How do you find the inverse of "A"=((5, 8), (17, 3))?

Mar 1, 2016

${\text{A}}^{-} 1 = \left(\begin{matrix}- \frac{3}{121} & \frac{8}{121} \\ \frac{17}{121} & - \frac{5}{121}\end{matrix}\right)$

Explanation:

The inverse of the matrix

$\text{A} = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$

is equal to

"A"^-1=1/("det A")((d,-b),(-c,a))

Note that $\text{det A}$ is the determinant of $\text{A}$ and is equal to $a d - b c$, so

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

For the matrix

$\text{A} = \left(\begin{matrix}5 & 8 \\ 17 & 3\end{matrix}\right)$

we have

$\left\{\begin{matrix}a = 5 \\ b = 8 \\ c = 17 \\ d = 3\end{matrix}\right.$

so

${\text{A}}^{-} 1 = \frac{1}{5 \left(3\right) - 17 \left(8\right)} \left(\begin{matrix}3 & - 8 \\ - 17 & 5\end{matrix}\right)$

${\text{A}}^{-} 1 = - \frac{1}{121} \left(\begin{matrix}3 & - 8 \\ - 17 & 5\end{matrix}\right)$

${\text{A}}^{-} 1 = \left(\begin{matrix}- \frac{3}{121} & \frac{8}{121} \\ \frac{17}{121} & - \frac{5}{121}\end{matrix}\right)$