Finite Math Examples

$[\begin{array}{c} 5 & 5 & 0 \\ 0 & - 5 & - 5 \\ 3 & 0 & - 3 \end{array}]$

Step 1

Find the determinant.

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Step 1.1

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 1.1.1

Consider the corresponding sign chart.

$| \begin{array}{c} + & - & + \\ - & + & - \\ + & - & + \end{array} |$

Step 1.1.2

The cofactor is the minor with the sign changed if the indices match a $-$ position on the sign chart.

Step 1.1.3

The minor for $a_{11}$ is the determinant with row $1$ and column $1$ deleted.

$| \begin{array}{c} - 5 & - 5 \\ 0 & - 3 \end{array} |$

Step 1.1.4

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

$5 | \begin{array}{c} - 5 & - 5 \\ 0 & - 3 \end{array} |$

Step 1.1.5

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

$| \begin{array}{c} 0 & - 5 \\ 3 & - 3 \end{array} |$

Step 1.1.6

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

$- 5 | \begin{array}{c} 0 & - 5 \\ 3 & - 3 \end{array} |$

Step 1.1.7

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

$| \begin{array}{c} 0 & - 5 \\ 3 & 0 \end{array} |$

Step 1.1.8

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

$0 | \begin{array}{c} 0 & - 5 \\ 3 & 0 \end{array} |$

Step 1.1.9

Add the terms together.

$5 | \begin{array}{c} - 5 & - 5 \\ 0 & - 3 \end{array} | - 5 | \begin{array}{c} 0 & - 5 \\ 3 & - 3 \end{array} | + 0 | \begin{array}{c} 0 & - 5 \\ 3 & 0 \end{array} |$

Step 1.2

Multiply $0$ by $| \begin{array}{c} 0 & - 5 \\ 3 & 0 \end{array} |$ .

$5 | \begin{array}{c} - 5 & - 5 \\ 0 & - 3 \end{array} | - 5 | \begin{array}{c} 0 & - 5 \\ 3 & - 3 \end{array} | + 0$

Step 1.3

Evaluate $| \begin{array}{c} - 5 & - 5 \\ 0 & - 3 \end{array} |$ .

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Step 1.3.1

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

$5 (- 5 \cdot - 3 + 0 \cdot - 5) - 5 | \begin{array}{c} 0 & - 5 \\ 3 & - 3 \end{array} | + 0$

Step 1.3.2

Simplify the determinant.

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Step 1.3.2.1

Simplify each term.

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Step 1.3.2.1.1

Multiply $- 5$ by $- 3$ .

$5 (15 + 0 \cdot - 5) - 5 | \begin{array}{c} 0 & - 5 \\ 3 & - 3 \end{array} | + 0$

Step 1.3.2.1.2

Multiply $0$ by $- 5$ .

$5 (15 + 0) - 5 | \begin{array}{c} 0 & - 5 \\ 3 & - 3 \end{array} | + 0$

Step 1.3.2.2

Add $15$ and $0$ .

$5 \cdot 15 - 5 | \begin{array}{c} 0 & - 5 \\ 3 & - 3 \end{array} | + 0$

Step 1.4

Evaluate $| \begin{array}{c} 0 & - 5 \\ 3 & - 3 \end{array} |$ .

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Step 1.4.1

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

$5 \cdot 15 - 5 (0 \cdot - 3 - 3 \cdot - 5) + 0$

Step 1.4.2

Simplify the determinant.

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Step 1.4.2.1

Simplify each term.

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Step 1.4.2.1.1

Multiply $0$ by $- 3$ .

$5 \cdot 15 - 5 (0 - 3 \cdot - 5) + 0$

Step 1.4.2.1.2

Multiply $- 3$ by $- 5$ .

$5 \cdot 15 - 5 (0 + 15) + 0$

Step 1.4.2.2

Add $0$ and $15$ .

$5 \cdot 15 - 5 \cdot 15 + 0$

Step 1.5

Simplify the determinant.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

Multiply $5$ by $15$ .

$75 - 5 \cdot 15 + 0$

Step 1.5.1.2

Multiply $- 5$ by $15$ .

$75 - 75 + 0$

Step 1.5.2

Subtract $75$ from $75$ .

$0 + 0$

Step 1.5.3

Add $0$ and $0$ .

$0$

Step 2

There is no inverse because the determinant is $0$ .