# How do you find the inverse of A=((1, -1, 0), (1, 0, -1), (-6, 2, 3))?

##### 1 Answer
Jan 2, 2017

The answer is $= \left(\begin{matrix}- 2 & - 3 & - 1 \\ - 3 & - 3 & - 1 \\ - 2 & - 4 & - 1\end{matrix}\right)$

#### Explanation:

$A = \left(\begin{matrix}1 & - 1 & 0 \\ 1 & 0 & - 1 \\ - 6 & 2 & 3\end{matrix}\right)$

We start by calculating the determinant

$\det A = | \left(1 , - 1 , 0\right) , \left(1 , 0 , - 1\right) , \left(- 6 , 2 , 3\right) |$

$= 1 | \left(0 , - 1\right) , \left(2 , 3\right) | + 1 | \left(1 , - 1\right) , \left(- 6 , 3\right) | + 0 | \left(1 , 0\right) , \left(- 6 , 2\right) |$

$= \left(2\right) + \left(- 3\right) = - 1$

As $\det A = - 1$, the matrix is invertible

To determine the inverse ${A}^{- 1}$, we start by calculating the matrix of cofactors

$C = \left(\begin{matrix}| \left(0 - 1\right) & \left(2 3\right) | & - | \left(1 - 1\right) & \left(- 6 3\right) | & | \left(1 0\right) & \left(- 6 2\right) | \\ - | \left(- 1 0\right) & \left(2 3\right) | & | \left(1 0\right) & \left(- 6 3\right) | & - | \left(1 - 1\right) & \left(- 6 2\right) | \\ | \left(- 1 0\right) & \left(0 - 1\right) | & - | \left(1 0\right) & \left(1 - 1\right) | & | \left(1 - 1\right) & \left(1 0\right) |\end{matrix}\right)$

$= \left(\begin{matrix}2 & 3 & 2 \\ 3 & 3 & 4 \\ 1 & 1 & 1\end{matrix}\right)$

Now we calculate the traspose matrix of $C$

${C}^{T} = \left(\begin{matrix}2 & 3 & 1 \\ 3 & 3 & 1 \\ 2 & 4 & 1\end{matrix}\right)$

To obtain the inverse of $A$, we divide ${C}^{T}$ by the determinant

${A}^{-} 1 = {C}^{T} / \det \left(A\right)$

$= \left(\begin{matrix}- 2 & - 3 & - 1 \\ - 3 & - 3 & - 1 \\ - 2 & - 4 & - 1\end{matrix}\right)$

Verification

$A \cdot {A}^{-} 1 = \left(\begin{matrix}1 & - 1 & 0 \\ 1 & 0 & - 1 \\ - 6 & 2 & 3\end{matrix}\right) \cdot \left(\begin{matrix}- 2 & - 3 & - 1 \\ - 3 & - 3 & - 1 \\ - 2 & - 4 & - 1\end{matrix}\right)$

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