# Thread: Find the particular matrix

1. ## Find the particular matrix

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!

2. Originally Posted by 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!

All you've done so far is correct! Now you need to check the last two conditions:

$\displaystyle Px=-x\,,\,Py=4y\Longrightarrow \begin{pmatrix}a&b\\4b&a+3b\end{pmatrix}\begin{pma trix}u\\\!\!\!-u\end{pmatrix}=\begin{pmatrix}\!\!\!-u\\u\end{pmatrix}$ $\displaystyle \,,\,\begin{pmatrix}a&b\\4b&a+3b\end{pmatrix}\begi n{pmatrix}s\\4s\end{pmatrix}=\begin{pmatrix}4s\\16 s\end{pmatrix}$ . You'll get a linear system of two equations in $\displaystyle a,\,b$ with a unique solution.

Tonio