Find the Inverse [[9,-2],[-10,9]]

$[\begin{array}{c} 9 & - 2 \\ - 10 & 9 \end{array}]$

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.

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

$9 \cdot 9 - (- 10 \cdot - 2)$

Simplify the determinant.

Simplify each term.

Multiply $9$ by $9$ .

$81 - (- 10 \cdot - 2)$

Multiply $- (- 10 \cdot - 2)$ .

Multiply $- 10$ by $- 2$ .

$81 - 1 \cdot 20$

Multiply $- 1$ by $20$ .

$81 - 20$

Subtract $20$ from $81$ .

$61$

Step 3

Substitute the known values into the formula for the inverse.

$\frac{1}{61} [\begin{array}{c} 9 & 2 \\ 10 & 9 \end{array}]$

Step 4

Multiply $\frac{1}{61}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{61} \cdot 9 & \frac{1}{61} \cdot 2 \\ \frac{1}{61} \cdot 10 & \frac{1}{61} \cdot 9 \end{array}]$

Step 5

Simplify each element in the matrix.

Combine $\frac{1}{61}$ and $9$ .

$[\begin{array}{c} \frac{9}{61} & \frac{1}{61} \cdot 2 \\ \frac{1}{61} \cdot 10 & \frac{1}{61} \cdot 9 \end{array}]$

Combine $\frac{1}{61}$ and $2$ .

$[\begin{array}{c} \frac{9}{61} & \frac{2}{61} \\ \frac{1}{61} \cdot 10 & \frac{1}{61} \cdot 9 \end{array}]$

Combine $\frac{1}{61}$ and $10$ .

$[\begin{array}{c} \frac{9}{61} & \frac{2}{61} \\ \frac{10}{61} & \frac{1}{61} \cdot 9 \end{array}]$

Combine $\frac{1}{61}$ and $9$ .

$[\begin{array}{c} \frac{9}{61} & \frac{2}{61} \\ \frac{10}{61} & \frac{9}{61} \end{array}]$

Find the Inverse [[9,-2],[-10,9]]

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