Find the Inverse [[1,7],[2,6]]

$[\begin{array}{c} 1 & 7 \\ 2 & 6 \end{array}]$

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

$1 \cdot 6 - 2 \cdot 7$

Simplify the determinant.

Simplify each term.

Multiply $6$ by $1$ .

$6 - 2 \cdot 7$

Multiply $- 2$ by $7$ .

$6 - 14$

Subtract $14$ from $6$ .

$- 8$

Step 3

Substitute the known values into the formula for the inverse.

$\frac{1}{- 8} [\begin{array}{c} 6 & - 7 \\ - 2 & 1 \end{array}]$

Step 4

Move the negative in front of the fraction.

$- \frac{1}{8} [\begin{array}{c} 6 & - 7 \\ - 2 & 1 \end{array}]$

Step 5

Multiply $- \frac{1}{8}$ by each element of the matrix.

$[\begin{array}{c} - \frac{1}{8} \cdot 6 & - \frac{1}{8} \cdot - 7 \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Step 6

Simplify each element in the matrix.

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$[\begin{array}{c} \frac{- 1}{8} \cdot 6 & - \frac{1}{8} \cdot - 7 \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Factor $2$ out of $8$ .

$[\begin{array}{c} \frac{- 1}{2 (4)} \cdot 6 & - \frac{1}{8} \cdot - 7 \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Factor $2$ out of $6$ .

$[\begin{array}{c} \frac{- 1}{2 \cdot 4} \cdot (2 \cdot 3) & - \frac{1}{8} \cdot - 7 \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} \frac{- 1}{2 \cdot 4} \cdot (2 \cdot 3) & - \frac{1}{8} \cdot - 7 \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} \frac{- 1}{4} \cdot 3 & - \frac{1}{8} \cdot - 7 \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Combine $\frac{- 1}{4}$ and $3$ .

$[\begin{array}{c} \frac{- 1 \cdot 3}{4} & - \frac{1}{8} \cdot - 7 \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Multiply $- 1$ by $3$ .

$[\begin{array}{c} \frac{- 3}{4} & - \frac{1}{8} \cdot - 7 \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Move the negative in front of the fraction.

$[\begin{array}{c} - \frac{3}{4} & - \frac{1}{8} \cdot - 7 \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Multiply $- \frac{1}{8} \cdot - 7$ .

Multiply $- 7$ by $- 1$ .

$[\begin{array}{c} - \frac{3}{4} & 7 (\frac{1}{8}) \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Combine $7$ and $\frac{1}{8}$ .

$[\begin{array}{c} - \frac{3}{4} & \frac{7}{8} \\ - \frac{1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$[\begin{array}{c} - \frac{3}{4} & \frac{7}{8} \\ \frac{- 1}{8} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Factor $2$ out of $8$ .

$[\begin{array}{c} - \frac{3}{4} & \frac{7}{8} \\ \frac{- 1}{2 (4)} \cdot - 2 & - \frac{1}{8} \cdot 1 \end{array}]$

Factor $2$ out of $- 2$ .

$[\begin{array}{c} - \frac{3}{4} & \frac{7}{8} \\ \frac{- 1}{2 \cdot 4} \cdot (2 \cdot - 1) & - \frac{1}{8} \cdot 1 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} - \frac{3}{4} & \frac{7}{8} \\ \frac{- 1}{2 \cdot 4} \cdot (2 \cdot - 1) & - \frac{1}{8} \cdot 1 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} - \frac{3}{4} & \frac{7}{8} \\ \frac{- 1}{4} \cdot - 1 & - \frac{1}{8} \cdot 1 \end{array}]$

Combine $\frac{- 1}{4}$ and $- 1$ .

$[\begin{array}{c} - \frac{3}{4} & \frac{7}{8} \\ \frac{- - 1}{4} & - \frac{1}{8} \cdot 1 \end{array}]$

Multiply $- 1$ by $- 1$ .

$[\begin{array}{c} - \frac{3}{4} & \frac{7}{8} \\ \frac{1}{4} & - \frac{1}{8} \cdot 1 \end{array}]$

Multiply $- 1$ by $1$ .

$[\begin{array}{c} - \frac{3}{4} & \frac{7}{8} \\ \frac{1}{4} & - \frac{1}{8} \end{array}]$

Find the Inverse [[1,7],[2,6]]

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