Find the Inverse C=[[12,-20],[36,-52]]

$C = [\begin{array}{c} 12 & - 20 \\ 36 & - 52 \end{array}]$

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 $[\begin{array}{c} 12 & - 20 \\ 36 & - 52 \end{array}]$ .

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

$determinant [\begin{array}{c} 12 & - 20 \\ 36 & - 52 \end{array}] = | \begin{array}{c} 12 & - 20 \\ 36 & - 52 \end{array} |$

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

$(12) (- 52) - 36 \cdot - 20$

Simplify the determinant.

Simplify each term.

Multiply $12$ by $- 52$ .

$- 624 - 36 \cdot - 20$

Multiply $- 36$ by $- 20$ .

$- 624 + 720$

Add $- 624$ and $720$ .

$96$

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

$\frac{1}{96} [\begin{array}{c} - 52 & - (- 20) \\ - (36) & 12 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (- 20)$ .

$\frac{1}{96} [\begin{array}{c} - 52 & 20 \\ - (36) & 12 \end{array}]$

Rearrange $- (36)$ .

$\frac{1}{96} [\begin{array}{c} - 52 & 20 \\ - 36 & 12 \end{array}]$

Multiply $\frac{1}{96}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{96} \cdot - 52 & \frac{1}{96} \cdot 20 \\ \frac{1}{96} \cdot - 36 & \frac{1}{96} \cdot 12 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{96} \cdot - 52$ .

$[\begin{array}{c} - \frac{13}{24} & \frac{1}{96} \cdot 20 \\ \frac{1}{96} \cdot - 36 & \frac{1}{96} \cdot 12 \end{array}]$

Rearrange $\frac{1}{96} \cdot 20$ .

$[\begin{array}{c} - \frac{13}{24} & \frac{5}{24} \\ \frac{1}{96} \cdot - 36 & \frac{1}{96} \cdot 12 \end{array}]$

Rearrange $\frac{1}{96} \cdot - 36$ .

$[\begin{array}{c} - \frac{13}{24} & \frac{5}{24} \\ - \frac{3}{8} & \frac{1}{96} \cdot 12 \end{array}]$

Rearrange $\frac{1}{96} \cdot 12$ .

$[\begin{array}{c} - \frac{13}{24} & \frac{5}{24} \\ - \frac{3}{8} & \frac{1}{8} \end{array}]$

Find the Inverse C=[[12,-20],[36,-52]]

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