Examples

$[\begin{array}{c} 2 & 1 & 2 \\ 0 & 3 & 2 \\ 3 & 4 & 2 \end{array}]$

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 column

$1$ 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.

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

Step 1.4

Multiply element

$a_{11}$ by its cofactor.

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

Step 1.5

The minor for

$a_{21}$ is the determinant with row

$2$ and column

$1$ deleted.

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

Step 1.6

Multiply element

$a_{21}$ by its cofactor.

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

Step 1.7

The minor for

$a_{31}$ is the determinant with row

$3$ and column

$1$ deleted.

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

Step 1.8

Multiply element

$a_{31}$ by its cofactor.

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

Step 1.9

Add the terms together.

$2 | \begin{array}{c} 3 & 2 \\ 4 & 2 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 4 & 2 \end{array} | + 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} |$

Step 2

Multiply

$0$ by

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

$2 | \begin{array}{c} 3 & 2 \\ 4 & 2 \end{array} | + 0 + 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} |$

Step 3

Evaluate

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

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Step 3.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$2 (3 \cdot 2 - 4 \cdot 2) + 0 + 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} |$

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

$3$ by

$2$ .

$2 (6 - 4 \cdot 2) + 0 + 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} |$

Step 3.2.1.2

Multiply

$- 4$ by

$2$ .

$2 (6 - 8) + 0 + 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} |$

Step 3.2.2

Subtract

$8$ from

$6$ .

$2 \cdot - 2 + 0 + 3 | \begin{array}{c} 1 & 2 \\ 3 & 2 \end{array} |$

Step 4

Evaluate

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

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Step 4.1

The determinant of a

$2 \times 2$ matrix can be found using the formula

$| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$2 \cdot - 2 + 0 + 3 (1 \cdot 2 - 3 \cdot 2)$

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

$1$ .

$2 \cdot - 2 + 0 + 3 (2 - 3 \cdot 2)$

Step 4.2.1.2

Multiply

$- 3$ by

$2$ .

$2 \cdot - 2 + 0 + 3 (2 - 6)$

Step 4.2.2

Subtract

$6$ from

$2$ .

$2 \cdot - 2 + 0 + 3 \cdot - 4$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply

$2$ by

$- 2$ .

$- 4 + 0 + 3 \cdot - 4$

Step 5.1.2

Multiply

$3$ by

$- 4$ .

$- 4 + 0 - 12$

Step 5.2

Add

$- 4$ and

$0$ .

$- 4 - 12$

Step 5.3

Subtract

$12$ from

$- 4$ .

$- 16$

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