\ \frac{2}{7} & - \frac{1}{7} & - \frac{1}{7} \\ - \frac{2}{7} & \frac{1}{7} & - \frac{6}{7} \end{array}]$