Find the Inverse [[-1,3],[-3,-1]]

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

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

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

$determinant [\begin{array}{c} - 1 & 3 \\ - 3 & - 1 \end{array}] = | \begin{array}{c} - 1 & 3 \\ - 3 & - 1 \end{array} |$

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

$(- 1) (- 1) + 3 \cdot 3$

Simplify the determinant.

Simplify each term.

Multiply $- 1$ by $- 1$ .

$1 + 3 \cdot 3$

Multiply $3$ by $3$ .

$1 + 9$

Add $1$ and $9$ .

$10$

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

$\frac{1}{10} [\begin{array}{c} - 1 & - (3) \\ - (- 3) & - 1 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (3)$ .

$\frac{1}{10} [\begin{array}{c} - 1 & - 3 \\ - (- 3) & - 1 \end{array}]$

Rearrange $- (- 3)$ .

$\frac{1}{10} [\begin{array}{c} - 1 & - 3 \\ 3 & - 1 \end{array}]$

Multiply $\frac{1}{10}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{10} \cdot - 1 & \frac{1}{10} \cdot - 3 \\ \frac{1}{10} \cdot 3 & \frac{1}{10} \cdot - 1 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{10} \cdot - 1$ .

$[\begin{array}{c} - \frac{1}{10} & \frac{1}{10} \cdot - 3 \\ \frac{1}{10} \cdot 3 & \frac{1}{10} \cdot - 1 \end{array}]$

Rearrange $\frac{1}{10} \cdot - 3$ .

$[\begin{array}{c} - \frac{1}{10} & - \frac{3}{10} \\ \frac{1}{10} \cdot 3 & \frac{1}{10} \cdot - 1 \end{array}]$

Rearrange $\frac{1}{10} \cdot 3$ .

$[\begin{array}{c} - \frac{1}{10} & - \frac{3}{10} \\ \frac{3}{10} & \frac{1}{10} \cdot - 1 \end{array}]$

Rearrange $\frac{1}{10} \cdot - 1$ .

$[\begin{array}{c} - \frac{1}{10} & - \frac{3}{10} \\ \frac{3}{10} & - \frac{1}{10} \end{array}]$

Find the Inverse [[-1,3],[-3,-1]]

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