Linear Algebra Examples

$[\begin{array}{c} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 0 \end{array}]$

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 $3$ 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 $a_{31}$ is the determinant with row $3$ and column $1$ deleted.

$| \begin{array}{c} 2 & 3 \\ 5 & 6 \end{array} |$

Step 1.4

Multiply element $a_{31}$ by its cofactor.

$7 | \begin{array}{c} 2 & 3 \\ 5 & 6 \end{array} |$

Step 1.5

The minor for $a_{32}$ is the determinant with row $3$ and column $2$ deleted.

$| \begin{array}{c} 1 & 3 \\ 4 & 6 \end{array} |$

Step 1.6

Multiply element $a_{32}$ by its cofactor.

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

Step 1.7

The minor for $a_{33}$ is the determinant with row $3$ and column $3$ deleted.

$| \begin{array}{c} 1 & 2 \\ 4 & 5 \end{array} |$

Step 1.8

Multiply element $a_{33}$ by its cofactor.

$0 | \begin{array}{c} 1 & 2 \\ 4 & 5 \end{array} |$

Step 1.9

Add the terms together.

$7 | \begin{array}{c} 2 & 3 \\ 5 & 6 \end{array} | - 8 | \begin{array}{c} 1 & 3 \\ 4 & 6 \end{array} | + 0 | \begin{array}{c} 1 & 2 \\ 4 & 5 \end{array} |$

Step 2

Multiply $0$ by $| \begin{array}{c} 1 & 2 \\ 4 & 5 \end{array} |$ .

$7 | \begin{array}{c} 2 & 3 \\ 5 & 6 \end{array} | - 8 | \begin{array}{c} 1 & 3 \\ 4 & 6 \end{array} | + 0$

Step 3

Evaluate $| \begin{array}{c} 2 & 3 \\ 5 & 6 \end{array} |$ .

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Step 3.1

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

$7 (2 \cdot 6 - 5 \cdot 3) - 8 | \begin{array}{c} 1 & 3 \\ 4 & 6 \end{array} | + 0$

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 $2$ by $6$ .

$7 (12 - 5 \cdot 3) - 8 | \begin{array}{c} 1 & 3 \\ 4 & 6 \end{array} | + 0$

Step 3.2.1.2

Multiply $- 5$ by $3$ .

$7 (12 - 15) - 8 | \begin{array}{c} 1 & 3 \\ 4 & 6 \end{array} | + 0$

Step 3.2.2

Subtract $15$ from $12$ .

$7 \cdot - 3 - 8 | \begin{array}{c} 1 & 3 \\ 4 & 6 \end{array} | + 0$

Step 4

Evaluate $| \begin{array}{c} 1 & 3 \\ 4 & 6 \end{array} |$ .

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Step 4.1

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

$7 \cdot - 3 - 8 (1 \cdot 6 - 4 \cdot 3) + 0$

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 $6$ by $1$ .

$7 \cdot - 3 - 8 (6 - 4 \cdot 3) + 0$

Step 4.2.1.2

Multiply $- 4$ by $3$ .

$7 \cdot - 3 - 8 (6 - 12) + 0$

Step 4.2.2

Subtract $12$ from $6$ .

$7 \cdot - 3 - 8 \cdot - 6 + 0$

Step 5

Simplify the determinant.

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Step 5.1

Simplify each term.

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Step 5.1.1

Multiply $7$ by $- 3$ .

$- 21 - 8 \cdot - 6 + 0$

Step 5.1.2

Multiply $- 8$ by $- 6$ .

$- 21 + 48 + 0$

Step 5.2

Add $- 21$ and $48$ .

$27 + 0$

Step 5.3

Add $27$ and $0$ .

$27$