Results 1 to 6 of 6

Math Help - Matrices

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    179

    Smile Matrices

    Let A=
    [ 2 3 ]
    [ 4 5 ],

    B=
    [ 5 5 ]
    [-2 1 ],

    C=
    [ 0 -1]
    [ 6 6 ]

    Find X if: A−1BX= C

    NOTE: A is one matrix , B is one and C is also one matrix. Sorry I don't know how to use latex. Any help will be appreciated. The 'A−1' in A−1BX= C is supposed to mean the multiplicative inverse of A.

    The answer is:
    1/15[-132 -144]
    ----[ 186 162 ]

    Note that I have also put "----" at left of the matrix to make the bottom numbers align the top ones. They do not mean anything.

    I want to see any working out. Thank you
    Last edited by Joker37; December 18th 2008 at 06:33 AM. Reason: The reveal the answer to question
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by Joker37 View Post
    Let A=
    [ 2 3 ]
    [ 4 5 ],

    B=
    [ 5 5 ]
    [-2 1 ],

    C=
    [ 0 -1]
    [ 6 6 ]

    Find X if: A−1BX= C

    NOTE: A is one matrix , B is one and C is also one matrix. Sorry I don't know how to use latex. Any help will be appreciated. The 'A−1' in A−1BX= C is supposed to mean the multiplicative inverse of A.
    Mutliply both sides by A, giving:

     A \times A^{-1} \times B \times X = C \times A

    Which gives:

     B \times X = C \times A

    Multiply both sides by the inverse of B

     B^{-1} \times B \times X = C \times A \times B^{-1}

    giving:

     X = C \times A \times B^{-1}

    And note that the inverse of matrix B is given by  B^{-1} = \frac{Adj(B)}{Det(B)}

    PS: for a 2x2 matrix:

    [a b]
    [c d]

    then the Adjacent (Adj) of that matrix is given by:

    [d -b]
    [-c a]

    And you should know that the determinant of said matrix is:

    ad-cb
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2008
    Posts
    179
    Sorry, I still don't understand how to get the answer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, Joker37!

    Let: . A\:=\:\begin{bmatrix}2&3\\4&5\end{bmatrix}\quad B \:=\:\begin{bmatrix}5&5\\\text{-}2&1\end{bmatrix} \quad C \:=\:\begin{bmatrix}0&\text{-}1\\6&6\end{bmatrix}

    Given: . A^{-1}BX \:=\:C.\qquad \text{Find X.}


    Answer: . \tfrac{1}{15}\begin{bmatrix}\text{-}132 & \text{-}144 \\ 186 & 162 \end{bmatrix} . . . . Why didn't they reduce?
    If you know the Fast Way to find a 2-by-2 inverse, it certainly saves time.

    Given: . M \:=\:\begin{bmatrix}a & b \\ c & d \end{bmatrix} . the inverse is: . M^{-1} \;=\;\frac{1}{ad-bc}\begin{bmatrix}d & \text{-}b \\ \text{-}c & a\end{bmatrix}

    In baby-talk:
    . . (1) Switch the two elements on the main diagonal,
    . . (2) Change the signs of the elements on the other diagonal,
    . . (3) Divide by the determinant of the matrix.


    We have: . A \:=\:\begin{bmatrix}2&3\\4&5\end{bmatrix}
    . . The determinant is: . \begin{vmatrix}2&3\\4&5\end{vmatrix} \:=\:(2)(5)-(3)(4) \:=\:-2
    . . The inverse is: . A^{-1} \;=\;\text{-}\tfrac{1}{2}\begin{bmatrix}5&\text{-}3\\\text{-}4&2\end{bmatrix}\;=\;\begin{bmatrix}\text{-}\frac{5}{2} & \frac{3}{2} \\ 2 & \text{-}1 \end{bmatrix}


    The equation is: . \overbrace{\begin{bmatrix}\text{-}\frac{5}{2} & \frac{3}{2}\\2&\text{-}1\end{bmatrix}}^{A^{-1}}\cdot\overbrace{\begin{bmatrix}5&5\\\text{-}2&1\end{bmatrix}}^B\cdot X \;=\;\overbrace{\begin{bmatrix}0&\text{-}1\\6&6\end{bmatrix}}^C

    . . . . . . . . . . . . . . . \underbrace{\begin{bmatrix}\text{-}\frac{31}{2}&\text{-}11\\ 12 & 9\end{bmatrix}}_D\cdot X \;=\;\begin{bmatrix}0 & \text{-}1 \\ 6&6 \end{bmatrix}\;\;{\color{blue}[1]}


    Then we find that: . D^{-1} \:=\:\begin{bmatrix}\text{-}\frac{6}{5} & \text{-}\frac{22}{15} \\ \\[-4mm] \frac{8}{5} & \frac{31}{15}\end{bmatrix}


    Left-multiply both side of [1] by D^{-1}

    . . \begin{bmatrix}\text{-}\frac{6}{5}&\text{-}\frac{22}{15} \\ \frac{8}{5}&\frac{31}{15}\end{bmatrix}\cdot\begin{  bmatrix}\text{-}\frac{31}{2} & \text{-}11 \\ 12 & 9 \end{bmatrix}\cdot X \;=\;\begin{bmatrix}\text{-}\frac{6}{5} & \text{-}\frac{22}{15} \\ \frac{8}{5} & \frac{31}{5}\end{bmatrix}\cdot \begin{bmatrix}0&\text{-}1\\6&6\end{bmatrix}

    . . . . . . . . . . . . . . . X \;=\;\begin{bmatrix}\text{-}\frac{44}{5} & \text{-}\frac{38}{5} \\ \\[-4mm] \frac{62}{5} & \frac{54}{5} \end{bmatrix} \quad\hdots . . . There!

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2008
    Posts
    179
    Quote Originally Posted by Soroban View Post

    Then we find that: . D^{-1} \:=\:\begin{bmatrix}\text{-}\frac{6}{5} & \text{-}\frac{22}{15} \\ \\[-4mm] \frac{8}{5} & \frac{31}{15}\end{bmatrix}
    How did you get this? It looks strange with all the fractions.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, Joker37!

    Use the Fast Way that I explained . . . or find it yourself.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 25th 2010, 06:34 PM
  2. Total matrices and Commutative matrices in GL(r,Zn)
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: August 16th 2010, 02:11 AM
  3. Matrices Help
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 24th 2009, 09:36 PM
  4. Matrices represented by Symmetric/Skew Symmetric Matrices
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: October 25th 2008, 05:06 PM
  5. matrices
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 13th 2007, 02:36 PM

Search Tags


/mathhelpforum @mathhelpforum