Results 1 to 6 of 6

Math Help - matrix inversion

  1. #1
    Member
    Joined
    May 2009
    Posts
    91

    Cool matrix inversion

    Hi guys,

    I am unable to get the answer to the following question: Find all the values of (x,y,z) which satisfy:

     \left( \begin{array}{ccc} <br />
7 & 2 & 4 \\ 4 & 1 & 2 \\ 3 & 1 & 1 \end{array} \right)<br />
\left( \begin{array}{c} x \\ y \\ z  \end{array} \right) =<br />
\left( \begin{array}{c} p \\ q \\ r \end{array} \right)<br />

    if (p,q,r) satisfies

     \left( \begin{array}{ccc} <br />
2 & -7 & 5 \\ 6 & -9 & -1 \\ -4 & 5 & 2 \end{array}  \right)<br />
\left( \begin{array}{c} p \\ q \\ r \end{array} \right) =<br />
\left( \begin{array}{c} 5 \\ 7 \\ -4  \end{array} \right)<br />

    So, we have:
    \mathbf{A}x = p

    detA = 7(-1) -2(-2) + 4(1) = 1. So it has an inverse. I used two methods to calculate the inverse, the adjoint and row reduction. Both gave the same result:

     A^{-1} =  \left( \begin{array}{ccc} <br />
 -1 & 2 & 0 \\ -2 & -5 & 1 \\ 0 & -1 & -1  \end{array}  \right)<br />

    which gives:
    x = 2q - p
    y = 2p - 5q + 2r
    z = p - q - r

    The second matrix B is singular i.e.

     det B = 2(-13) + 7(8) + 5(-6) = 0 But we have the following relationships:

    2p - 7q + 5r = 5
    6p - 9q - r = 7
    -4p + 5q + 2r = -4

    I messed around and got:

    p = 7x + 2y + 4z
    q = 4x + y + 2z
    r = 3x + y + z

    but this doesn't help. The answers are given in terms of \lambda! Am I missing something here?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by s_ingram View Post
    Hi guys,

    I am unable to get the answer to the following question: Find all the values of (x,y,z) which satisfy:

     \left( \begin{array}{ccc} <br />
7 & 2 & 4 \\ 4 & 1 & 2 \\ 3 & 1 & 1 \end{array} \right)<br />
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =<br />
\left( \begin{array}{c} p \\ q \\ r \end{array} \right)<br />

    if (p,q,r) satisfies

     \left( \begin{array}{ccc} <br />
2 & -7 & 5 \\ 6 & -9 & -1 \\ -4 & 5 & 2 \end{array} \right)<br />
\left( \begin{array}{c} p \\ q \\ r \end{array} \right) =<br />
\left( \begin{array}{c} 5 \\ 7 \\ -4 \end{array} \right)<br />

    So, we have:
    \mathbf{A}x = p

    detA = 7(-1) -2(-2) + 4(1) = 1. So it has an inverse. I used two methods to calculate the inverse, the adjoint and row reduction. Both gave the same result:

     A^{-1} = \left( \begin{array}{ccc} <br />
-1 & 2 & 0 \\ -2 & -5 & 1 \\ 0 & -1 & -1 \end{array} \right)<br />

    which gives:
    x = 2q - p
    y = 2p - 5q + 2r
    z = p - q - r

    The second matrix B is singular i.e.

     det B = 2(-13) + 7(8) + 5(-6) = 0 But we have the following relationships:

    2p - 7q + 5r = 5
    6p - 9q - r = 7
    -4p + 5q + 2r = -4

    I messed around and got:

    p = 7x + 2y + 4z
    q = 4x + y + 2z
    r = 3x + y + z

    but this doesn't help. The answers are given in terms of \lambda! Am I missing something here?
    It follows from the first two matrix equations that  \left( \begin{array}{ccc} <br />
2 & -7 & 5 \\ 6 & -9 & -1 \\ -4 & 5 & 2 \end{array} \right)<br />
\left( \begin{array}{ccc} <br />
7 & 2 & 4 \\ 4 & 1 & 2 \\ 3 & 1 & 1 \end{array} \right)<br />
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =<br />
\left( \begin{array}{c} 5 \\ 7 \\ -4 \end{array} \right)<br />
.

