Matrix and Equations:: The cost of $3$ copies and $5$ pens is Rs. $21$. The cost of $2$ copies and $3$ pens is Rs. $13$. Find the cost of each copy and pen using matrix method.
My Attempt:
Let the cost of each copy and each pen be $x$ and $y$ respectively.
Then
$$3x+5y=21$$
$$2x+3y=13$$
I have made two equations which can be easily solved by either substitution, elimination or graphical method. But the question asks to solve by matrix method. Please some body help me in solving with matrix method..
 A: The system is equivalent to the expression:
$$\begin{pmatrix}
3&5\\
2&3
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}=\begin{pmatrix}
21\\
13
\end{pmatrix}$$
And applying the inverse on both sides we get:
$$\begin{pmatrix}
x\\
y
\end{pmatrix}=\begin{pmatrix}
3&5\\
2&3
\end{pmatrix}^{-1}\begin{pmatrix}
21\\
13
\end{pmatrix}$$
A: You must reduce it to row-echoleon form.
These system of equations are equivalent to the augmented matrix:
\begin{bmatrix}
 3& 5 & |21\\ 
 2&3  & |13
\end{bmatrix}
Perform a series of row operations now.
\begin{bmatrix}
 1& \frac{5}{3} & |7\\
 2&3  & |13
\end{bmatrix}
$$\text{Divided the first row by 3. Now let's mess with the second row. We will do $R_2 = -2R_1 + R2$}$$
\begin{bmatrix}
 1& \frac{5}{3} & |7\\
 0&-\frac{1}{3}  & |-1
\end{bmatrix}
Now we have an equation! We have:
$$-\frac{1}{3}y = -1 \to y = 3$$
$$\text{Using substitution now, we can sub in $y=3$ in the first row}$$
$$1x + \frac{5}{3}y = 7 \to x + \frac{5}{3}*3 = 7 \to x = 2$$
So our solutions are: $$x=2, y=3$$
