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Math Help - [SOLVED] Solving this matrix for C

  1. #1
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    [SOLVED] Solving this matrix for C

    Hi I am trying to solve NC=E to find C. I was thinking I need the inverse of N but before I do that I wanted to check that there is not an easier way to solve for C?

    Matrices are:

    5x2 matrix
    C=
    R1(c11,c12)
    R2(c21,c22)
    R3(c31,c32)
    R4(c41,c42)
    R5(c51,c52)

    5x2 matrix
    E=
    R1(43,7)
    R2(41,4)
    R3(29,5)
    R4(57,10)
    R5(20,6)

    5X5 Matrix
    N=
    R1(1,2,2,3,3)
    R2(1,1,3,2,2)
    R3(2,1,2,1,2)
    R4(1,3,3,2,1)
    R5(3,1,1,1,3)
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  2. #2
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    Quote Originally Posted by Neverquit View Post
    Hi I am trying to solve NC=E to find C. I was thinking I need the inverse of N but before I do that I wanted to check that there is not an easier way to solve for C?

    Matrices are:

    5x2 matrix
    C=
    R1(c11,c12)
    R2(c21,c22)
    R3(c31,c32)
    R4(c41,c42)
    R5(c51,c52)

    5x2 matrix
    E=
    R1(43,7)
    R2(41,4)
    R3(29,5)
    R4(57,10)
    R5(20,6)

    5X5 Matrix
    N=
    R1(1,2,2,3,3)
    R2(1,1,3,2,2)
    R3(2,1,2,1,2)
    R4(1,3,3,2,1)
    R5(3,1,1,1,3)
    To make it easier to read, you are asked to solve

    \left[\begin{matrix}1 & 2 & 2 & 3 & 3 \\ 1 & 1 & 3 & 2 & 2 \\ 2 & 1 & 2 & 1 & 2 \\ 1 & 3 & 3 & 2 & 1 \\ 3 & 1 & 1 & 1 & 3\end{matrix}\right]\left[\begin{matrix}c_{11} & c_{12} \\ c_{21} & c_{22}\\ c_{31} & c_{32} \\ c_{41} & c_{42} \\ c_{51} & c_{52}\end{matrix}\right] = \left[\begin{matrix}43 & 7 \\ 41 & 4 \\ 29 & 5\\ 57 & 10 \\ 20 & 6 \end{matrix}\right]


    You can premultiply both sides by N^{-1}, or you can do row operations.

    Apply R_2 - R_1 \to R_2, R_3 - 2R_1 \to R_3, R_4 - R_1 \to R_4, R_5 - 3R_1 \to R_1

    \left[\begin{matrix}\phantom{-}1 & \phantom{-}2 & \phantom{-}2 & \phantom{-}3 & \phantom{-}3 \\ \phantom{-}0 & -1  & \phantom{-}1 & -1 & -1 \\ \phantom{-}0 & -3 & -2 & -5 & -4 \\ \phantom{-}0 & \phantom{-}1 & \phantom{-}1 & -1 & -2 \\ \phantom{-}0 & -5 & -5 & -8 & -6\end{matrix}\right]\left[\begin{matrix}c_{11} & c_{12} \\ c_{21}  & c_{22}\\ c_{31} & c_{32} \\ c_{41} & c_{42} \\ c_{51}  & c_{52}\end{matrix}\right] =\left[\begin{matrix}43 & \phantom{-}7 \\ -2 & -3 \\ -57 & -9 \\ \phantom{-}14 & \phantom{-}3 \\ -106 & -13\end{matrix}\right]

    Go from here to try and upper triangularise the system. Then you can evaluate C by solving and back substituting.
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