Precalculus Examples

$[\begin{array}{c} 4 & 1 \\ 1 & 1 \end{array}]$

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

$4 \cdot 1 - 1 \cdot 1$

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 $4$ by $1$ .

$4 - 1 \cdot 1$

Step 2.2.1.2

Multiply $- 1$ by $1$ .

$4 - 1$

Step 2.2.2

Subtract $1$ from $4$ .

$3$

Step 3

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

Step 4

Substitute the known values into the formula for the inverse.

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

Step 5

Multiply $\frac{1}{3}$ by each element of the matrix.

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

Step 6

Simplify each element in the matrix.

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Step 6.1

Multiply $\frac{1}{3}$ by $1$ .

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

Step 6.2

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

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

Step 6.3

Move the negative in front of the fraction.

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

Step 6.4

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

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

Step 6.5

Move the negative in front of the fraction.

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

Step 6.6

Combine $\frac{1}{3}$ and $4$ .

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