Algebra Examples

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

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 $1$ 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.

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

Step 1.1.4

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

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

Step 1.1.5

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

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

Step 1.1.6

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

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

Step 1.1.7

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

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

Step 1.1.8

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

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

Step 1.1.9

Add the terms together.

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

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 $2$ .

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

Step 1.3.2.1.2

Multiply $- 3$ by $0$ .

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

Step 1.3.2.2

Add $4$ and $0$ .

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

Step 1.4

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

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Step 1.4.1

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

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

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 $2$ .

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

Step 1.4.2.1.2

Multiply $- 3$ by $2$ .

$1 \cdot 4 - 2 (4 - 6) + 0$

Step 1.4.2.2

Subtract $6$ from $4$ .

$1 \cdot 4 - 2 \cdot - 2 + 0$

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 $4$ by $1$ .

$4 - 2 \cdot - 2 + 0$

Step 1.5.1.2

Multiply $- 2$ by $- 2$ .

$4 + 4 + 0$

Step 1.5.2

Add $4$ and $4$ .

$8 + 0$

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.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

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 4.1.1

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

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

Step 4.1.2

Simplify $R_{2}$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Simplify $R_{3}$ .

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

Step 4.4

Multiply each element of $R_{3}$ by $- \frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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Step 4.4.1

Multiply each element of $R_{3}$ by $- \frac{1}{4}$ to make the entry at $3, 3$ a $1$ .

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

Step 4.4.2

Simplify $R_{3}$ .

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

Step 4.5

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

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Step 4.5.1

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

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

Step 4.5.2

Simplify $R_{2}$ .

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

Step 4.6

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

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Step 4.6.1

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

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

Step 4.6.2

Simplify $R_{1}$ .

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

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 $0$ .

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

Step 4.7.2

Simplify $R_{1}$ .

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

Step 5

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

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

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