Precalculus Examples

$[\begin{array}{c} 4 & 8 & 9 \\ 6 & 6 & 8 \\ 1 & 1 & 1 \end{array}]$

Step 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

Consider the corresponding sign chart.

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

Step 1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

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

Step 1.4

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

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

Step 1.5

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

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

Step 1.6

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

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

Step 1.7

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

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

Step 1.8

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

$9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 1.9

Add the terms together.

$4 | \begin{array}{c} 6 & 8 \\ 1 & 1 \end{array} | - 8 | \begin{array}{c} 6 & 8 \\ 1 & 1 \end{array} | + 9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 2

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

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Step 2.1

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

$4 (6 \cdot 1 - 1 \cdot 8) - 8 | \begin{array}{c} 6 & 8 \\ 1 & 1 \end{array} | + 9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 2.2

Simplify the determinant.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Multiply $6$ by $1$ .

$4 (6 - 1 \cdot 8) - 8 | \begin{array}{c} 6 & 8 \\ 1 & 1 \end{array} | + 9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 2.2.1.2

Multiply $- 1$ by $8$ .

$4 (6 - 8) - 8 | \begin{array}{c} 6 & 8 \\ 1 & 1 \end{array} | + 9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 2.2.2

Subtract $8$ from $6$ .

$4 \cdot - 2 - 8 | \begin{array}{c} 6 & 8 \\ 1 & 1 \end{array} | + 9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 3

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

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Step 3.1

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

$4 \cdot - 2 - 8 (6 \cdot 1 - 1 \cdot 8) + 9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 3.2

Simplify the determinant.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply $6$ by $1$ .

$4 \cdot - 2 - 8 (6 - 1 \cdot 8) + 9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 3.2.1.2

Multiply $- 1$ by $8$ .

$4 \cdot - 2 - 8 (6 - 8) + 9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 3.2.2

Subtract $8$ from $6$ .

$4 \cdot - 2 - 8 \cdot - 2 + 9 | \begin{array}{c} 6 & 6 \\ 1 & 1 \end{array} |$

Step 4

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

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Step 4.1

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

$4 \cdot - 2 - 8 \cdot - 2 + 9 (6 \cdot 1 - 1 \cdot 6)$

Step 4.2

Simplify the determinant.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Multiply $6$ by $1$ .

$4 \cdot - 2 - 8 \cdot - 2 + 9 (6 - 1 \cdot 6)$

Step 4.2.1.2

Multiply $- 1$ by $6$ .

$4 \cdot - 2 - 8 \cdot - 2 + 9 (6 - 6)$

Step 4.2.2

Subtract $6$ from $6$ .

$4 \cdot - 2 - 8 \cdot - 2 + 9 \cdot 0$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply $4$ by $- 2$ .

$- 8 - 8 \cdot - 2 + 9 \cdot 0$

Step 5.1.2

Multiply $- 8$ by $- 2$ .

$- 8 + 16 + 9 \cdot 0$

Step 5.1.3

Multiply $9$ by $0$ .

$- 8 + 16 + 0$

Step 5.2

Add $- 8$ and $16$ .

$8 + 0$

Step 5.3

Add $8$ and $0$ .

$8$