Linear Algebra Examples

$[\begin{array}{c} a & b \\ c & d \end{array}]$

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

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$a d - c b$

Step 3

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

Step 4

Substitute the known values into the formula for the inverse.

$\frac{1}{a d - c b} [\begin{array}{c} d & - b \\ - c & a \end{array}]$

Step 5

Multiply $\frac{1}{a d - c b}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{a d - c b} d & \frac{1}{a d - c b} (- b) \\ \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{array}]$

Step 6

Simplify each element in the matrix.

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Step 6.1

Combine $\frac{1}{a d - c b}$ and $d$ .

$[\begin{array}{c} \frac{d}{a d - c b} & \frac{1}{a d - c b} (- b) \\ \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{array}]$

Step 6.2

Rewrite using the commutative property of multiplication.

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{1}{a d - c b} b \\ \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{array}]$

Step 6.3

Combine $b$ and $\frac{1}{a d - c b}$ .

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \\ \frac{1}{a d - c b} (- c) & \frac{1}{a d - c b} a \end{array}]$

Step 6.4

Rewrite using the commutative property of multiplication.

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \\ - \frac{1}{a d - c b} c & \frac{1}{a d - c b} a \end{array}]$

Step 6.5

Combine $c$ and $\frac{1}{a d - c b}$ .

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \\ - \frac{c}{a d - c b} & \frac{1}{a d - c b} a \end{array}]$

Step 6.6

Combine $\frac{1}{a d - c b}$ and $a$ .

$[\begin{array}{c} \frac{d}{a d - c b} & - \frac{b}{a d - c b} \\ - \frac{c}{a d - c b} & \frac{a}{a d - c b} \end{array}]$