Precalculus Examples

$[\begin{array}{c} 8 & 23 \\ 2 & 6 \end{array}]$

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

$8 \cdot 6 - 2 \cdot 23$

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 $8$ by $6$ .

$48 - 2 \cdot 23$

Step 2.2.1.2

Multiply $- 2$ by $23$ .

$48 - 46$

Step 2.2.2

Subtract $46$ from $48$ .

$2$

Step 3

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

Step 4

Substitute the known values into the formula for the inverse.

$\frac{1}{2} [\begin{array}{c} 6 & - 23 \\ - 2 & 8 \end{array}]$

Step 5

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

$[\begin{array}{c} \frac{1}{2} \cdot 6 & \frac{1}{2} \cdot - 23 \\ \frac{1}{2} \cdot - 2 & \frac{1}{2} \cdot 8 \end{array}]$

Step 6

Simplify each element in the matrix.

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Step 6.1

Cancel the common factor of $2$ .

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Step 6.1.1

Factor $2$ out of $6$ .

$[\begin{array}{c} \frac{1}{2} \cdot (2 (3)) & \frac{1}{2} \cdot - 23 \\ \frac{1}{2} \cdot - 2 & \frac{1}{2} \cdot 8 \end{array}]$

Step 6.1.2

Cancel the common factor.

$[\begin{array}{c} \frac{1}{2} \cdot (2 \cdot 3) & \frac{1}{2} \cdot - 23 \\ \frac{1}{2} \cdot - 2 & \frac{1}{2} \cdot 8 \end{array}]$

Step 6.1.3

Rewrite the expression.

$[\begin{array}{c} 3 & \frac{1}{2} \cdot - 23 \\ \frac{1}{2} \cdot - 2 & \frac{1}{2} \cdot 8 \end{array}]$

Step 6.2

Combine $\frac{1}{2}$ and $- 23$ .

$[\begin{array}{c} 3 & \frac{- 23}{2} \\ \frac{1}{2} \cdot - 2 & \frac{1}{2} \cdot 8 \end{array}]$

Step 6.3

Move the negative in front of the fraction.

$[\begin{array}{c} 3 & - \frac{23}{2} \\ \frac{1}{2} \cdot - 2 & \frac{1}{2} \cdot 8 \end{array}]$

Step 6.4

Cancel the common factor of $2$ .

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Step 6.4.1

Factor $2$ out of $- 2$ .

$[\begin{array}{c} 3 & - \frac{23}{2} \\ \frac{1}{2} \cdot (2 (- 1)) & \frac{1}{2} \cdot 8 \end{array}]$

Step 6.4.2

Cancel the common factor.

$[\begin{array}{c} 3 & - \frac{23}{2} \\ \frac{1}{2} \cdot (2 \cdot - 1) & \frac{1}{2} \cdot 8 \end{array}]$

Step 6.4.3

Rewrite the expression.

$[\begin{array}{c} 3 & - \frac{23}{2} \\ - 1 & \frac{1}{2} \cdot 8 \end{array}]$

Step 6.5

Cancel the common factor of $2$ .

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Step 6.5.1

Factor $2$ out of $8$ .

$[\begin{array}{c} 3 & - \frac{23}{2} \\ - 1 & \frac{1}{2} \cdot (2 (4)) \end{array}]$

Step 6.5.2

Cancel the common factor.

$[\begin{array}{c} 3 & - \frac{23}{2} \\ - 1 & \frac{1}{2} \cdot (2 \cdot 4) \end{array}]$

Step 6.5.3

Rewrite the expression.

$[\begin{array}{c} 3 & - \frac{23}{2} \\ - 1 & 4 \end{array}]$