Linear Algebra Examples

[\begin{matrix} 2 & 3 1 & 1 \end{matrix}]

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

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

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

$2 \cdot 1 - 1 \cdot 3$

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

$1$ .

$2 - 1 \cdot 3$

Step 2.2.1.2

Multiply

$- 1$ by

$3$ .

$2 - 3$

Step 2.2.2

Subtract

$3$ from

$2$ .

$- 1$

Step 3

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

Step 4

Substitute the known values into the formula for the inverse.

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

Step 5

Divide

$1$ by

$- 1$ .

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

Step 6

Multiply

$- 1$ by each element of the matrix.

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

Step 7

Simplify each element in the matrix.

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Step 7.1

Multiply

$- 1$ by

$1$ .

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

Step 7.2

Multiply

$- 1$ by

$- 3$ .

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

Step 7.3

Multiply

$- 1$ by

$- 1$ .

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

Step 7.4

Multiply

$- 1$ by

$2$ .

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