Find the Inverse [[6,3],[8,6]]

$[\begin{array}{c} 6 & 3 \\ 8 & 6 \end{array}]$

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

If $A = [\begin{array}{c} a & b \\ c & d \end{array}]$ then $A^{- 1} = \frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$

Find the determinant of $[\begin{array}{c} 6 & 3 \\ 8 & 6 \end{array}]$ .

These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 6 & 3 \\ 8 & 6 \end{array}] = | \begin{array}{c} 6 & 3 \\ 8 & 6 \end{array} |$

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

$(6) (6) - 8 \cdot 3$

Simplify the determinant.

Simplify each term.

Multiply $6$ by $6$ .

$36 - 8 \cdot 3$

Multiply $- 8$ by $3$ .

$36 - 24$

Subtract $24$ from $36$ .

$12$

Substitute the known values into the formula for the inverse of a matrix.

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

Simplify each element in the matrix.

Rearrange $- (3)$ .

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

Rearrange $- (8)$ .

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

Multiply $\frac{1}{12}$ by each element of the matrix.

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

Simplify each element in the matrix.

Rearrange $\frac{1}{12} \cdot 6$ .

$[\begin{array}{c} \frac{1}{2} & \frac{1}{12} \cdot - 3 \\ \frac{1}{12} \cdot - 8 & \frac{1}{12} \cdot 6 \end{array}]$

Rearrange $\frac{1}{12} \cdot - 3$ .

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{4} \\ \frac{1}{12} \cdot - 8 & \frac{1}{12} \cdot 6 \end{array}]$

Rearrange $\frac{1}{12} \cdot - 8$ .

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{4} \\ - \frac{2}{3} & \frac{1}{12} \cdot 6 \end{array}]$

Rearrange $\frac{1}{12} \cdot 6$ .

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

Find the Inverse [[6,3],[8,6]]

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