# How do you find the inverse of A=((-3, -3, -4), (0, 1, 1), (4, 3, 4))?

Feb 2, 2017

The answer is $= \left(\begin{matrix}1 & 0 & 1 \\ 4 & 4 & 3 \\ - 4 & - 3 & - 3\end{matrix}\right)$

#### Explanation:

We start by calculating the determinant of matrix $A$ to see if it's invertible

$\det A = | \left(- 3 , - 3 , - 4\right) , \left(0 , 1 , 1\right) , \left(4 , 3 , 4\right) |$

$= - 3 \cdot | \left(1 , 1\right) , \left(3 , 4\right) | + 3 \cdot | \left(0 , 1\right) , \left(4 , 4\right) | - 4 \cdot | \left(0 , 1\right) , \left(4 , 3\right) |$

$= \left(- 3 \cdot 1\right) + \left(3 \cdot - 4\right) - 4 \left(- 4\right)$

$= - 3 - 12 + 16 = 1$

Therefore,

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

We start by calculating the matrix of cofactors

$C = \left(\begin{matrix}| \left(1 1\right) & \left(3 4\right) | & - | \left(0 1\right) & \left(4 4\right) | & | \left(0 1\right) & \left(4 4\right) | \\ - | \left(- 3 - 4\right) & \left(3 4\right) | & | \left(- 3 - 4\right) & \left(4 4\right) | & - | \left(- 3 - 3\right) & \left(4 3\right) | \\ | \left(- 3 - 4\right) & \left(1 1\right) | & - | \left(- 3 - 4\right) & \left(0 1\right) | & | \left(- 3 - 3\right) & \left(0 1\right) |\end{matrix}\right)$

$C = \left(\begin{matrix}1 & 4 & - 4 \\ 0 & 4 & - 3 \\ 1 & 3 & - 3\end{matrix}\right)$

Now, we determine the transpose of $C$

${C}^{T} = \left(\begin{matrix}1 & 0 & 1 \\ 4 & 4 & 3 \\ - 4 & - 3 & - 3\end{matrix}\right)$

So,

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

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

Verification

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

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