Precalculus Examples

$[\begin{array}{c} - 3 & - 7 \\ 2 & 5 \end{array}]$

Step 1

The inverse of a $2 \times 2$ matrix can be found using the formula $\frac{1}{a d - b c} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where $a d - b c$ is the determinant.

Step 2

Find the determinant.

Tap for more steps...

Step 2.1

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

$- 3 \cdot 5 - 2 \cdot - 7$

Step 2.2

Simplify the determinant.

Tap for more steps...

Step 2.2.1

Simplify each term.

Tap for more steps...

Step 2.2.1.1

Multiply $- 3$ by $5$ .

$- 15 - 2 \cdot - 7$

Step 2.2.1.2

Multiply $- 2$ by $- 7$ .

$- 15 + 14$

Step 2.2.2

Add $- 15$ and $14$ .

$- 1$

Step 3

Since the determinant is non-zero, the inverse exists.

Step 4

Substitute the known values into the formula for the inverse.

$\frac{1}{- 1} [\begin{array}{c} 5 & 7 \\ - 2 & - 3 \end{array}]$

Step 5

Divide $1$ by $- 1$ .

$- [\begin{array}{c} 5 & 7 \\ - 2 & - 3 \end{array}]$

Step 6

Multiply $- 1$ by each element of the matrix.

$[\begin{array}{c} - 1 \cdot 5 & - 1 \cdot 7 \\ - - 2 & - - 3 \end{array}]$

Step 7

Simplify each element in the matrix.

Tap for more steps...

Step 7.1

Multiply $- 1$ by $5$ .

$[\begin{array}{c} - 5 & - 1 \cdot 7 \\ - - 2 & - - 3 \end{array}]$

Step 7.2

Multiply $- 1$ by $7$ .

$[\begin{array}{c} - 5 & - 7 \\ - - 2 & - - 3 \end{array}]$

Step 7.3

Multiply $- 1$ by $- 2$ .

$[\begin{array}{c} - 5 & - 7 \\ 2 & - - 3 \end{array}]$

Step 7.4

Multiply $- 1$ by $- 3$ .

$[\begin{array}{c} - 5 & - 7 \\ 2 & 3 \end{array}]$