Find the Inverse [[1,2],[1,3]]^-1

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

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.

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

$1 \cdot 3 - 1 \cdot 2$

Simplify the determinant.

Simplify each term.

Multiply $3$ by $1$ .

$3 - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$3 - 2$

Subtract $2$ from $3$ .

$1$

Step 3

Substitute the known values into the formula for the inverse.

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

Step 4

Divide $1$ by $1$ .

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

Step 5

Multiply $1$ by each element of the matrix.

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

Step 6

Simplify each element in the matrix.

Multiply $3$ by $1$ .

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

Multiply $- 2$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Multiply $1$ by $1$ .

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

Find the Inverse [[1,2],[1,3]]^-1

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