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The matrix $P$ is given as $\pmatrix{1& -1\\ 2&1}$

The fact that it asked for $2\times 2$ matrices which it implies there are other matrices, confused me. I know that $Q$ may be the identity matrix. I also tried to give $Q$ random unknown letters and equated it in $PQ = QP$ but I did not manage to work it out.

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    $\begingroup$ Two months, five questions asked: time's passed to write mathematics properly in this site. $\endgroup$ – DonAntonio Apr 22 '17 at 12:43
  • $\begingroup$ Weird that you couldn't work it out: it is a system of four linear equations in four variables. $\endgroup$ – user228113 Apr 22 '17 at 12:45
  • $\begingroup$ You can see, for instance, here for how to write mathematics on this site. See specifically this paragraph for how to write matrices. $\endgroup$ – Arthur Apr 22 '17 at 12:48
  • $\begingroup$ In this type of question it's clear that $Q=I$ is one solution, $Q=P$ is another and linear combinations of solutions are solutions so that $Q=aI+bP$ is a solution. The question remains: are these all the solutions? $\endgroup$ – Lord Shark the Unknown Apr 22 '17 at 12:58
  • $\begingroup$ @DonAntonio Yes I tried that process but I am left with just 2 equations and I got stuck in solving them. I am left with 2b=-c and a=d I am not sure how to continue to get numerical values $\endgroup$ – Nicole Tabone Apr 22 '17 at 13:04
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$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&\!-1\\2&1\end{pmatrix}=\begin{pmatrix}1&\!-1\\2&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\iff$$

$$\begin{pmatrix}a+2b&-a+b\\c+2d&-c+d\end{pmatrix}=\begin{pmatrix}a-c&b-d\\2a+c&2b+d\end{pmatrix}$$

Now solve the corresponding system of linear equations. For example, taking the $\;1,1\;$ entry, we get

$$2b=-c\;,\;\text{and taking the entry}\;1,2\;:\;\;\;a=d\;,\;\;etc.$$

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