Precalculus Examples

$[\begin{array}{c} 6 & 1 & 7 \\ 2 & - 3 & 3 \\ 4 & - 1 & 2 \end{array}]$

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.

$| \begin{array}{c} - 3 & 3 \\ - 1 & 2 \end{array} |$

Step 1.4

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

$6 | \begin{array}{c} - 3 & 3 \\ - 1 & 2 \end{array} |$

Step 1.5

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

$| \begin{array}{c} 2 & 3 \\ 4 & 2 \end{array} |$

Step 1.6

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

$- 1 | \begin{array}{c} 2 & 3 \\ 4 & 2 \end{array} |$

Step 1.7

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

$| \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 1.8

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

$7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 1.9

Add the terms together.

$6 | \begin{array}{c} - 3 & 3 \\ - 1 & 2 \end{array} | - 1 | \begin{array}{c} 2 & 3 \\ 4 & 2 \end{array} | + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 2

Evaluate $| \begin{array}{c} - 3 & 3 \\ - 1 & 2 \end{array} |$ .

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Step 2.1

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

$6 (- 3 \cdot 2 - (- 1 \cdot 3)) - 1 | \begin{array}{c} 2 & 3 \\ 4 & 2 \end{array} | + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

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 $- 3$ by $2$ .

$6 (- 6 - (- 1 \cdot 3)) - 1 | \begin{array}{c} 2 & 3 \\ 4 & 2 \end{array} | + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 2.2.1.2

Multiply $- (- 1 \cdot 3)$ .

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Step 2.2.1.2.1

Multiply $- 1$ by $3$ .

$6 (- 6 - - 3) - 1 | \begin{array}{c} 2 & 3 \\ 4 & 2 \end{array} | + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 2.2.1.2.2

Multiply $- 1$ by $- 3$ .

$6 (- 6 + 3) - 1 | \begin{array}{c} 2 & 3 \\ 4 & 2 \end{array} | + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 2.2.2

Add $- 6$ and $3$ .

$6 \cdot - 3 - 1 | \begin{array}{c} 2 & 3 \\ 4 & 2 \end{array} | + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 3

Evaluate $| \begin{array}{c} 2 & 3 \\ 4 & 2 \end{array} |$ .

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Step 3.1

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

$6 \cdot - 3 - 1 (2 \cdot 2 - 4 \cdot 3) + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

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 $2$ by $2$ .

$6 \cdot - 3 - 1 (4 - 4 \cdot 3) + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 3.2.1.2

Multiply $- 4$ by $3$ .

$6 \cdot - 3 - 1 (4 - 12) + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 3.2.2

Subtract $12$ from $4$ .

$6 \cdot - 3 - 1 \cdot - 8 + 7 | \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$

Step 4

Evaluate $| \begin{array}{c} 2 & - 3 \\ 4 & - 1 \end{array} |$ .

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Step 4.1

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

$6 \cdot - 3 - 1 \cdot - 8 + 7 (2 \cdot - 1 - 4 \cdot - 3)$

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 $2$ by $- 1$ .

$6 \cdot - 3 - 1 \cdot - 8 + 7 (- 2 - 4 \cdot - 3)$

Step 4.2.1.2

Multiply $- 4$ by $- 3$ .

$6 \cdot - 3 - 1 \cdot - 8 + 7 (- 2 + 12)$

Step 4.2.2

Add $- 2$ and $12$ .

$6 \cdot - 3 - 1 \cdot - 8 + 7 \cdot 10$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply $6$ by $- 3$ .

$- 18 - 1 \cdot - 8 + 7 \cdot 10$

Step 5.1.2

Multiply $- 1$ by $- 8$ .

$- 18 + 8 + 7 \cdot 10$

Step 5.1.3

Multiply $7$ by $10$ .

$- 18 + 8 + 70$

Step 5.2

Add $- 18$ and $8$ .

$- 10 + 70$

Step 5.3

Add $- 10$ and $70$ .

$60$