# 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...

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$A^{-1} = \frac{1}{14}\begin{bmatrix} 3 & 1 \\ -2 & 4\end{bmatrix}$.

$A^{-1} \binom{1}{3} = \frac{1}{14} \binom{6}{10} = \frac{1}{7} \binom{3}{5}$.

$A^{-1} \binom{-3}{3} = \frac{1}{14} \binom{-6}{18} = \frac{1}{7} \binom{-3}{9}$.

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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.

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