Find the Inverse A=[[27,-9],[-15,24]]

$A = [\begin{array}{c} 27 & - 9 \\ - 15 & 24 \end{array}]$

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

$27 \cdot 24 - (- 15 \cdot - 9)$

Simplify the determinant.

Simplify each term.

Multiply $27$ by $24$ .

$648 - (- 15 \cdot - 9)$

Multiply $- (- 15 \cdot - 9)$ .

Multiply $- 15$ by $- 9$ .

$648 - 1 \cdot 135$

Multiply $- 1$ by $135$ .

$648 - 135$

Subtract $135$ from $648$ .

$513$

Step 3

Substitute the known values into the formula for the inverse.

$\frac{1}{513} [\begin{array}{c} 24 & 9 \\ 15 & 27 \end{array}]$

Step 4

Multiply $\frac{1}{513}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{513} \cdot 24 & \frac{1}{513} \cdot 9 \\ \frac{1}{513} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Step 5

Simplify each element in the matrix.

Cancel the common factor of $3$ .

Factor $3$ out of $513$ .

$[\begin{array}{c} \frac{1}{3 (171)} \cdot 24 & \frac{1}{513} \cdot 9 \\ \frac{1}{513} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Factor $3$ out of $24$ .

$[\begin{array}{c} \frac{1}{3 \cdot 171} \cdot (3 \cdot 8) & \frac{1}{513} \cdot 9 \\ \frac{1}{513} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} \frac{1}{3 \cdot 171} \cdot (3 \cdot 8) & \frac{1}{513} \cdot 9 \\ \frac{1}{513} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} \frac{1}{171} \cdot 8 & \frac{1}{513} \cdot 9 \\ \frac{1}{513} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Combine $\frac{1}{171}$ and $8$ .

$[\begin{array}{c} \frac{8}{171} & \frac{1}{513} \cdot 9 \\ \frac{1}{513} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Cancel the common factor of $9$ .

Factor $9$ out of $513$ .

$[\begin{array}{c} \frac{8}{171} & \frac{1}{9 (57)} \cdot 9 \\ \frac{1}{513} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} \frac{8}{171} & \frac{1}{9 \cdot 57} \cdot 9 \\ \frac{1}{513} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} \frac{8}{171} & \frac{1}{57} \\ \frac{1}{513} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Cancel the common factor of $3$ .

Factor $3$ out of $513$ .

$[\begin{array}{c} \frac{8}{171} & \frac{1}{57} \\ \frac{1}{3 (171)} \cdot 15 & \frac{1}{513} \cdot 27 \end{array}]$

Factor $3$ out of $15$ .

$[\begin{array}{c} \frac{8}{171} & \frac{1}{57} \\ \frac{1}{3 \cdot 171} \cdot (3 \cdot 5) & \frac{1}{513} \cdot 27 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} \frac{8}{171} & \frac{1}{57} \\ \frac{1}{3 \cdot 171} \cdot (3 \cdot 5) & \frac{1}{513} \cdot 27 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} \frac{8}{171} & \frac{1}{57} \\ \frac{1}{171} \cdot 5 & \frac{1}{513} \cdot 27 \end{array}]$

Combine $\frac{1}{171}$ and $5$ .

$[\begin{array}{c} \frac{8}{171} & \frac{1}{57} \\ \frac{5}{171} & \frac{1}{513} \cdot 27 \end{array}]$

Cancel the common factor of $27$ .

Factor $27$ out of $513$ .

$[\begin{array}{c} \frac{8}{171} & \frac{1}{57} \\ \frac{5}{171} & \frac{1}{27 (19)} \cdot 27 \end{array}]$

Cancel the common factor.

$[\begin{array}{c} \frac{8}{171} & \frac{1}{57} \\ \frac{5}{171} & \frac{1}{27 \cdot 19} \cdot 27 \end{array}]$

Rewrite the expression.

$[\begin{array}{c} \frac{8}{171} & \frac{1}{57} \\ \frac{5}{171} & \frac{1}{19} \end{array}]$

Find the Inverse A=[[27,-9],[-15,24]]

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