Linear Algebra Examples

$[\begin{array}{c} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{array}]$

Step 1

Find the determinant.

Tap for more steps...

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.

Tap for more steps...

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.

$| \begin{array}{c} 4 & 5 \\ 0 & 6 \end{array} |$

Step 1.1.4

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

$1 | \begin{array}{c} 4 & 5 \\ 0 & 6 \end{array} |$

Step 1.1.5

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

$| \begin{array}{c} 2 & 3 \\ 0 & 6 \end{array} |$

Step 1.1.6

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

$0 | \begin{array}{c} 2 & 3 \\ 0 & 6 \end{array} |$

Step 1.1.7

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

$| \begin{array}{c} 2 & 3 \\ 4 & 5 \end{array} |$

Step 1.1.8

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

$1 | \begin{array}{c} 2 & 3 \\ 4 & 5 \end{array} |$

Step 1.1.9

Add the terms together.

$1 | \begin{array}{c} 4 & 5 \\ 0 & 6 \end{array} | + 0 | \begin{array}{c} 2 & 3 \\ 0 & 6 \end{array} | + 1 | \begin{array}{c} 2 & 3 \\ 4 & 5 \end{array} |$

Step 1.2

Multiply $0$ by $| \begin{array}{c} 2 & 3 \\ 0 & 6 \end{array} |$ .

$1 | \begin{array}{c} 4 & 5 \\ 0 & 6 \end{array} | + 0 + 1 | \begin{array}{c} 2 & 3 \\ 4 & 5 \end{array} |$

Step 1.3

Evaluate $| \begin{array}{c} 4 & 5 \\ 0 & 6 \end{array} |$ .

Tap for more steps...

Step 1.3.1

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

$1 (4 \cdot 6 + 0 \cdot 5) + 0 + 1 | \begin{array}{c} 2 & 3 \\ 4 & 5 \end{array} |$

Step 1.3.2

Simplify the determinant.

Tap for more steps...

Step 1.3.2.1

Simplify each term.

Tap for more steps...

Step 1.3.2.1.1

Multiply $4$ by $6$ .

$1 (24 + 0 \cdot 5) + 0 + 1 | \begin{array}{c} 2 & 3 \\ 4 & 5 \end{array} |$

Step 1.3.2.1.2

Multiply $0$ by $5$ .

$1 (24 + 0) + 0 + 1 | \begin{array}{c} 2 & 3 \\ 4 & 5 \end{array} |$

Step 1.3.2.2

Add $24$ and $0$ .

$1 \cdot 24 + 0 + 1 | \begin{array}{c} 2 & 3 \\ 4 & 5 \end{array} |$

Step 1.4

Evaluate $| \begin{array}{c} 2 & 3 \\ 4 & 5 \end{array} |$ .

Tap for more steps...

Step 1.4.1

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

$1 \cdot 24 + 0 + 1 (2 \cdot 5 - 4 \cdot 3)$

Step 1.4.2

Simplify the determinant.

Tap for more steps...

Step 1.4.2.1

Simplify each term.

Tap for more steps...

Step 1.4.2.1.1

Multiply $2$ by $5$ .

$1 \cdot 24 + 0 + 1 (10 - 4 \cdot 3)$

Step 1.4.2.1.2

Multiply $- 4$ by $3$ .

$1 \cdot 24 + 0 + 1 (10 - 12)$

Step 1.4.2.2

Subtract $12$ from $10$ .

$1 \cdot 24 + 0 + 1 \cdot - 2$

Step 1.5

Simplify the determinant.

Tap for more steps...

Step 1.5.1

Simplify each term.

Tap for more steps...

Step 1.5.1.1

Multiply $24$ by $1$ .

$24 + 0 + 1 \cdot - 2$

Step 1.5.1.2

Multiply $- 2$ by $1$ .

$24 + 0 - 2$

Step 1.5.2

Add $24$ and $0$ .

$24 - 2$

Step 1.5.3

Subtract $2$ from $24$ .

$22$

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.

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 4 & 5 & 0 & 1 & 0 \\ 1 & 0 & 6 & 0 & 0 & 1 \end{array}]$

Step 4

Find the reduced row echelon form.

Tap for more steps...

Step 4.1

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

Tap for more steps...

