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Math Help - Solving for a Matrix

  1. #1
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    Solving for a Matrix

    The question is: Which of the following matrices, when substitued for T, satisfies the matrix equation  T\begin{bmatrix}-3 \\ 5\end{bmatrix}= \begin{bmatrix}5 \\ 3\end{bmatrix}

    I know that the answer is the matrix \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix} but how do you obtain it without guessing/checking the other choices they had? I mean you can't divide both sides by T like any other equation, so I'm stuck..
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  2. #2
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    There are an infinite number of possibilities for T.

    For example: \begin{bmatrix}-5/3 & 0 \\ 0 & 3/5\end{bmatrix} or \begin{bmatrix}-2/3 & 3/5 \\ 0 & 3/5\end{bmatrix} or \begin{bmatrix}-2/3 & 3/5 \\ 2/3 & 1\end{bmatrix}

    Clearly, this only makes sense for a multiple choice question, where there is only one right answer.
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  3. #3
    Super Member Quacky's Avatar
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    "Which of the following matrices"

    What are the possible options? It's obviously intended as a multiple choice.
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  4. #4
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    Quote Originally Posted by mjoshua View Post
    The question is: Which of the following matrices, when substitued for T, satisfies the matrix equation  T\begin{bmatrix}-3 \\ 5\end{bmatrix}= \begin{bmatrix}5 \\ 3\end{bmatrix}

    I know that the answer is the matrix \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix} but how do you obtain it without guessing/checking the other choices they had? I mean you can't divide both sides by T like any other equation, so I'm stuck..
    Since T maps a 2-vector to a 2-vector it must be a 2 by 2 matrix- you can write it as T= \begin{bmatrix}a & b \\ c & d\end{bmatrix} and then the equation becomes
    \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}-3 \\ 5\end{bmatrix}= \begin{bmatrix}-3a+ 5b \\ -3c+ 5d\end{bmatrix}= \begin{bmatrix}5 \\ 3\end{bmatrix}.

    That is, you have two equations to solve for four unknown numbers. That is, as others have told you,
    \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}
    is NOT "the" solution- there are an infinite number of solutions. a= 0, b=1, c= -1, d= 0 do satisfy -3(0)+ 5(1)= 5 and -3(1)+ 5(0)= -3 but, for example, a= -5/3, b= 0, c= 0, d= 3/5 also satisfy those equations so
    \begin{bmatrix}-\frac{5}{3} & 0 \\ 0 & \frac{3}{5}\end{bmatrix}
    also satisfies the given matrix equation.
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  5. #5
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    Hi everyone,
    This is a practice ACT type question--all of the choices that were included consisted of different variations of 0s and 1s in the 2x2 matrix. So I guess they wanted us to guess and check? I guess I was hoping for a more algebraic way to solve it than by substitution of their choices and seeing which one worked..
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