Finite Math Examples

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

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.

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

Step 1.1.4

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

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

Step 1.1.5

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

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

Step 1.1.6

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

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

Step 1.1.7

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

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

Step 1.1.8

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

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

Step 1.1.9

Add the terms together.

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

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

$1 (1 \cdot 1 + 0 \cdot 2) + 0 + d | \begin{array}{c} 1 & 1 \\ 0 & 0 \end{array} |$

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 $1$ .

$1 (1 + 0 \cdot 2) + 0 + d | \begin{array}{c} 1 & 1 \\ 0 & 0 \end{array} |$

Step 1.3.2.1.2

Multiply $0$ by $2$ .

$1 (1 + 0) + 0 + d | \begin{array}{c} 1 & 1 \\ 0 & 0 \end{array} |$

Step 1.3.2.2

Add $1$ and $0$ .

$1 \cdot 1 + 0 + d | \begin{array}{c} 1 & 1 \\ 0 & 0 \end{array} |$

Step 1.4

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

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Step 1.4.1

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

$1 \cdot 1 + 0 + d (1 \cdot 0 + 0 \cdot 1)$

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 $0$ by $1$ .

$1 \cdot 1 + 0 + d (0 + 0 \cdot 1)$

Step 1.4.2.1.2

Multiply $0$ by $1$ .

$1 \cdot 1 + 0 + d (0 + 0)$

Step 1.4.2.2

Add $0$ and $0$ .

$1 \cdot 1 + 0 + d \cdot 0$

Step 1.5

Simplify the determinant.

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Step 1.5.1

Add $1 \cdot 1$ and $0$ .

$1 \cdot 1 + d \cdot 0$

Step 1.5.2

Simplify each term.

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Step 1.5.2.1

Multiply $1$ by $1$ .

$1 + d \cdot 0$

Step 1.5.2.2

Multiply $d$ by $0$ .

$1 + 0$

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.

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

Step 4

Find the reduced row echelon form.

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

Simplify $R_{2}$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

Simplify $R_{2}$ .

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

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Simplify $R_{1}$ .

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

Step 5

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

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