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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:

Better layout here

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

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  • $\begingroup$ What are your trying to do precisely? What are the roles of the two matrices? The second has no inverse. $\endgroup$ – quid Feb 29 '16 at 18:46
  • $\begingroup$ How did the second matrix turn into a $3\times4$ matrix? $\endgroup$ – GoodDeeds Feb 29 '16 at 18:48
  • $\begingroup$ oops i missed 1 row $\endgroup$ – 33ted Feb 29 '16 at 18:50
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    $\begingroup$ @33ted Why would you write an equal sign between two matrices which are obviously unequal? I bet your class uses some sort of augmented matrix instead. $\endgroup$ – rschwieb Feb 29 '16 at 18:52
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    $\begingroup$ I understand it now, there isn't an equal sign its A multiplied by this matrix yields the one on the right where as the roles are reversed in the second part. You just need to show it as multiplication instead of equality $\endgroup$ – Triatticus Feb 29 '16 at 19:39
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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 $KC = A^{-1}$.

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  • $\begingroup$ Thank you and sorry for the confusion. $\endgroup$ – 33ted Feb 29 '16 at 21:06

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