Linear Algebra Examples

$[\begin{array}{c} 2 & 0 \\ 3 & 1 \end{array}]$

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

$2 \cdot 1 - 3 \cdot 0$

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 $2$ by $1$ .

$2 - 3 \cdot 0$

Step 2.2.1.2

Multiply $- 3$ by $0$ .

$2 + 0$

Step 2.2.2

Add $2$ and $0$ .

$2$

Step 3

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

Step 4

Substitute the known values into the formula for the inverse.

$\frac{1}{2} [\begin{array}{c} 1 & 0 \\ - 3 & 2 \end{array}]$

Step 5

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

$[\begin{array}{c} \frac{1}{2} \cdot 1 & \frac{1}{2} \cdot 0 \\ \frac{1}{2} \cdot - 3 & \frac{1}{2} \cdot 2 \end{array}]$

Step 6

Simplify each element in the matrix.

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Step 6.1

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

$[\begin{array}{c} \frac{1}{2} & \frac{1}{2} \cdot 0 \\ \frac{1}{2} \cdot - 3 & \frac{1}{2} \cdot 2 \end{array}]$

Step 6.2

Multiply $\frac{1}{2}$ by $0$ .

$[\begin{array}{c} \frac{1}{2} & 0 \\ \frac{1}{2} \cdot - 3 & \frac{1}{2} \cdot 2 \end{array}]$

Step 6.3

Combine $\frac{1}{2}$ and $- 3$ .

$[\begin{array}{c} \frac{1}{2} & 0 \\ \frac{- 3}{2} & \frac{1}{2} \cdot 2 \end{array}]$

Step 6.4

Move the negative in front of the fraction.

$[\begin{array}{c} \frac{1}{2} & 0 \\ - \frac{3}{2} & \frac{1}{2} \cdot 2 \end{array}]$

Step 6.5

Cancel the common factor of $2$ .

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Step 6.5.1

Cancel the common factor.

$[\begin{array}{c} \frac{1}{2} & 0 \\ - \frac{3}{2} & \frac{1}{2} \cdot 2 \end{array}]$

Step 6.5.2

Rewrite the expression.

$[\begin{array}{c} \frac{1}{2} & 0 \\ - \frac{3}{2} & 1 \end{array}]$

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