Precalculus Examples

$[\begin{array}{c} 5 & 4 \\ 3 & 5 \end{array}]$

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.

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Step 2.1

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

$5 \cdot 5 - 3 \cdot 4$

Step 2.2

Simplify the determinant.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Multiply $5$ by $5$ .

$25 - 3 \cdot 4$

Step 2.2.1.2

Multiply $- 3$ by $4$ .

$25 - 12$

Step 2.2.2

Subtract $12$ from $25$ .

$13$

Step 3

Since the determinant is non-zero, the inverse exists.

Step 4

Substitute the known values into the formula for the inverse.

$\frac{1}{13} [\begin{array}{c} 5 & - 4 \\ - 3 & 5 \end{array}]$

Step 5

Multiply $\frac{1}{13}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{13} \cdot 5 & \frac{1}{13} \cdot - 4 \\ \frac{1}{13} \cdot - 3 & \frac{1}{13} \cdot 5 \end{array}]$

Step 6

Simplify each element in the matrix.

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Step 6.1

Combine $\frac{1}{13}$ and $5$ .

$[\begin{array}{c} \frac{5}{13} & \frac{1}{13} \cdot - 4 \\ \frac{1}{13} \cdot - 3 & \frac{1}{13} \cdot 5 \end{array}]$

Step 6.2

Combine $\frac{1}{13}$ and $- 4$ .

$[\begin{array}{c} \frac{5}{13} & \frac{- 4}{13} \\ \frac{1}{13} \cdot - 3 & \frac{1}{13} \cdot 5 \end{array}]$

Step 6.3

Move the negative in front of the fraction.

$[\begin{array}{c} \frac{5}{13} & - \frac{4}{13} \\ \frac{1}{13} \cdot - 3 & \frac{1}{13} \cdot 5 \end{array}]$

Step 6.4

Combine $\frac{1}{13}$ and $- 3$ .

$[\begin{array}{c} \frac{5}{13} & - \frac{4}{13} \\ \frac{- 3}{13} & \frac{1}{13} \cdot 5 \end{array}]$

Step 6.5

Move the negative in front of the fraction.

$[\begin{array}{c} \frac{5}{13} & - \frac{4}{13} \\ - \frac{3}{13} & \frac{1}{13} \cdot 5 \end{array}]$

Step 6.6

Combine $\frac{1}{13}$ and $5$ .

$[\begin{array}{c} \frac{5}{13} & - \frac{4}{13} \\ - \frac{3}{13} & \frac{5}{13} \end{array}]$