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I have 4x4 matrices:

$$A\begin{bmatrix} 3&1&3&-4\\ 6&4&8&10\\ 3&2&5&-1\\ -9&5&-2&-4 \end{bmatrix} = \begin{bmatrix} 0&1&0&0\\ 0&0&0&1\\ 0&0&1&0\\ 1&0&0&0 \end{bmatrix} $$

and

$$\begin{bmatrix} 3&1&3&-4\\ 6&4&8&10\\ 3&2&5&-1\\ -9&5&-2&-4 \end{bmatrix} B = \begin{bmatrix} 0&1&0&0\\ 0&0&0&1\\ 0&0&1&0\\ 1&0&0&0 \end{bmatrix}$$

and I need to find $A^{-1}$ and $B^{-1}$.

Now, there's the long way of doing it (augmented matrix with identity on the right side and solving for A, then finding A inverse, B, then B inverse), but I thought this was too much work and there must be an efficient way of doing it.

I've noticed that the right side of the equation for both A[] and []B look like they are just another arrangement of identity matrix, which must provide some useful clue to this problem.

I have a gut feeling that we must do something with elementary row matrices but don't have a concrete idea to get me started..

Can anyone give me some insight or introduce efficient way of solving this question? (Without determinant or long mechanical computation)

Thank you.

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The right hand matrix is known as permutation matrix and its inverse is its transpose, i.e. $$ P^{-1}=\begin{bmatrix} 0&1&0&0\\ 0&0&0&1\\ 0&0&1&0\\ 1&0&0&0 \end{bmatrix}^T=\begin{bmatrix} 0&0&0&1\\ 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0 \end{bmatrix}=[e_2\: e_4\: e_3 \:e_1]=\begin{bmatrix}e_4'\\e_1'\\e_3'\\e_2'\end{bmatrix} $$ where $e_i$ is elementary column vector and $e_i'$ is elementary row vector. Also $Ae_i$ is the $i$th column of $A$ and $e_i'A$ is the $i$th row of $A$.

So $$ A^{-1}=\begin{bmatrix} 3&1&3&-4\\ 6&4&8&10\\ 3&2&5&-1\\ -9&5&-2&-4 \end{bmatrix}[e_2\: e_4\: e_3 \:e_1]=\begin{bmatrix} 1&-4&3&3\\ 4&10&8&6\\ 2&-1&5&3\\ 5&-4&-2&-9 \end{bmatrix} $$ And $$ B^{-1}=\begin{bmatrix}e_4'\\e_1'\\e_3'\\e_2'\end{bmatrix}\begin{bmatrix} 3&1&3&-4\\ 6&4&8&10\\ 3&2&5&-1\\ -9&5&-2&-4 \end{bmatrix}=\begin{bmatrix} -9&5&-2&-4\\ 3&1&3&-4\\ 3&2&5&-1\\ 6&4&8&10 \end{bmatrix} $$

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  • $\begingroup$ Thank you! I was missing the property of permutation matrix, which you reminded me. $\endgroup$ – Corp. and Ltd. Nov 4 '15 at 7:04

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