Find the Inverse [[3,-2],[6,4]]

$[\begin{array}{c} 3 & - 2 \\ 6 & 4 \end{array}]$

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

$3 \cdot 4 - 6 \cdot - 2$

Simplify the determinant.

Simplify each term.

Multiply $3$ by $4$ .

$12 - 6 \cdot - 2$

Multiply $- 6$ by $- 2$ .

$12 + 12$

Add $12$ and $12$ .

$24$

Step 3

Substitute the known values into the formula for the inverse.

$\frac{1}{24} [\begin{array}{c} 4 & 2 \\ - 6 & 3 \end{array}]$

Step 4

Multiply $\frac{1}{24}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{24} \cdot 4 & \frac{1}{24} \cdot 2 \\ \frac{1}{24} \cdot - 6 & \frac{1}{24} \cdot 3 \end{array}]$

Step 5

Simplify each element in the matrix.

Cancel the common factor of $4$ .

Factor $4$ out of $24$ .

$[\begin{array}{c} \frac{1}{4 (6)} \cdot 4 & \frac{1}{24} \cdot 2 \\ \frac{1}{24} \cdot - 6 & \frac{1}{24} \cdot 3 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} \frac{1}{4 \cdot 6} \cdot 4 & \frac{1}{24} \cdot 2 \\ \frac{1}{24} \cdot - 6 & \frac{1}{24} \cdot 3 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} \frac{1}{6} & \frac{1}{24} \cdot 2 \\ \frac{1}{24} \cdot - 6 & \frac{1}{24} \cdot 3 \end{array}]$

Cancel the common factor of $2$ .

Factor $2$ out of $24$ .

$[\begin{array}{c} \frac{1}{6} & \frac{1}{2 (12)} \cdot 2 \\ \frac{1}{24} \cdot - 6 & \frac{1}{24} \cdot 3 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} \frac{1}{6} & \frac{1}{2 \cdot 12} \cdot 2 \\ \frac{1}{24} \cdot - 6 & \frac{1}{24} \cdot 3 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ \frac{1}{24} \cdot - 6 & \frac{1}{24} \cdot 3 \end{array}]$

Cancel the common factor of $6$ .

Factor $6$ out of $24$ .

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ \frac{1}{6 (4)} \cdot - 6 & \frac{1}{24} \cdot 3 \end{array}]$

Factor $6$ out of $- 6$ .

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ \frac{1}{6 \cdot 4} \cdot (6 \cdot - 1) & \frac{1}{24} \cdot 3 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ \frac{1}{6 \cdot 4} \cdot (6 \cdot - 1) & \frac{1}{24} \cdot 3 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ \frac{1}{4} \cdot - 1 & \frac{1}{24} \cdot 3 \end{array}]$

Combine $\frac{1}{4}$ and $- 1$ .

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ \frac{- 1}{4} & \frac{1}{24} \cdot 3 \end{array}]$

Move the negative in front of the fraction.

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ - \frac{1}{4} & \frac{1}{24} \cdot 3 \end{array}]$

Cancel the common factor of $3$ .

Factor $3$ out of $24$ .

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ - \frac{1}{4} & \frac{1}{3 (8)} \cdot 3 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ - \frac{1}{4} & \frac{1}{3 \cdot 8} \cdot 3 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} \frac{1}{6} & \frac{1}{12} \\ - \frac{1}{4} & \frac{1}{8} \end{array}]$

Find the Inverse [[3,-2],[6,4]]

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