Finite Math Examples

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

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

$| \begin{array}{c} 0 & 5 \\ 2 & 7 \end{array} |$

Step 1.1.4

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

$0 | \begin{array}{c} 0 & 5 \\ 2 & 7 \end{array} |$

Step 1.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

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

Step 1.1.6

Multiply element $a_{22}$ by its cofactor.

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

Step 1.1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

$| \begin{array}{c} 2 & 0 \\ 3 & 2 \end{array} |$

Step 1.1.8

Multiply element $a_{23}$ by its cofactor.

$0 | \begin{array}{c} 2 & 0 \\ 3 & 2 \end{array} |$

Step 1.1.9

Add the terms together.

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

Step 1.2

Multiply $0$ by $| \begin{array}{c} 0 & 5 \\ 2 & 7 \end{array} |$ .

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

Step 1.3

Multiply $0$ by $| \begin{array}{c} 2 & 0 \\ 3 & 2 \end{array} |$ .

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

Step 1.4

Evaluate $| \begin{array}{c} 2 & 5 \\ 3 & 7 \end{array} |$ .

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Step 1.4.1

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

$0 - 1 (2 \cdot 7 - 3 \cdot 5) + 0$

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 $7$ .

$0 - 1 (14 - 3 \cdot 5) + 0$

Step 1.4.2.1.2

Multiply $- 3$ by $5$ .

$0 - 1 (14 - 15) + 0$

Step 1.4.2.2

Subtract $15$ from $14$ .

$0 - 1 \cdot - 1 + 0$

Step 1.5

Simplify the determinant.

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Step 1.5.1

Multiply $- 1$ by $- 1$ .

$0 + 1 + 0$

Step 1.5.2

Add $0$ and $1$ .

$1 + 0$

Step 1.5.3

Add $1$ and $0$ .

$1$

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.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $\frac{1}{2}$ 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 $1$ .

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

Step 4.1.2

Simplify $R_{1}$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{3}$ .

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

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Simplify $R_{2}$ .

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

Step 4.4

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

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Step 4.4.1

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

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

Step 4.4.2

Simplify $R_{3}$ .

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

Step 4.5

Multiply each element of $R_{3}$ by $- 2$ to make the entry at $3, 3$ a $1$ .

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Step 4.5.1

Multiply each element of $R_{3}$ by $- 2$ to make the entry at $3, 3$ a $1$ .

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

Step 4.5.2

Simplify $R_{3}$ .

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

Step 4.6

Perform the row operation $R_{1} = R_{1} - \frac{5}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

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Step 4.6.1

Perform the row operation $R_{1} = R_{1} - \frac{5}{2} R_{3}$ to make the entry at $1, 3$ a $0$ .

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

Step 4.6.2

Simplify $R_{1}$ .

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

Step 5

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

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