# Thread: Solving for a Matrix

1. ## Solving for a Matrix

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

I know that the answer is the matrix $\displaystyle \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$ 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..

2. There are an infinite number of possibilities for T.

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

Clearly, this only makes sense for a multiple choice question, where there is only one right answer.

3. "Which of the following matrices"

What are the possible options? It's obviously intended as a multiple choice.

4. Originally Posted by mjoshua
The question is: Which of the following matrices, when substitued for T, satisfies the matrix equation $\displaystyle T\begin{bmatrix}-3 \\ 5\end{bmatrix}= \begin{bmatrix}5 \\ 3\end{bmatrix}$

I know that the answer is the matrix $\displaystyle \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$ 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 $\displaystyle T= \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and then the equation becomes
$\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}-3 \\ 5\end{bmatrix}= \begin{bmatrix}-3a+ 5b \\ -3c+ 5d\end{bmatrix}= \begin{bmatrix}5 \\ 3\end{bmatrix}$.

That is, you have two equations to solve for four unknown numbers. That is, as others have told you,
$\displaystyle \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$
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
$\displaystyle \begin{bmatrix}-\frac{5}{3} & 0 \\ 0 & \frac{3}{5}\end{bmatrix}$
also satisfies the given matrix equation.

5. 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..