Usually when we want to find the inverse we row reduce a matrix along with the identity matrix on the right side until we're done and the inverse would be the one on the right side.
I'm not sure about how to find the inverse of this one though as the right side doesn't look like identity matrix. What approach should I use to find the inverse?
UPDATE: This is Full Question:
suppose A and B are 4 by 4 matrices such that
A=\begin{bmatrix}3 & 1 & 3 & -4\\ 6 & 4 & 8 & 10 \\ 3 & 2 & 5 & -1\\-9 & 5 & -2 & -2\end{bmatrix} =\begin{bmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\1 & 0 & 0 & 0\end{bmatrix}
And
\begin{bmatrix}3 & 1 & 3 & -4\\ 6 & 4 & 8 & 10 \\ 3 & 2 & 5 & -1\\-9 & 5 & -2 & -2\end{bmatrix} B=\begin{bmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\1 & 0 & 0 & 0\end{bmatrix}
a) find A inverse
b) find B inverse
This is confusing cause they look the same
Best Answer
From your attached picture it looks like you should have $$AK = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$$ and $$KB = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$$ where $$K = \begin{bmatrix} 3 & 1 & 3 & -4 \\ 6 & 4 & 8 & 10 \\3 & 2 & 5 & -1 \\-9 & 5 & -2 & -2 \end{bmatrix}$$ (Not that A = K = B.) So if you can find a matrix $C$ such that $$C \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} = I_4$$ then $$(CK)B = C \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} = I_4$$ so $CK = B^{-1}$. And since $$C \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}C = I_4$$ $$A(KC) = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}C = I_4$$ and $KC = A^{-1}$.