Algebra Examples

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

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

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

Step 1.1.4

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

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

Step 1.1.5

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

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

Step 1.1.6

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

$1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} |$

Step 1.1.7

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

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

Step 1.1.8

Multiply element $a_{32}$ by its cofactor.

$0 | \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |$

Step 1.1.9

Add the terms together.

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

Step 1.2

Multiply $0$ by $| \begin{array}{c} 4 & 4 \\ 1 & 2 \end{array} |$ .

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

Step 1.3

Evaluate $| \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} |$ .

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Step 1.3.1

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

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

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 $1$ .

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

Step 1.3.2.1.2

Multiply $- 3$ by $2$ .

$- 3 (2 - 6) + 1 | \begin{array}{c} 4 & 4 \\ 3 & 2 \end{array} | + 0$

Step 1.3.2.2

Subtract $6$ from $2$ .

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

Step 1.4

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

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Step 1.4.1

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

$- 3 \cdot - 4 + 1 (4 \cdot 2 - 3 \cdot 4) + 0$

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 $4$ by $2$ .

$- 3 \cdot - 4 + 1 (8 - 3 \cdot 4) + 0$

Step 1.4.2.1.2

Multiply $- 3$ by $4$ .

$- 3 \cdot - 4 + 1 (8 - 12) + 0$

Step 1.4.2.2

Subtract $12$ from $8$ .

$- 3 \cdot - 4 + 1 \cdot - 4 + 0$

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 $- 3$ by $- 4$ .

$12 + 1 \cdot - 4 + 0$

Step 1.5.1.2

Multiply $- 4$ by $1$ .

$12 - 4 + 0$

Step 1.5.2

Subtract $4$ from $12$ .

$8 + 0$

Step 1.5.3

Add $8$ and $0$ .

$8$

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.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify $R_{1}$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

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 4.3.1

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

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

Step 4.3.2

Simplify $R_{3}$ .

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

Step 4.4

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

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Step 4.4.1

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

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

Step 4.4.2

Simplify $R_{2}$ .

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

Step 4.5

Perform the row operation $R_{3} = R_{3} + \frac{9}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 4.5.1

Perform the row operation $R_{3} = R_{3} + \frac{9}{4} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{3}{4} & 1 & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 4 & - 1 & 4 & 0 \\ 0 + \frac{9}{4} \cdot 0 & - \frac{9}{4} + \frac{9}{4} \cdot 1 & - 1 + \frac{9}{4} \cdot 4 & - \frac{3}{4} + \frac{9}{4} \cdot - 1 & 0 + \frac{9}{4} \cdot 4 & 1 + \frac{9}{4} \cdot 0 \end{array}]$

Step 4.5.2

Simplify $R_{3}$ .

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

Step 4.6

Multiply each element of $R_{3}$ by $\frac{1}{8}$ to make the entry at $3, 3$ a $1$ .

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Step 4.6.1

Multiply each element of $R_{3}$ by $\frac{1}{8}$ to make the entry at $3, 3$ a $1$ .

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

Step 4.6.2

Simplify $R_{3}$ .

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

Step 4.7

Perform the row operation $R_{2} = R_{2} - 4 R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 4.7.1

Perform the row operation $R_{2} = R_{2} - 4 R_{3}$ to make the entry at $2, 3$ a $0$ .

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

Step 4.7.2

Simplify $R_{2}$ .

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

Step 4.8

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

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Step 4.8.1

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

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

Step 4.8.2

Simplify $R_{1}$ .

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

Step 4.9

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 4.9.1

Perform the row operation $R_{1} = R_{1} - \frac{3}{4} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

Step 4.9.2

Simplify $R_{1}$ .

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

Step 5

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

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

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