I have 4x4 matrices:

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


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

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.


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

  • $\begingroup$ 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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