Algebra Examples

$[\begin{array}{c} 5 & x \\ 3 & 6 \end{array}]$

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

$5 \cdot 6 - 3 x$

Step 2.2

Multiply $5$ by $6$ .

$30 - 3 x$

Step 3

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

Step 4

Substitute the known values into the formula for the inverse.

$\frac{1}{30 - 3 x} [\begin{array}{c} 6 & - x \\ - 3 & 5 \end{array}]$

Step 5

Factor $3$ out of $30 - 3 x$ .

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Step 5.1

Factor $3$ out of $30$ .

$\frac{1}{3 (10) - 3 x} [\begin{array}{c} 6 & - x \\ - 3 & 5 \end{array}]$

Step 5.2

Factor $3$ out of $- 3 x$ .

$\frac{1}{3 (10) + 3 (- x)} [\begin{array}{c} 6 & - x \\ - 3 & 5 \end{array}]$

Step 5.3

Factor $3$ out of $3 (10) + 3 (- x)$ .

$\frac{1}{3 (10 - x)} [\begin{array}{c} 6 & - x \\ - 3 & 5 \end{array}]$

Step 6

Multiply $\frac{1}{3 (10 - x)}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{3 (10 - x)} \cdot 6 & \frac{1}{3 (10 - x)} (- x) \\ \frac{1}{3 (10 - x)} \cdot - 3 & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7

Simplify each element in the matrix.

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Step 7.1

Cancel the common factor of $3$ .

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Step 7.1.1

Factor $3$ out of $6$ .

$[\begin{array}{c} \frac{1}{3 (10 - x)} \cdot (3 (2)) & \frac{1}{3 (10 - x)} (- x) \\ \frac{1}{3 (10 - x)} \cdot - 3 & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.1.2

Cancel the common factor.

$[\begin{array}{c} \frac{1}{3 (10 - x)} \cdot (3 \cdot 2) & \frac{1}{3 (10 - x)} (- x) \\ \frac{1}{3 (10 - x)} \cdot - 3 & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.1.3

Rewrite the expression.

$[\begin{array}{c} \frac{1}{10 - x} \cdot 2 & \frac{1}{3 (10 - x)} (- x) \\ \frac{1}{3 (10 - x)} \cdot - 3 & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.2

Combine $\frac{1}{10 - x}$ and $2$ .

$[\begin{array}{c} \frac{2}{10 - x} & \frac{1}{3 (10 - x)} (- x) \\ \frac{1}{3 (10 - x)} \cdot - 3 & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.3

Rewrite using the commutative property of multiplication.

$[\begin{array}{c} \frac{2}{10 - x} & - \frac{1}{3 (10 - x)} x \\ \frac{1}{3 (10 - x)} \cdot - 3 & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.4

Combine $x$ and $\frac{1}{3 (10 - x)}$ .

$[\begin{array}{c} \frac{2}{10 - x} & - \frac{x}{3 (10 - x)} \\ \frac{1}{3 (10 - x)} \cdot - 3 & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.5

Cancel the common factor of $3$ .

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Step 7.5.1

Factor $3$ out of $- 3$ .

$[\begin{array}{c} \frac{2}{10 - x} & - \frac{x}{3 (10 - x)} \\ \frac{1}{3 (10 - x)} \cdot (3 (- 1)) & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.5.2

Cancel the common factor.

$[\begin{array}{c} \frac{2}{10 - x} & - \frac{x}{3 (10 - x)} \\ \frac{1}{3 (10 - x)} \cdot (3 \cdot - 1) & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.5.3

Rewrite the expression.

$[\begin{array}{c} \frac{2}{10 - x} & - \frac{x}{3 (10 - x)} \\ \frac{1}{10 - x} \cdot - 1 & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.6

Combine $\frac{1}{10 - x}$ and $- 1$ .

$[\begin{array}{c} \frac{2}{10 - x} & - \frac{x}{3 (10 - x)} \\ \frac{- 1}{10 - x} & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.7

Move the negative in front of the fraction.

$[\begin{array}{c} \frac{2}{10 - x} & - \frac{x}{3 (10 - x)} \\ - \frac{1}{10 - x} & \frac{1}{3 (10 - x)} \cdot 5 \end{array}]$

Step 7.8

Combine $\frac{1}{3 (10 - x)}$ and $5$ .

$[\begin{array}{c} \frac{2}{10 - x} & - \frac{x}{3 (10 - x)} \\ - \frac{1}{10 - x} & \frac{5}{3 (10 - x)} \end{array}]$