Step 4.1.1

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

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 4 & 5 & 0 & 1 & 0 \\ 1 - 1 & 0 - 2 & 6 - 3 & 0 - 1 & 0 - 0 & 1 - 0 \end{array}]$

Step 4.1.2

Simplify $R_{3}$ .

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

Step 4.2

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

Tap for more steps...

Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

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

Tap for more steps...

Step 4.3.1

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

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

Step 4.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & \frac{5}{4} & 0 & \frac{1}{4} & 0 \\ 0 & 0 & \frac{11}{2} & - 1 & \frac{1}{2} & 1 \end{array}]$

Step 4.4

Multiply each element of $R_{3}$ by $\frac{2}{11}$ to make the entry at $3, 3$ a $1$ .

Tap for more steps...

Step 4.4.1

Multiply each element of $R_{3}$ by $\frac{2}{11}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & \frac{5}{4} & 0 & \frac{1}{4} & 0 \\ \frac{2}{11} \cdot 0 & \frac{2}{11} \cdot 0 & \frac{2}{11} \cdot \frac{11}{2} & \frac{2}{11} \cdot - 1 & \frac{2}{11} \cdot \frac{1}{2} & \frac{2}{11} \cdot 1 \end{array}]$

Step 4.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & \frac{5}{4} & 0 & \frac{1}{4} & 0 \\ 0 & 0 & 1 & - \frac{2}{11} & \frac{1}{11} & \frac{2}{11} \end{array}]$

Step 4.5

Perform the row operation $R_{2} = R_{2} - \frac{5}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

Tap for more steps...

Step 4.5.1

Perform the row operation $R_{2} = R_{2} - \frac{5}{4} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 - \frac{5}{4} \cdot 0 & 1 - \frac{5}{4} \cdot 0 & \frac{5}{4} - \frac{5}{4} \cdot 1 & 0 - \frac{5}{4} (- \frac{2}{11}) & \frac{1}{4} - \frac{5}{4} \cdot \frac{1}{11} & 0 - \frac{5}{4} \cdot \frac{2}{11} \\ 0 & 0 & 1 & - \frac{2}{11} & \frac{1}{11} & \frac{2}{11} \end{array}]$

Step 4.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{5}{22} & \frac{3}{22} & - \frac{5}{22} \\ 0 & 0 & 1 & - \frac{2}{11} & \frac{1}{11} & \frac{2}{11} \end{array}]$

Step 4.6

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

Tap for more steps...

Step 4.6.1

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

$[\begin{array}{c} 1 - 3 \cdot 0 & 2 - 3 \cdot 0 & 3 - 3 \cdot 1 & 1 - 3 (- \frac{2}{11}) & 0 - 3 (\frac{1}{11}) & 0 - 3 (\frac{2}{11}) \\ 0 & 1 & 0 & \frac{5}{22} & \frac{3}{22} & - \frac{5}{22} \\ 0 & 0 & 1 & - \frac{2}{11} & \frac{1}{11} & \frac{2}{11} \end{array}]$

Step 4.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 2 & 0 & \frac{17}{11} & - \frac{3}{11} & - \frac{6}{11} \\ 0 & 1 & 0 & \frac{5}{22} & \frac{3}{22} & - \frac{5}{22} \\ 0 & 0 & 1 & - \frac{2}{11} & \frac{1}{11} & \frac{2}{11} \end{array}]$

Step 4.7

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

Tap for more steps...

Step 4.7.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & \frac{17}{11} - 2 (\frac{5}{22}) & - \frac{3}{11} - 2 (\frac{3}{22}) & - \frac{6}{11} - 2 (- \frac{5}{22}) \\ 0 & 1 & 0 & \frac{5}{22} & \frac{3}{22} & - \frac{5}{22} \\ 0 & 0 & 1 & - \frac{2}{11} & \frac{1}{11} & \frac{2}{11} \end{array}]$

Step 4.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{12}{11} & - \frac{6}{11} & - \frac{1}{11} \\ 0 & 1 & 0 & \frac{5}{22} & \frac{3}{22} & - \frac{5}{22} \\ 0 & 0 & 1 & - \frac{2}{11} & \frac{1}{11} & \frac{2}{11} \end{array}]$

Step 5

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

$[\begin{array}{c} \frac{12}{11} & - \frac{6}{11} & - \frac{1}{11} \\ \frac{5}{22} & \frac{3}{22} & - \frac{5}{22} \\ - \frac{2}{11} & \frac{1}{11} & \frac{2}{11} \end{array}]$