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Math Help - Find the particular matrix

  1. #1
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    Find the particular matrix

    Find the most general form of the matrix P if PQ=QP, where Q is \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 \left(\begin{array}{cc}a&b\\4b&{3b+a}\end{array}\r  ight)

    I named x and y as \left(\begin{array}{cc}u\\v\end{array}\right) and \left(\begin{array}{cc}s\\t\end{array}\right) respectively. Then I found Qx and Qy
    they are \left(\begin{array}{cc}{2u+v}\\{4u+5v}\end{array}\  right) and \left(\begin{array}{cc}{2s+t}\\{4s+5t}\end{array}\  right) respectively.
    so i get the equations:
    2u+v=u__1
    2s+t=6s__2
    which simplify to u=-v and s=\frac{t}{4}
    So x= \left(\begin{array}{cc}u\\-u\end{array}\right)
    and y= \left(\begin{array}{cc}s\\4s\end{array}\right)
    Now I don't know what to do next.
    Thanks!
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  2. #2
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    Quote Originally Posted by arze View Post
    Find the most general form of the matrix P if PQ=QP, where Q is \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 \left(\begin{array}{cc}a&b\\4b&{3b+a}\end{array}\r  ight)

    I named x and y as \left(\begin{array}{cc}u\\v\end{array}\right) and \left(\begin{array}{cc}s\\t\end{array}\right) respectively. Then I found Qx and Qy
    they are \left(\begin{array}{cc}{2u+v}\\{4u+5v}\end{array}\  right) and \left(\begin{array}{cc}{2s+t}\\{4s+5t}\end{array}\  right) respectively.
    so i get the equations:
    2u+v=u__1
    2s+t=6s__2
    which simplify to u=-v and s=\frac{t}{4}
    So x= \left(\begin{array}{cc}u\\-u\end{array}\right)
    and y= \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:

    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} \,,\,\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 a,\,b with a unique solution.

    Tonio
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