Precalculus Examples

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

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 $2$ 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 $a_{21}$ is the determinant with row $2$ and column $1$ deleted.

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

Step 1.4

Multiply element $a_{21}$ by its cofactor.

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

Step 1.5

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

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

Step 1.6

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

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

Step 1.7

The minor for $a_{23}$ is the determinant with row $2$ and column $3$ deleted.

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

Step 1.8

Multiply element $a_{23}$ by its cofactor.

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

Step 1.9

Add the terms together.

$- 1 | \begin{array}{c} - 4 & 6 \\ 2 & 4 \end{array} | + 4 | \begin{array}{c} 2 & 6 \\ - 1 & 4 \end{array} | + 0 | \begin{array}{c} 2 & - 4 \\ - 1 & 2 \end{array} |$

Step 2

Multiply $0$ by $| \begin{array}{c} 2 & - 4 \\ - 1 & 2 \end{array} |$ .

$- 1 | \begin{array}{c} - 4 & 6 \\ 2 & 4 \end{array} | + 4 | \begin{array}{c} 2 & 6 \\ - 1 & 4 \end{array} | + 0$

Step 3

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

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Step 3.1

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

$- 1 (- 4 \cdot 4 - 2 \cdot 6) + 4 | \begin{array}{c} 2 & 6 \\ - 1 & 4 \end{array} | + 0$

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 $- 4$ by $4$ .

$- 1 (- 16 - 2 \cdot 6) + 4 | \begin{array}{c} 2 & 6 \\ - 1 & 4 \end{array} | + 0$

Step 3.2.1.2

Multiply $- 2$ by $6$ .

$- 1 (- 16 - 12) + 4 | \begin{array}{c} 2 & 6 \\ - 1 & 4 \end{array} | + 0$

Step 3.2.2

Subtract $12$ from $- 16$ .

$- 1 \cdot - 28 + 4 | \begin{array}{c} 2 & 6 \\ - 1 & 4 \end{array} | + 0$

Step 4

Evaluate $| \begin{array}{c} 2 & 6 \\ - 1 & 4 \end{array} |$ .

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Step 4.1

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

$- 1 \cdot - 28 + 4 (2 \cdot 4 - (- 1 \cdot 6)) + 0$

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 $4$ .

$- 1 \cdot - 28 + 4 (8 - (- 1 \cdot 6)) + 0$

Step 4.2.1.2

Multiply $- (- 1 \cdot 6)$ .

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Step 4.2.1.2.1

Multiply $- 1$ by $6$ .

$- 1 \cdot - 28 + 4 (8 - - 6) + 0$

Step 4.2.1.2.2

Multiply $- 1$ by $- 6$ .

$- 1 \cdot - 28 + 4 (8 + 6) + 0$

Step 4.2.2

Add $8$ and $6$ .

$- 1 \cdot - 28 + 4 \cdot 14 + 0$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply $- 1$ by $- 28$ .

$28 + 4 \cdot 14 + 0$

Step 5.1.2

Multiply $4$ by $14$ .

$28 + 56 + 0$

Step 5.2

Add $28$ and $56$ .

$84 + 0$

Step 5.3

Add $84$ and $0$ .

$84$