Linear Algebra Examples

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

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.

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

Step 1.1.4

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

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

Step 1.1.5

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

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

Step 1.1.6

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

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

Step 1.1.7

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

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

Step 1.1.8

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

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

Step 1.1.9

Add the terms together.

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

Simplify the determinant.

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Step 1.2.2.1

Simplify each term.

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Step 1.2.2.1.1

Multiply $4$ by $1$ .

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

Step 1.2.2.1.2

Multiply $- 4$ by $4$ .

$4 (4 - 16) - 1 | \begin{array}{c} 1 & 4 \\ 4 & 1 \end{array} | + 3 | \begin{array}{c} 1 & 4 \\ 4 & 4 \end{array} |$

Step 1.2.2.2

Subtract $16$ from $4$ .

$4 \cdot - 12 - 1 | \begin{array}{c} 1 & 4 \\ 4 & 1 \end{array} | + 3 | \begin{array}{c} 1 & 4 \\ 4 & 4 \end{array} |$

Step 1.3

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

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Step 1.3.1

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

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

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 $1$ by $1$ .

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

Step 1.3.2.1.2

Multiply $- 4$ by $4$ .

$4 \cdot - 12 - 1 (1 - 16) + 3 | \begin{array}{c} 1 & 4 \\ 4 & 4 \end{array} |$

Step 1.3.2.2

Subtract $16$ from $1$ .

$4 \cdot - 12 - 1 \cdot - 15 + 3 | \begin{array}{c} 1 & 4 \\ 4 & 4 \end{array} |$

Step 1.4

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

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Step 1.4.1

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

$4 \cdot - 12 - 1 \cdot - 15 + 3 (1 \cdot 4 - 4 \cdot 4)$

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 $1$ .

$4 \cdot - 12 - 1 \cdot - 15 + 3 (4 - 4 \cdot 4)$

Step 1.4.2.1.2

Multiply $- 4$ by $4$ .

$4 \cdot - 12 - 1 \cdot - 15 + 3 (4 - 16)$

Step 1.4.2.2

Subtract $16$ from $4$ .

$4 \cdot - 12 - 1 \cdot - 15 + 3 \cdot - 12$

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 $- 12$ .

$- 48 - 1 \cdot - 15 + 3 \cdot - 12$

Step 1.5.1.2

Multiply $- 1$ by $- 15$ .

$- 48 + 15 + 3 \cdot - 12$

Step 1.5.1.3

Multiply $3$ by $- 12$ .

$- 48 + 15 - 36$

Step 1.5.2

Add $- 48$ and $15$ .

$- 33 - 36$

Step 1.5.3

Subtract $36$ from $- 33$ .

$- 69$

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.

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

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 $1$ .

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

Step 4.1.2

Simplify $R_{1}$ .

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

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 $0$ .

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

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 & \frac{15}{4} & \frac{13}{4} & - \frac{1}{4} & 1 & 0 \\ 4 & 4 & 1 & 0 & 0 & 1 \end{array}]$

Step 4.3

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 4.3.1

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

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 & \frac{15}{4} & \frac{13}{4} & - \frac{1}{4} & 1 & 0 \\ 4 - 4 \cdot 1 & 4 - 4 (\frac{1}{4}) & 1 - 4 (\frac{3}{4}) & 0 - 4 (\frac{1}{4}) & 0 - 4 \cdot 0 & 1 - 4 \cdot 0 \end{array}]$

Step 4.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 & \frac{15}{4} & \frac{13}{4} & - \frac{1}{4} & 1 & 0 \\ 0 & 3 & - 2 & - 1 & 0 & 1 \end{array}]$

Step 4.4

Multiply each element of $R_{2}$ by $\frac{4}{15}$ to make the entry at $2, 2$ a $1$ .

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Step 4.4.1

Multiply each element of $R_{2}$ by $\frac{4}{15}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ \frac{4}{15} \cdot 0 & \frac{4}{15} \cdot \frac{15}{4} & \frac{4}{15} \cdot \frac{13}{4} & \frac{4}{15} (- \frac{1}{4}) & \frac{4}{15} \cdot 1 & \frac{4}{15} \cdot 0 \\ 0 & 3 & - 2 & - 1 & 0 & 1 \end{array}]$

Step 4.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & \frac{13}{15} & - \frac{1}{15} & \frac{4}{15} & 0 \\ 0 & 3 & - 2 & - 1 & 0 & 1 \end{array}]$

Step 4.5

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 4.5.1

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

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & \frac{13}{15} & - \frac{1}{15} & \frac{4}{15} & 0 \\ 0 - 3 \cdot 0 & 3 - 3 \cdot 1 & - 2 - 3 (\frac{13}{15}) & - 1 - 3 (- \frac{1}{15}) & 0 - 3 (\frac{4}{15}) & 1 - 3 \cdot 0 \end{array}]$

Step 4.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & \frac{13}{15} & - \frac{1}{15} & \frac{4}{15} & 0 \\ 0 & 0 & - \frac{23}{5} & - \frac{4}{5} & - \frac{4}{5} & 1 \end{array}]$

Step 4.6

Multiply each element of $R_{3}$ by $- \frac{5}{23}$ to make the entry at $3, 3$ a $1$ .

