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I am a little confused by this question the matrix A is given as, $$ \begin{pmatrix} 2&1\\ 5&3 \end{pmatrix} $$ Matrix B $$ \begin{pmatrix} 2&4\\ 1&3 \end{pmatrix} $$

I found AX=B, but YA=B wouldn't that give me the same answer ? I dont understand why would i get a different answer.

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In general $X\neq Y$ where $AX=B$ and $YA=B$ since matrix multiplcation is not commutative. Since $\det A$ and $\det B$ are not zero you can compute $X$ and $Y$ by $$ X= A^{-1} B \qquad \text{and} \qquad Y= BA^{-1}.$$ You will see that $ A^{-1} B\neq BA^{-1}$ what shows $X\neq Y$.

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    @Julian answered my question correctly as well but since you have the exact answer ill select this , yea in my mind I was thinking of matrix being commutative. – 2017-02-10
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    You can accept whatever you like :) We just want to help you and make things clear! – 2017-02-10
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It's not about getting a different matrix $B$, but about the concept that the multiplication of matrices is not commutative, so it is not necessarily $AX = XA$. You are obligated to find a different matrix $Y$ such that $AX = YA$, but that doesn't mean that $X = Y$. Since you already solved $AX = B$, try to calculate $XA$, and you should find that $AX \neq XA$.

This tells you that, in order to solve the excercise, you need to define $Y$ differently.