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**arze** Find the most general form of the matrix **P** if **PQ=QP**, where **Q** is $\displaystyle \left(\begin{array}{cc}2&1\\4&5\end{array}\right)$.

If the non-zero column vectors **x **and **y** are such the **Qx=x** and **Qy=6y**, obtain the particular matrix **P** such that **PQ=QP**, **Px=-x**, **Py=4y**.

I have found the most general form of **P** to be $\displaystyle \left(\begin{array}{cc}a&b\\4b&{3b+a}\end{array}\r ight)$

I named **x** and **y** as $\displaystyle \left(\begin{array}{cc}u\\v\end{array}\right)$ and $\displaystyle \left(\begin{array}{cc}s\\t\end{array}\right)$ respectively. Then I found **Qx** and **Qy**

they are $\displaystyle \left(\begin{array}{cc}{2u+v}\\{4u+5v}\end{array}\ right)$ and $\displaystyle \left(\begin{array}{cc}{2s+t}\\{4s+5t}\end{array}\ right)$ respectively.

so i get the equations:

$\displaystyle 2u+v=u$__1

$\displaystyle 2s+t=6s$__2

which simplify to $\displaystyle u=-v$ and $\displaystyle s=\frac{t}{4}$

So **x**=$\displaystyle \left(\begin{array}{cc}u\\-u\end{array}\right)$

and **y**=$\displaystyle \left(\begin{array}{cc}s\\4s\end{array}\right)$

Now I don't know what to do next.

Thanks!