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Math Help - Matrix multiplication/addition problem

  1. #1
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    Matrix multiplication/addition problem

    Hi, I need to solve for a matrix X in the equation below:

    \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X + \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right]

    What I tried to do was first minus the \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] to the other side, then I took the inverse of \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right] and multiplied that on both sides. I thought that taking the inverse and multiplying with it would have a division type of effect. Unfortunately I didn't get the correct answer. Would anyone mind telling me what I've done wrong?

    Thanks!

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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cb220 View Post
    Hi, I need to solve for a matrix X in the equation below:

    \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X + \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right]

    What I tried to do was first minus the \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] to the other side, then I took the inverse of \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right] and multiplied that on both sides. I thought that taking the inverse and multiplying with it would have a division type of effect. Unfortunately I didn't get the correct answer. Would anyone mind telling me what I've done wrong?

    Thanks!

    What did you get for the inverse of \begin{bmatrix}-7 & -6\\ -9 & 5\end{bmatrix} ??
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  3. #3
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    I got:

      \begin{bmatrix}-5/19 & -6/19\\ -9/19 & -7/19\end{bmatrix}
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by cb220 View Post
    I got:

      \begin{bmatrix}-5/19 & -6/19\\ -9/19 & -7/19\end{bmatrix}
    I don't get those values...

    I get \begin{bmatrix}-5/89 & -6/89\\ -9/89 & 7/89\end{bmatrix} for the inverse matrix.
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  5. #5
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    Hmm, throwing it into a matrix calculator online seems to give these values:

    -0.263157894736842 -0.315789473684210
    -0.473684210526316 -0.368421052631579

    Those are consistent with my values, unfortunately after trying to solve the problem with both my own values and your values, I still got the answer wrong in both cases .

    I think I must be missing something here *frustration*!
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  6. #6
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    Quote Originally Posted by cb220 View Post
    Hmm, throwing it into a matrix calculator online seems to give these values:

    -0.263157894736842 -0.315789473684210
    -0.473684210526316 -0.368421052631579

    Those are consistent with my values, unfortunately after trying to solve the problem with both my own values and your values, I still got the answer wrong in both cases .

    I think I must be missing something here *frustration*!
    Why don't you show us exactly what you have right now, step by step. Then we can pick out any mistakes that you might be making.

    -Dan
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  7. #7
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    Okay, starting with:

    \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X + \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right]

    Moving over the second matrix on the left side to the right side:

    \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right] - \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right]

    Should simplify to:

    \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X = \left[ \begin{array}{cc} -6 & 3 \\ 1 & 1 \end{array} \right]

    Then I took the inverse of the matrix attached to X and multiplied both sides by it, I'll use the solution that I arrived at for the inverse, though I don't know whether it is right or not at this point:

    X = \left[ \begin{array}{cc} -6 & 3 \\ 1 & 1 \end{array} \right] * \left[ \begin{array}{cc} -5/19 & -6/19 \\ -9/19 & -7/19 \end{array} \right]

    And my final answer for X came out to be:

    X = \left[ \begin{array}{cc} -3/19 & 15/19 \\ -14/19 & -13/19 \end{array} \right]


    If you want me to expand on any of what I've done let me know.


    EDIT: Made a correction in my typing up of the math.
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  8. #8
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    Quote Originally Posted by cb220 View Post
    Hi, I need to solve for a matrix X in the equation below:

    \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X + \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] = \left[ \begin{array}{cc} -8 & 4 \\ 5 & 3 \end{array} \right]

    What I tried to do was first minus the \left[ \begin{array}{cc} -2 & 1 \\ 4 & 2 \end{array} \right] to the other side, then I took the inverse of \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right] and multiplied that on both sides. I thought that taking the inverse and multiplying with it would have a division type of effect. Unfortunately I didn't get the correct answer. Would anyone mind telling me what I've done wrong?

    Thanks!

    i get, as a first step:

    \left[ \begin{array}{cc} -7 & -6 \\ -9 & 5 \end{array} \right]X = \left[ \begin{array}{cc} -6 & 3 \\ 1 & 1 \end{array} \right]

    i concur with Chris L that the inverse of the first matrix is:

    \begin{bmatrix}-5/89 & -6/89\\ -9/89 & 7/89\end{bmatrix}

    applying this to the RHS after the first step checks out in the original equation. i suggest you double-check your arithmetic, i made several errors verifyng this myself.

    one more thing you need to be aware of: matrix multiplcation is NOT commutative.

    if AX = B, then X = A^-1B, NOT BA^-1.

    the determinant of your original matrix (the one you inverted) is (-7)(5) - (-6)(-9) = -35 - 54 = -89. by a well-known rule, to invert a 2x2 matrix, you swap the upper-left and lower-right corners, and change the sign of the other two entries (leaving them where they are), and then divide each one by the determinant.
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  9. #9
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    Ahh, I see now. I messed up on the inversion and somehow managed to even punch it into the calculator in such a way that produced the same wrong results. Also, I messed up with the order of multiplication and did BA^-1. Thanks for the help guys!
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