Finite Math Examples

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

Step 1

Find the determinant.

Tap for more steps...

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 column $1$ by its cofactor and add.

Tap for more steps...

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.

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

Step 1.1.4

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

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

Step 1.1.5

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

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

Step 1.1.6

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

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

Step 1.1.7

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

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

Step 1.1.8

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

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

Step 1.1.9

Add the terms together.

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

Step 1.2

Multiply $0$ by $| \begin{array}{c} 3 & - 2 \\ 4 & 0 \end{array} |$ .

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

Step 1.3

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

Tap for more steps...

Step 1.3.1

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

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

Step 1.3.2

Simplify the determinant.

Tap for more steps...

Step 1.3.2.1

Multiply $4$ by $2$ .

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

Step 1.3.2.2

Subtract $0$ from $8$ .

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

Step 1.4

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

Tap for more steps...

Step 1.4.1

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

$2 \cdot 8 - 2 (3 \cdot 2 - 1 \cdot - 2) + 0$

Step 1.4.2

Simplify the determinant.

Tap for more steps...

Step 1.4.2.1

Simplify each term.

Tap for more steps...

Step 1.4.2.1.1

Multiply $3$ by $2$ .

$2 \cdot 8 - 2 (6 - 1 \cdot - 2) + 0$

Step 1.4.2.1.2

Multiply $- 1$ by $- 2$ .

$2 \cdot 8 - 2 (6 + 2) + 0$

Step 1.4.2.2

Add $6$ and $2$ .

$2 \cdot 8 - 2 \cdot 8 + 0$

Step 1.5

Simplify the determinant.

Tap for more steps...

Step 1.5.1

Simplify each term.

Tap for more steps...

Step 1.5.1.1

Multiply $2$ by $8$ .

$16 - 2 \cdot 8 + 0$

Step 1.5.1.2

Multiply $- 2$ by $8$ .

$16 - 16 + 0$

Step 1.5.2

Subtract $16$ from $16$ .

$0 + 0$

Step 1.5.3

Add $0$ and $0$ .

$0$

Step 2

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