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Step 4.6.1

Multiply each element of $R_{3}$ by $- \frac{5}{23}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & \frac{13}{15} & - \frac{1}{15} & \frac{4}{15} & 0 \\ - \frac{5}{23} \cdot 0 & - \frac{5}{23} \cdot 0 & - \frac{5}{23} (- \frac{23}{5}) & - \frac{5}{23} (- \frac{4}{5}) & - \frac{5}{23} (- \frac{4}{5}) & - \frac{5}{23} \cdot 1 \end{array}]$

Step 4.6.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & \frac{13}{15} & - \frac{1}{15} & \frac{4}{15} & 0 \\ 0 & 0 & 1 & \frac{4}{23} & \frac{4}{23} & - \frac{5}{23} \end{array}]$

Step 4.7

Perform the row operation $R_{2} = R_{2} - \frac{13}{15} R_{3}$ to make the entry at $2, 3$ a $0$ .

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Step 4.7.1

Perform the row operation $R_{2} = R_{2} - \frac{13}{15} R_{3}$ to make the entry at $2, 3$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 - \frac{13}{15} \cdot 0 & 1 - \frac{13}{15} \cdot 0 & \frac{13}{15} - \frac{13}{15} \cdot 1 & - \frac{1}{15} - \frac{13}{15} \cdot \frac{4}{23} & \frac{4}{15} - \frac{13}{15} \cdot \frac{4}{23} & 0 - \frac{13}{15} (- \frac{5}{23}) \\ 0 & 0 & 1 & \frac{4}{23} & \frac{4}{23} & - \frac{5}{23} \end{array}]$

Step 4.7.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & \frac{3}{4} & \frac{1}{4} & 0 & 0 \\ 0 & 1 & 0 & - \frac{5}{23} & \frac{8}{69} & \frac{13}{69} \\ 0 & 0 & 1 & \frac{4}{23} & \frac{4}{23} & - \frac{5}{23} \end{array}]$

Step 4.8

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

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Step 4.8.1

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

$[\begin{array}{c} 1 - \frac{3}{4} \cdot 0 & \frac{1}{4} - \frac{3}{4} \cdot 0 & \frac{3}{4} - \frac{3}{4} \cdot 1 & \frac{1}{4} - \frac{3}{4} \cdot \frac{4}{23} & 0 - \frac{3}{4} \cdot \frac{4}{23} & 0 - \frac{3}{4} (- \frac{5}{23}) \\ 0 & 1 & 0 & - \frac{5}{23} & \frac{8}{69} & \frac{13}{69} \\ 0 & 0 & 1 & \frac{4}{23} & \frac{4}{23} & - \frac{5}{23} \end{array}]$

Step 4.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{4} & 0 & \frac{11}{92} & - \frac{3}{23} & \frac{15}{92} \\ 0 & 1 & 0 & - \frac{5}{23} & \frac{8}{69} & \frac{13}{69} \\ 0 & 0 & 1 & \frac{4}{23} & \frac{4}{23} & - \frac{5}{23} \end{array}]$

Step 4.9

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

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Step 4.9.1

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

$[\begin{array}{c} 1 - \frac{1}{4} \cdot 0 & \frac{1}{4} - \frac{1}{4} \cdot 1 & 0 - \frac{1}{4} \cdot 0 & \frac{11}{92} - \frac{1}{4} (- \frac{5}{23}) & - \frac{3}{23} - \frac{1}{4} \cdot \frac{8}{69} & \frac{15}{92} - \frac{1}{4} \cdot \frac{13}{69} \\ 0 & 1 & 0 & - \frac{5}{23} & \frac{8}{69} & \frac{13}{69} \\ 0 & 0 & 1 & \frac{4}{23} & \frac{4}{23} & - \frac{5}{23} \end{array}]$

Step 4.9.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & \frac{4}{23} & - \frac{11}{69} & \frac{8}{69} \\ 0 & 1 & 0 & - \frac{5}{23} & \frac{8}{69} & \frac{13}{69} \\ 0 & 0 & 1 & \frac{4}{23} & \frac{4}{23} & - \frac{5}{23} \end{array}]$

Step 5

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

$[\begin{array}{c} \frac{4}{23} & - \frac{11}{69} & \frac{8}{69} \\ - \frac{5}{23} & \frac{8}{69} & \frac{13}{69} \\ \frac{4}{23} & \frac{4}{23} & - \frac{5}{23} \end{array}]$

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