Precalculus Examples

$[\begin{array}{c} 2 & x \\ x & x^{2} \end{array}]$

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

$2 x^{2} - x \cdot x$

Step 2.2

Simplify the determinant.

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Step 2.2.1

Rewrite.

$1 x^{2}$

Step 2.2.2

Multiply $x^{2}$ by $1$ .

$x^{2}$

Step 3

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

Step 4

Substitute the known values into the formula for the inverse.

$\frac{1}{x^{2}} [\begin{array}{c} x^{2} & - x \\ - x & 2 \end{array}]$

Step 5

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

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

Step 6

Simplify each element in the matrix.

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Step 6.1

Cancel the common factor of $x^{2}$ .

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Step 6.1.1

Cancel the common factor.

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

Step 6.1.2

Rewrite the expression.

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

Step 6.2

Rewrite using the commutative property of multiplication.

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

Step 6.3

Cancel the common factor of $x$ .

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Step 6.3.1

Move the leading negative in $- \frac{1}{x^{2}}$ into the numerator.

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

Step 6.3.2

Factor $x$ out of $x^{2}$ .

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

Step 6.3.3

Cancel the common factor.

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

Step 6.3.4

Rewrite the expression.

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

Step 6.4

Move the negative in front of the fraction.

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

Step 6.5

Rewrite using the commutative property of multiplication.

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

Step 6.6

Cancel the common factor of $x$ .

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Step 6.6.1

Move the leading negative in $- \frac{1}{x^{2}}$ into the numerator.

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

Step 6.6.2

Factor $x$ out of $x^{2}$ .

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

Step 6.6.3

Cancel the common factor.

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

Step 6.6.4

Rewrite the expression.

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

Step 6.7

Move the negative in front of the fraction.

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

Step 6.8

Combine $\frac{1}{x^{2}}$ and $2$ .

$[\begin{array}{c} 1 & - \frac{1}{x} \\ - \frac{1}{x} & \frac{2}{x^{2}} \end{array}]$