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Determinants and inverses of matrices

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data
    P=\begin{pmatrix}3 & -1\\
    2 & 4
    \end{pmatrix}

    Q=\begin{pmatrix}4 & -1\\
    -2 & 1
    \end{pmatrix}

    R=\begin{pmatrix}3 & -3\\
    2 & 4
    \end{pmatrix}

    S=\begin{pmatrix}4 & 7\\
    9 & 1
    \end{pmatrix}

    PX = Q
    QY = R
    RZ = S

    Find Matrices X, Y, and Z.
    2. Relevant equations
    ad - bc = det

    How do I use things regarding the topic of determinants and inverses of matrices to solve this question? :confused:

    Cheers,
    Thorn
     
  2. jcsd
  3. Sep 8, 2012 #2

    ehild

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    If you multiply the equation PX=Q with the inverse of P from the left, you get :

    P-1PX=X=P-1Q.

    First find the inverse of P.

    ehild
     
    Last edited: Sep 8, 2012
  4. Sep 8, 2012 #3

    HallsofIvy

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    Though it is not necessary to actually find any determinants and inverse matrices. I would use "row reduction" instead. Have you studied row reduction yet and how have you learned to find determinants and inverses?
     
  5. Sep 8, 2012 #4
    Not yet, I'm still new to this topic. In fact, I am still rather rusty with matrix multiplication, although I am getting better at it.
     
  6. Sep 8, 2012 #5
    Is that a rule of some sort?
     
  7. Sep 8, 2012 #6

    HallsofIvy

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    It is pretty much the definition of "inverse"! If A is a matrix, [itex]A^{-1}[/itex] is its inverse if and only if [tex]A^{-1}A= AA^{-1}= I[/tex]

    If Ax= b then [itex]A^{-1}(Ax)= A^{-1}b[/itex], [itex](A^{-1}A)x= A^{-1}b[/itex] (associative law for matrix multiplication), [itex]Ix= x= A^{-1}b[/itex].
     
  8. Sep 8, 2012 #7
    Yes, it most certainly is,
    This is the rule of manipulating equations
    you have A=B, therefore (whatever I do to A)=(the same thing I do to B)
    So, if PX=Q, then multiplying the left side by P-1 will be still equal to multiplying the right side of the equation by also P-1
    hence, P-1.P.X=P-1.Q
    then, since P-1 is by definition such that P-1P=I (identity)
    then it follows that X(=P-1.P).X=P-1Q
    You would do he same thing if it was 'regular' algebra
    ax=y => x=y/a, , that would come from ax=y => ax/a=y/a => x=y/a, and you have to take the same care here: is a!=0 ? (for matrices, this will translate to, is a inversible ?) with the additional dificulty that multiplication is not commutative, so you must be careful with your handling of the equations
     
  9. Sep 8, 2012 #8

    ehild

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    Then find the matrix T[itex]=\begin{pmatrix}a & b\\
    c & d
    \end{pmatrix}[/itex] so as the product of T and P is the unit matrix U[itex]=\begin{pmatrix}1 & 0\\
    0 & 1 \end{pmatrix}[/itex]: TP=U.
    Show the multiplication in detail.
     
  10. Sep 8, 2012 #9

    HallsofIvy

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    Well, you probably will have it soon. It is a much simpler way to find inverses and solve systems, especially for very large matrices. Here these matrices are all "2 by 2", especially easy. You probably already know that for the matrix
    [tex]\begin{pmatrix}a & b \\ c & d\end{pmatrix}[/tex]
    the determinant is ad- bc. And its inverse matrix is
    [tex]\frac{1}{ad- bc}\begin{pmatrix}d & -b \\ -c & a\end{pmatrix}[/tex]
     
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