# How do you find the inverse of A=((1, 2, 0), (-2, -2, 1), (0, -1, -1))?

Oct 25, 2016

The inverse is

${A}^{- 1} = \left(\begin{matrix}- 3 & - 2 & - 2 \\ 2 & 1 & 1 \\ - 2 & - 1 & - 2\end{matrix}\right)$

#### Explanation:

We start by calculating the determinant of A
$\det A = 1 \cdot \left(2 + 1\right) - 2 \cdot \left(2\right) \cdot 0 = 3 - 4 = - 1$
The determinant of A is $\ne 0$ , so the matrix has an inverse.

$A = \left(\begin{matrix}1 & 2 & 0 \\ - 2 & - 2 & 1 \\ 0 & - 1 & - 1\end{matrix}\right)$

Now we calculate the 9 minors and cofactors

${M}_{11} = 3$$\textcolor{w h i t e}{a a a a a a a}$ ${C}_{11} = 3$
${M}_{21} = - 2$$\textcolor{w h i t e}{a a a a a}$ ${C}_{21} = 2$
${M}_{31} = 2$$\textcolor{w h i t e}{a a a a a a a}$ ${C}_{31} = 2$
${M}_{12} = 2$$\textcolor{w h i t e}{a a a a a a a}$ ${C}_{12} = - 2$
${M}_{22} = - 1$$\textcolor{w h i t e}{a a a a a}$ ${C}_{22} = - 1$
${M}_{32} = 1$$\textcolor{w h i t e}{a a a a a a a}$ ${C}_{32} = - 1$
${M}_{13} = 2$$\textcolor{w h i t e}{a a a a a a a}$ ${C}_{13} = 2$
${M}_{23} = - 1$$\textcolor{w h i t e}{a a a a a a}$ ${C}_{23} = - 1$
${M}_{33} = 2$$\textcolor{w h i t e}{a a a a a a a}$ ${C}_{33} = 2$

matrix of co factors = $\left(\begin{matrix}3 & - 2 & 2 \\ 2 & - 1 & 1 \\ 2 & - 1 & 2\end{matrix}\right)$

Then the transpose = $\left(\begin{matrix}3 & 2 & 2 \\ - 2 & - 1 & - 1 \\ 2 & 1 & 2\end{matrix}\right)$

Then we divide by the determinant to obtain the inverse

=$\left(\begin{matrix}- 3 & - 2 & - 2 \\ 2 & 1 & 1 \\ - 2 & - 1 & - 2\end{matrix}\right)$

Check by mltiplying the matrix and its inverse

$\left(\begin{matrix}- 3 & - 2 & - 2 \\ 2 & 1 & 1 \\ - 2 & - 1 & - 2\end{matrix}\right)$$\left(\begin{matrix}1 & 2 & 0 \\ - 2 & - 2 & 1 \\ 0 & - 1 & - 1\end{matrix}\right)$*$= \left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$