Find the Inverse A^-1=[[2,-3],[-7,11]]

$A^{- 1} = [\begin{array}{c} 2 & - 3 \\ - 7 & 11 \end{array}]$

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.

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

$2 \cdot 11 - (- 7 \cdot - 3)$

Simplify the determinant.

Simplify each term.

Multiply $2$ by $11$ .

$22 - (- 7 \cdot - 3)$

Multiply $- (- 7 \cdot - 3)$ .

Multiply $- 7$ by $- 3$ .

$22 - 1 \cdot 21$

Multiply $- 1$ by $21$ .

$22 - 21$

Subtract $21$ from $22$ .

$1$

Step 3

Substitute the known values into the formula for the inverse.

$\frac{1}{1} [\begin{array}{c} 11 & 3 \\ 7 & 2 \end{array}]$

Step 4

Divide $1$ by $1$ .

$1 [\begin{array}{c} 11 & 3 \\ 7 & 2 \end{array}]$

Step 5

Multiply $1$ by each element of the matrix.

$[\begin{array}{c} 1 \cdot 11 & 1 \cdot 3 \\ 1 \cdot 7 & 1 \cdot 2 \end{array}]$

Step 6

Simplify each element in the matrix.

Multiply $11$ by $1$ .

$[\begin{array}{c} 11 & 1 \cdot 3 \\ 1 \cdot 7 & 1 \cdot 2 \end{array}]$

Multiply $3$ by $1$ .

$[\begin{array}{c} 11 & 3 \\ 1 \cdot 7 & 1 \cdot 2 \end{array}]$

Multiply $7$ by $1$ .

$[\begin{array}{c} 11 & 3 \\ 7 & 1 \cdot 2 \end{array}]$

Multiply $2$ by $1$ .

$[\begin{array}{c} 11 & 3 \\ 7 & 2 \end{array}]$

Find the Inverse A^-1=[[2,-3],[-7,11]]

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