Finding the Solutions of the two systems by using the inverse. I am having a difficult time understanding where I went wrong with the following:
$$\begin{matrix}4x-y = 1 \\ 2x+3y = 3 \end{matrix} $$
$$\begin{matrix}4x-y = -3 \\ 2x+3y = 3 \end{matrix} $$
I found the inverse of the common coefficient matrix of the systems:
$$A^{-1} \begin{cases} \frac3{14}, \frac1{14} \\ \\ -\frac17, \frac27 \end{cases} $$
The issue is this question:
Find the solutions to the two systems by using the inverse, i.e. by evaluating $A^{-1}B$  where $B$ represents the right hand side (i.e. $B = \begin{bmatrix}1 \\ 3 \end{bmatrix}$  for system (a) and $B = \begin{bmatrix}-3 \\ 3 \end{bmatrix}$   for system (b)). 
Now whatever I find for x and y for both solutions keep coming as wrong. I am thinking, I might of read the question wrongly....I am using the negative values to find the x and y. Not too sure how to go at this now...
 A: $A^{-1} = \frac{1}{14}\begin{bmatrix} 3 & 1 \\ -2 & 4\end{bmatrix}$.
$A^{-1} \binom{1}{3} = \frac{1}{14} \binom{6}{10} = \frac{1}{7} \binom{3}{5}$.
$A^{-1} \binom{-3}{3} = \frac{1}{14} \binom{-6}{18} = \frac{1}{7} \binom{-3}{9}$.
A: You can rewrite the given equations by defining the coefficient matrix A, variable matrix X, and constant matrix B as
$$
A =\left[\begin{matrix}
4&-1\\
2&3
\end{matrix}\right]
\\
X = \left[\begin{matrix}
x\\
y
\end{matrix}\right]
\\
B=\left[\begin{matrix}
1&-3\\
3&3
\end{matrix}\right]
$$
Then, your two systems can be represented simply by $$ AX = B, $$ and to solve the system, you can left multiply both sides by the inverse of A, which is 
$$
A^{-1} = \left[\begin{matrix}
\frac {3}{14}&\frac {1}{14}\\
-\frac {1}{7}&\frac {2}{7}
\end{matrix}\right]
$$
This gives 
$$
A^{-1}AX=A^{-1}B\\
X = \left[\begin{matrix}
\frac {3}{7}&\frac {-3}{7}\\
-\frac {5}{7}&\frac {9}{7}
\end{matrix}\right]
$$
So $(x,y) = (\frac3 7,\frac5 7)$ for the first equation, and $(x,y) = (-\frac37,\frac97)$ for the second.
