Find the Inverse [[3,0],[0,3]]

$[\begin{array}{c} 3 & 0 \\ 0 & 3 \end{array}]$

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 $[\begin{array}{c} 3 & 0 \\ 0 & 3 \end{array}]$ .

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

$determinant [\begin{array}{c} 3 & 0 \\ 0 & 3 \end{array}] = | \begin{array}{c} 3 & 0 \\ 0 & 3 \end{array} |$

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

$(3) (3) + 0 \cdot 0$

Simplify the determinant.

Simplify each term.

Multiply $3$ by $3$ .

$9 + 0 \cdot 0$

Multiply $0$ by $0$ .

$9 + 0$

Add $9$ and $0$ .

$9$

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

$\frac{1}{9} [\begin{array}{c} 3 & - (0) \\ - (0) & 3 \end{array}]$

Simplify each element in the matrix.

Rearrange $- (0)$ .

$\frac{1}{9} [\begin{array}{c} 3 & 0 \\ - (0) & 3 \end{array}]$

Rearrange $- (0)$ .

$\frac{1}{9} [\begin{array}{c} 3 & 0 \\ 0 & 3 \end{array}]$

Multiply $\frac{1}{9}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{9} \cdot 3 & \frac{1}{9} \cdot 0 \\ \frac{1}{9} \cdot 0 & \frac{1}{9} \cdot 3 \end{array}]$

Simplify each element in the matrix.

Rearrange $\frac{1}{9} \cdot 3$ .

$[\begin{array}{c} \frac{1}{3} & \frac{1}{9} \cdot 0 \\ \frac{1}{9} \cdot 0 & \frac{1}{9} \cdot 3 \end{array}]$

Rearrange $\frac{1}{9} \cdot 0$ .

$[\begin{array}{c} \frac{1}{3} & 0 \\ \frac{1}{9} \cdot 0 & \frac{1}{9} \cdot 3 \end{array}]$

Rearrange $\frac{1}{9} \cdot 0$ .

$[\begin{array}{c} \frac{1}{3} & 0 \\ 0 & \frac{1}{9} \cdot 3 \end{array}]$

Rearrange $\frac{1}{9} \cdot 3$ .

$[\begin{array}{c} \frac{1}{3} & 0 \\ 0 & \frac{1}{3} \end{array}]$

Find the Inverse [[3,0],[0,3]]

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