# How do you find the inverse of ((3, 7), (4, 17))?

Feb 22, 2017

Inverse: $\frac{1}{23} \left(\begin{matrix}17 & - 7 \\ - 4 & 3\end{matrix}\right) = \left(\begin{matrix}\frac{17}{23} & - \frac{7}{23} \\ - \frac{4}{23} & \frac{3}{23}\end{matrix}\right)$

#### Explanation:

Since we have a square matrix, you can use Cramer's rule for a $2 \times 2$ matrix where $A = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ :

${A}^{-} 1 = \frac{1}{a d - b c} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$

After we substitute: ${A}^{-} 1 = \frac{1}{3 \cdot 17 - 7 \cdot 4} \left(\begin{matrix}17 & - 7 \\ - 4 & 3\end{matrix}\right)$

So ${A}^{-} 1 = \frac{1}{23} \left(\begin{matrix}17 & - 7 \\ - 4 & 3\end{matrix}\right) = \left(\begin{matrix}\frac{17}{23} & - \frac{7}{23} \\ - \frac{4}{23} & \frac{3}{23}\end{matrix}\right)$

Feb 22, 2017

The inverse is $\left(\begin{matrix}\frac{17}{23} & - \frac{7}{23} \\ - \frac{4}{23} & \frac{3}{23}\end{matrix}\right)$

#### Explanation:

Let $A = \left(\begin{matrix}3 & 7 \\ 4 & 17\end{matrix}\right)$

To determine if matrix $A$ is invertible, we calculate the determinant

$\det A = | \left(3 , 7\right) , \left(4 , 17\right) | = - 3 \cdot 17 - \left(7\right) \left(4\right) = 51 - 28 = 23$

As, $\det A \ne 0$, the matrix is invertible

We start, by calculating the matrix of co-factors

$C = \left(\begin{matrix}17 & - 4 \\ - 7 & 3\end{matrix}\right)$

Then, we calculate the transpose of matrix $C$

${C}^{T} = \left(\begin{matrix}17 & - 7 \\ - 4 & 3\end{matrix}\right)$

And the inverse is

${A}^{-} 1 = \frac{1}{\det} A \cdot {C}^{T}$

$= - \frac{1}{23} \cdot \left(\begin{matrix}17 & - 7 \\ - 4 & 3\end{matrix}\right)$

$= \left(\begin{matrix}\frac{17}{23} & - \frac{7}{23} \\ - \frac{4}{23} & \frac{3}{23}\end{matrix}\right)$

Verification,

$A \cdot {A}^{-} 1 = \left(\begin{matrix}3 & 7 \\ 4 & 17\end{matrix}\right) \cdot \left(\begin{matrix}\frac{17}{23} & - \frac{7}{23} \\ - \frac{4}{23} & \frac{3}{23}\end{matrix}\right)$

$= \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$

$= I$