Algebra Examples

$[\begin{array}{c} 1 & 3 & - 1 \\ 3 & 6 & 1 \\ 6 & 9 & 6 \end{array}]$

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.

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

Step 1.4

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

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

Step 1.5

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

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

Step 1.6

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

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

Step 1.7

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

$| \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

Step 1.8

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

$- 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

Step 1.9

Add the terms together.

$1 | \begin{array}{c} 6 & 1 \\ 9 & 6 \end{array} | - 3 | \begin{array}{c} 3 & 1 \\ 6 & 6 \end{array} | - 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

Step 2

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

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Step 2.1

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

$1 (6 \cdot 6 - 9 \cdot 1) - 3 | \begin{array}{c} 3 & 1 \\ 6 & 6 \end{array} | - 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

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 $6$ .

$1 (36 - 9 \cdot 1) - 3 | \begin{array}{c} 3 & 1 \\ 6 & 6 \end{array} | - 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

Step 2.2.1.2

Multiply $- 9$ by $1$ .

$1 (36 - 9) - 3 | \begin{array}{c} 3 & 1 \\ 6 & 6 \end{array} | - 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

Step 2.2.2

Subtract $9$ from $36$ .

$1 \cdot 27 - 3 | \begin{array}{c} 3 & 1 \\ 6 & 6 \end{array} | - 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

Step 3

Evaluate $| \begin{array}{c} 3 & 1 \\ 6 & 6 \end{array} |$ .

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Step 3.1

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

$1 \cdot 27 - 3 (3 \cdot 6 - 6 \cdot 1) - 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

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 $6$ .

$1 \cdot 27 - 3 (18 - 6 \cdot 1) - 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

Step 3.2.1.2

Multiply $- 6$ by $1$ .

$1 \cdot 27 - 3 (18 - 6) - 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

Step 3.2.2

Subtract $6$ from $18$ .

$1 \cdot 27 - 3 \cdot 12 - 1 | \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$

Step 4

Evaluate $| \begin{array}{c} 3 & 6 \\ 6 & 9 \end{array} |$ .

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Step 4.1

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

$1 \cdot 27 - 3 \cdot 12 - 1 (3 \cdot 9 - 6 \cdot 6)$

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 $3$ by $9$ .

$1 \cdot 27 - 3 \cdot 12 - 1 (27 - 6 \cdot 6)$

Step 4.2.1.2

Multiply $- 6$ by $6$ .

$1 \cdot 27 - 3 \cdot 12 - 1 (27 - 36)$

Step 4.2.2

Subtract $36$ from $27$ .

$1 \cdot 27 - 3 \cdot 12 - 1 \cdot - 9$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply $27$ by $1$ .

$27 - 3 \cdot 12 - 1 \cdot - 9$

Step 5.1.2

Multiply $- 3$ by $12$ .

$27 - 36 - 1 \cdot - 9$

Step 5.1.3

Multiply $- 1$ by $- 9$ .

$27 - 36 + 9$

Step 5.2

Subtract $36$ from $27$ .

$- 9 + 9$

Step 5.3

Add $- 9$ and $9$ .

$0$