    Do the matrix multiplication and then solve the resulting system.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    Posts
    91
    Thanks for your reply, but.. when I do multiply the matrices I get:

    \left( \begin{array}{ccc} <br />
1 & 2 & -1 \\ 3 & 2 & 5 \\ -2  & -1 & -4 \end{array} \right)<br />
\left( \begin{array}{c} x \\ y  \\ z \end{array} \right) =<br />
\left( \begin{array}{c} 5 \\ 7 \\ -4  \end{array} \right)

    and the matrix is singular.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by s_ingram View Post
    Thanks for your reply, but.. when I do multiply the matrices I get:

    \left( \begin{array}{ccc} <br />
1 & 2 & -1 \\ 3 & 2 & 5 \\ -2 & -1 & -4 \end{array} \right)<br />
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =<br />
\left( \begin{array}{c} 5 \\ 7 \\ -4 \end{array} \right)

    and the matrix is singular.
    det = 0 => either no solution or infinite solutions. So, deal with this situation in the way you have been taught (refer to class notes or textbook if necessary). If you need more help, please say where exactly you're stuck.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1
    Quote Originally Posted by s_ingram View Post
    Hi guys,

    I am unable to get the answer to the following question: Find all the values of (x,y,z) which satisfy:

    A= \left( \begin{array}{ccc} <br />
7 & 2 & 4 \\ 4 & 1 & 2 \\ 3 & 1 & 1 \end{array} \right)<br />
\vec x= \left( \begin{array}{c} x \\ y \\ z  \end{array} \right)<br />
\vec {soln}= \left( \begin{array}{c} p \\ q \\ r \end{array} \right)<br />
    This may help you to check your work, or maybe on a test.

    A\vec x= \vec {soln}

    A^{-1}A\vec x = A^{-1}\vec {soln}

    \vec x = A^{-1}\vec {soln}

    Now via Texas-instrument

    [2nd] [matrix] then scroll over to edit.

    Create the two arrays A and \vec {soln}.

    Then, (A^{-1})*\vec {soln}. gives the desired result
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2009
    Posts
    91
    Thanks guys.

    I took the matrix:

    <br />
\left( \begin{array}{ccc} <br />
1 & 2 & -1 \\ 3 & 2 & 5 \\  -2 & -1 & -4 \end{array} \right)<br />
\left( \begin{array}{c} x  \\ y \\ z \end{array} \right) =<br />
\left( \begin{array}{c} 5 \\ 7 \\ -4  \end{array} \right)<br />

    Then using row reduction, I converted it into an equivalent triangular matrix:
    <br />
\left( \begin{array}{ccc} <br />
1 & 2 & -1 \\ 0 & -4 & 8 \\  0 & 0 & 0 \end{array} \right)<br />
\left( \begin{array}{c} x  \\ y \\ z \end{array} \right) =<br />
\left( \begin{array}{c} 5 \\ -8 \\ 0  \end{array} \right)<br />


    which gives a consistent result:

    x + 2y - z = 5 (i)
    -4y + 8z = -8 (ii)
    0z = 0 (iii)

    The last equation is true for any z, so I can set z = \lambda and everything is fine.

    Previously I made an arithmetic error so that the equations appeared inconsistent with no solutions. I checked the result once and then twice and then posted. I think the best advice when doing matrix questions is not to post after a late night. Try again in the morning and things are usually a lot clearer! Anyway, thanks again guys.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Matrix Inversion
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 5th 2011, 11:40 AM
  2. matrix inversion
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 8th 2010, 09:44 AM
  3. Matrix Inversion Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 11th 2010, 11:30 PM
  4. inversion of a matrix
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 18th 2009, 02:00 PM
  5. matrix inversion
    Posted in the Advanced Algebra Forum
    Replies: 26
    Last Post: January 26th 2008, 11:02 PM

Search Tags


/mathhelpforum @mathhelpforum