Finite Math Examples

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

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

$| \begin{array}{c} 0 & - 2 \\ 2 & 5 \end{array} |$

Step 1.1.4

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

$0 | \begin{array}{c} 0 & - 2 \\ 2 & 5 \end{array} |$

Step 1.1.5

The minor for $a_{22}$ is the determinant with row $2$ and column $2$ deleted.

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

Step 1.1.6

Multiply element $a_{22}$ by its cofactor.

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

Step 1.1.7

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

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

Step 1.1.8

Multiply element $a_{32}$ by its cofactor.

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

Step 1.1.9

Add the terms together.

$0 | \begin{array}{c} 0 & - 2 \\ 2 & 5 \end{array} | + 1 | \begin{array}{c} 1 & 2 \\ 2 & 5 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 0 & - 2 \end{array} |$

Step 1.2

Multiply $0$ by $| \begin{array}{c} 0 & - 2 \\ 2 & 5 \end{array} |$ .

$0 + 1 | \begin{array}{c} 1 & 2 \\ 2 & 5 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 0 & - 2 \end{array} |$

Step 1.3

Multiply $0$ by $| \begin{array}{c} 1 & 2 \\ 0 & - 2 \end{array} |$ .

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

Step 1.4

Evaluate $| \begin{array}{c} 1 & 2 \\ 2 & 5 \end{array} |$ .

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Step 1.4.1

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

$0 + 1 (1 \cdot 5 - 2 \cdot 2) + 0$

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 $5$ by $1$ .

$0 + 1 (5 - 2 \cdot 2) + 0$

Step 1.4.2.1.2

Multiply $- 2$ by $2$ .

$0 + 1 (5 - 4) + 0$

Step 1.4.2.2

Subtract $4$ from $5$ .

$0 + 1 \cdot 1 + 0$

Step 1.5

Simplify the determinant.

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Step 1.5.1

Multiply $1$ by $1$ .

$0 + 1 + 0$

Step 1.5.2

Add $0$ and $1$ .

$1 + 0$

Step 1.5.3

Add $1$ and $0$ .

$1$

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.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify $R_{3}$ .

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

Step 4.2

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 4.2.1

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

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

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 4.3.1

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

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

Step 4.3.2

Simplify $R_{1}$ .

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

Step 5

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

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