Linear Algebra Examples

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

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.

$| \begin{array}{c} 0 & 3 \\ - 3 & 8 \end{array} |$

Step 1.1.4

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

$0 | \begin{array}{c} 0 & 3 \\ - 3 & 8 \end{array} |$

Step 1.1.5

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

$| \begin{array}{c} 1 & 3 \\ 4 & 8 \end{array} |$

Step 1.1.6

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

$- 1 | \begin{array}{c} 1 & 3 \\ 4 & 8 \end{array} |$

Step 1.1.7

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

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

Step 1.1.8

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

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

Step 1.1.9

Add the terms together.

$0 | \begin{array}{c} 0 & 3 \\ - 3 & 8 \end{array} | - 1 | \begin{array}{c} 1 & 3 \\ 4 & 8 \end{array} | + 2 | \begin{array}{c} 1 & 0 \\ 4 & - 3 \end{array} |$

Step 1.2

Multiply $0$ by $| \begin{array}{c} 0 & 3 \\ - 3 & 8 \end{array} |$ .

$0 - 1 | \begin{array}{c} 1 & 3 \\ 4 & 8 \end{array} | + 2 | \begin{array}{c} 1 & 0 \\ 4 & - 3 \end{array} |$

Step 1.3

Evaluate $| \begin{array}{c} 1 & 3 \\ 4 & 8 \end{array} |$ .

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Step 1.3.1

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

$0 - 1 (1 \cdot 8 - 4 \cdot 3) + 2 | \begin{array}{c} 1 & 0 \\ 4 & - 3 \end{array} |$

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 $8$ by $1$ .

$0 - 1 (8 - 4 \cdot 3) + 2 | \begin{array}{c} 1 & 0 \\ 4 & - 3 \end{array} |$

Step 1.3.2.1.2

Multiply $- 4$ by $3$ .

$0 - 1 (8 - 12) + 2 | \begin{array}{c} 1 & 0 \\ 4 & - 3 \end{array} |$

Step 1.3.2.2

Subtract $12$ from $8$ .

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

Step 1.4

Evaluate $| \begin{array}{c} 1 & 0 \\ 4 & - 3 \end{array} |$ .

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Step 1.4.1

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

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

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 $- 3$ by $1$ .

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

Step 1.4.2.1.2

Multiply $- 4$ by $0$ .

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

Step 1.4.2.2

Add $- 3$ and $0$ .

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

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 $- 1$ by $- 4$ .

$0 + 4 + 2 \cdot - 3$

Step 1.5.1.2

Multiply $2$ by $- 3$ .

$0 + 4 - 6$

Step 1.5.2

Add $0$ and $4$ .

$4 - 6$

Step 1.5.3

Subtract $6$ from $4$ .

$- 2$

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.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{3}$ .

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

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 $0$ .

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

Step 4.3.2

Simplify $R_{3}$ .

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

Step 4.4

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

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Step 4.4.1

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

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

Step 4.4.2

Simplify $R_{3}$ .

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

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 $0$ .

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

Step 4.5.2

Simplify $R_{2}$ .

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

Step 4.6

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

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Step 4.6.1

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

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

Step 4.6.2

Simplify $R_{1}$ .

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

Step 5

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

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