# How do you find the inverse of A=((-3, -1), (-7, 8))?

Mar 12, 2018

${A}^{- 1} =$$\left(\begin{matrix}- \frac{8}{31} & - \frac{1}{31} \\ - \frac{7}{31} & \frac{3}{31}\end{matrix}\right)$

#### Explanation:

$A =$$\left(\begin{matrix}- 3 & - 1 \\ - 7 & 8\end{matrix}\right)$

$C o f \left(A\right) =$$\left(\begin{matrix}8 & 7 \\ 1 & - 3\end{matrix}\right)$

$A \mathrm{dj} \left(A\right) =$$\left(\begin{matrix}8 & 1 \\ 7 & - 3\end{matrix}\right)$

$D e t \left(A\right) = \left(- 3\right) \cdot 8 - \left(- 1\right) \left(- 7\right) = - 31$

Thus, ${A}^{- 1} = \frac{A \mathrm{dj} \left(A\right)}{D e t \left(A\right)}$

${A}^{- 1} =$$\left(\begin{matrix}- \frac{8}{31} & - \frac{1}{31} \\ - \frac{7}{31} & \frac{3}{31}\end{matrix}\right)$

Jul 14, 2018

Answer: $\left[\begin{matrix}- \frac{8}{31} & - \frac{1}{31} \\ - \frac{7}{31} & \frac{3}{31}\end{matrix}\right]$

#### Explanation:

An easy way to find the inverse of a 2x2 square matrix is to apply the following formula*:
Let $A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$, then
${A}^{- 1} = \frac{1}{\det \left(A\right)} \left[\begin{matrix}d & - b \\ - c & a\end{matrix}\right]$

For this problem, we have
$A = \left[\begin{matrix}- 3 & - 1 \\ - 7 & 8\end{matrix}\right]$, so $a = - 3 , b = - 1 , c = - 7 , d = 8$

We can find the determinant:
$\det \left(A\right) = a d - b c = - 24 - 7 = - 31$

Plugging in our values, we have:
${A}^{- 1} = \frac{1}{- 31} \left[\begin{matrix}8 & 1 \\ 7 & - 3\end{matrix}\right] = \left[\begin{matrix}- \frac{8}{31} & - \frac{1}{31} \\ - \frac{7}{31} & \frac{3}{31}\end{matrix}\right]$, which is our answer

*Note: This formula only works for 2x2 matrices and does not work for larger matrices