Precalculus Examples

$[\begin{array}{c} - 5 & 1 \\ - 7 & 1 \end{array}]$

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 $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$- 5 \cdot 1 - (- 7 \cdot 1)$

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 $1$ .

$- 5 - (- 7 \cdot 1)$

Step 2.2.1.2

Multiply $- (- 7 \cdot 1)$ .

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Step 2.2.1.2.1

Multiply $- 7$ by $1$ .

$- 5 - - 7$

Step 2.2.1.2.2

Multiply $- 1$ by $- 7$ .

$- 5 + 7$

Step 2.2.2

Add $- 5$ and $7$ .

$2$

Step 3

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

Step 4

Substitute the known values into the formula for the inverse.

$\frac{1}{2} [\begin{array}{c} 1 & - 1 \\ 7 & - 5 \end{array}]$

Step 5

Multiply $\frac{1}{2}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{2} \cdot 1 & \frac{1}{2} \cdot - 1 \\ \frac{1}{2} \cdot 7 & \frac{1}{2} \cdot - 5 \end{array}]$

Step 6

Simplify each element in the matrix.

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Step 6.1

Multiply $\frac{1}{2}$ by $1$ .

$[\begin{array}{c} \frac{1}{2} & \frac{1}{2} \cdot - 1 \\ \frac{1}{2} \cdot 7 & \frac{1}{2} \cdot - 5 \end{array}]$

Step 6.2

Combine $\frac{1}{2}$ and $- 1$ .

$[\begin{array}{c} \frac{1}{2} & \frac{- 1}{2} \\ \frac{1}{2} \cdot 7 & \frac{1}{2} \cdot - 5 \end{array}]$

Step 6.3

Move the negative in front of the fraction.

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{2} \\ \frac{1}{2} \cdot 7 & \frac{1}{2} \cdot - 5 \end{array}]$

Step 6.4

Combine $\frac{1}{2}$ and $7$ .

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{2} \\ \frac{7}{2} & \frac{1}{2} \cdot - 5 \end{array}]$

Step 6.5

Combine $\frac{1}{2}$ and $- 5$ .

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{2} \\ \frac{7}{2} & \frac{- 5}{2} \end{array}]$

Step 6.6

Move the negative in front of the fraction.

$[\begin{array}{c} \frac{1}{2} & - \frac{1}{2} \\ \frac{7}{2} & - \frac{5}{2} \end{array}]$