Find the Inverse [[2,3],[2,2]]

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

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 $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$2 \cdot 2 - 2 \cdot 3$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $2$ .

$4 - 2 \cdot 3$

Multiply $- 2$ by $3$ .

$4 - 6$

Subtract $6$ from $4$ .

$- 2$

Step 3

Substitute the known values into the formula for the inverse.

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

Step 4

Move the negative in front of the fraction.

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

Step 5

Multiply $- \frac{1}{2}$ by each element of the matrix.

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

Step 6

Simplify each element in the matrix.

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- \frac{1}{2} \cdot - 3$ .

Multiply $- 3$ by $- 1$ .

$[\begin{array}{c} - 1 & 3 (\frac{1}{2}) \\ - \frac{1}{2} \cdot - 2 & - \frac{1}{2} \cdot 2 \end{array}]$

Combine $3$ and $\frac{1}{2}$ .

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

$[\begin{array}{c} - 1 & \frac{3}{2} \\ \frac{- 1}{2} \cdot (2 (- 1)) & - \frac{1}{2} \cdot 2 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} - 1 & \frac{3}{2} \\ \frac{- 1}{2} \cdot (2 \cdot - 1) & - \frac{1}{2} \cdot 2 \end{array}]$

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Find the Inverse [[2,3],[2,2]]

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