Finite Math Examples

$[\begin{array}{c} 2 & - 1 & - 2 \\ - 4 & 2 & 8 \\ 6 & - 2 & - 1 \end{array}]$

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.

$| \begin{array}{c} 2 & 8 \\ - 2 & - 1 \end{array} |$

Step 1.1.4

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

$2 | \begin{array}{c} 2 & 8 \\ - 2 & - 1 \end{array} |$

Step 1.1.5

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

$| \begin{array}{c} - 4 & 8 \\ 6 & - 1 \end{array} |$

Step 1.1.6

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

$1 | \begin{array}{c} - 4 & 8 \\ 6 & - 1 \end{array} |$

Step 1.1.7

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

$| \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.1.8

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

$- 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.1.9

Add the terms together.

$2 | \begin{array}{c} 2 & 8 \\ - 2 & - 1 \end{array} | + 1 | \begin{array}{c} - 4 & 8 \\ 6 & - 1 \end{array} | - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.2

Evaluate $| \begin{array}{c} 2 & 8 \\ - 2 & - 1 \end{array} |$ .

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Step 1.2.1

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

$2 (2 \cdot - 1 - (- 2 \cdot 8)) + 1 | \begin{array}{c} - 4 & 8 \\ 6 & - 1 \end{array} | - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

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 $2$ by $- 1$ .

$2 (- 2 - (- 2 \cdot 8)) + 1 | \begin{array}{c} - 4 & 8 \\ 6 & - 1 \end{array} | - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.2.2.1.2

Multiply $- (- 2 \cdot 8)$ .

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Step 1.2.2.1.2.1

Multiply $- 2$ by $8$ .

$2 (- 2 - - 16) + 1 | \begin{array}{c} - 4 & 8 \\ 6 & - 1 \end{array} | - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.2.2.1.2.2

Multiply $- 1$ by $- 16$ .

$2 (- 2 + 16) + 1 | \begin{array}{c} - 4 & 8 \\ 6 & - 1 \end{array} | - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.2.2.2

Add $- 2$ and $16$ .

$2 \cdot 14 + 1 | \begin{array}{c} - 4 & 8 \\ 6 & - 1 \end{array} | - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.3

Evaluate $| \begin{array}{c} - 4 & 8 \\ 6 & - 1 \end{array} |$ .

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Step 1.3.1

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

$2 \cdot 14 + 1 (- 4 \cdot - 1 - 6 \cdot 8) - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

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 $- 4$ by $- 1$ .

$2 \cdot 14 + 1 (4 - 6 \cdot 8) - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.3.2.1.2

Multiply $- 6$ by $8$ .

$2 \cdot 14 + 1 (4 - 48) - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.3.2.2

Subtract $48$ from $4$ .

$2 \cdot 14 + 1 \cdot - 44 - 2 | \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$

Step 1.4

Evaluate $| \begin{array}{c} - 4 & 2 \\ 6 & - 2 \end{array} |$ .

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Step 1.4.1

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

$2 \cdot 14 + 1 \cdot - 44 - 2 (- 4 \cdot - 2 - 6 \cdot 2)$

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 $- 2$ .

$2 \cdot 14 + 1 \cdot - 44 - 2 (8 - 6 \cdot 2)$

Step 1.4.2.1.2

Multiply $- 6$ by $2$ .

$2 \cdot 14 + 1 \cdot - 44 - 2 (8 - 12)$

Step 1.4.2.2

Subtract $12$ from $8$ .

$2 \cdot 14 + 1 \cdot - 44 - 2 \cdot - 4$

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 $2$ by $14$ .

$28 + 1 \cdot - 44 - 2 \cdot - 4$

Step 1.5.1.2

Multiply $- 44$ by $1$ .

$28 - 44 - 2 \cdot - 4$

Step 1.5.1.3

Multiply $- 2$ by $- 4$ .

$28 - 44 + 8$

Step 1.5.2

Subtract $44$ from $28$ .

$- 16 + 8$

Step 1.5.3

Add $- 16$ and $8$ .

$- 8$

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.

$[\begin{array}{c} 2 & - 1 & - 2 & 1 & 0 & 0 \\ - 4 & 2 & 8 & 0 & 1 & 0 \\ 6 & - 2 & - 1 & 0 & 0 & 1 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

$[\begin{array}{c} \frac{2}{2} & \frac{- 1}{2} & \frac{- 2}{2} & \frac{1}{2} & \frac{0}{2} & \frac{0}{2} \\ - 4 & 2 & 8 & 0 & 1 & 0 \\ 6 & - 2 & - 1 & 0 & 0 & 1 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - 1 & \frac{1}{2} & 0 & 0 \\ - 4 & 2 & 8 & 0 & 1 & 0 \\ 6 & - 2 & - 1 & 0 & 0 & 1 \end{array}]$

Step 4.2

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

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Step 4.2.1

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

$[\begin{array}{c} 1 & - \frac{1}{2} & - 1 & \frac{1}{2} & 0 & 0 \\ - 4 + 4 \cdot 1 & 2 + 4 (- \frac{1}{2}) & 8 + 4 \cdot - 1 & 0 + 4 (\frac{1}{2}) & 1 + 4 \cdot 0 & 0 + 4 \cdot 0 \\ 6 & - 2 & - 1 & 0 & 0 & 1 \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

Perform the row operation $R_{3} = R_{3} - 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 4.3.1

Perform the row operation $R_{3} = R_{3} - 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{1}{2} & - 1 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 4 & 2 & 1 & 0 \\ 6 - 6 \cdot 1 & - 2 - 6 (- \frac{1}{2}) & - 1 - 6 \cdot - 1 & 0 - 6 (\frac{1}{2}) & 0 - 6 \cdot 0 & 1 - 6 \cdot 0 \end{array}]$

Step 4.3.2

Simplify $R_{3}$ .

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

Step 4.4

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

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

Step 4.5

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

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Step 4.5.1

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

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

Step 4.5.2

Simplify $R_{3}$ .

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

Step 4.6

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

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Step 4.6.1

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

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

Step 4.6.2

Simplify $R_{2}$ .

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

Step 4.7

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

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Step 4.7.1

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

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

Step 4.7.2

Simplify $R_{1}$ .

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

Step 4.8

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

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Step 4.8.1

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

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

Step 4.8.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & - \frac{7}{4} & - \frac{3}{8} & \frac{1}{2} \\ 0 & 1 & 0 & - \frac{11}{2} & - \frac{5}{4} & 1 \\ 0 & 0 & 1 & \frac{1}{2} & \frac{1}{4} & 0 \end{array}]$

Step 5

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

$[\begin{array}{c} - \frac{7}{4} & - \frac{3}{8} & \frac{1}{2} \\ - \frac{11}{2} & - \frac{5}{4} & 1 \\ \frac{1}{2} & \frac{1}{4} & 0 \end{array}]$