Solving an augumented matrix (A|B) with same matrix coeff I am taking a linear algebra course at my university. We just have the first day of class but I am reading by myself, so I decided to solve some problems. I do not understand how to solve this problem:
"The two systems:
$a)$$2x + y  = 3\tag{1}$
  $4x + 3y= 5 \tag{2}$
$b)$ $2x + y  = -1\tag{3}$
   $4x + 3y= 1 \tag{4}$ 
have the same coefficient matrix but different right-hand sides. Solve both systems simultaneously by eliminating the $(2,1)$ entry of the augmented matrix
$$\Bigg(\begin{array}{cc|cc}
2 & 1 & 3 &-1\\ 
4 & 4 & 5 &-1\\
\end{array}\Bigg)
$$
and then performing back substitutions for each columns corresponding to the right hand sides."
What does the$(2,1)$ entry mean? Can someone explain me how to solve this?
I know that the problem must be easy, but I haven't seen an example yet.  
 A: The $(i,j)$ entry is the entry in row $i$, column $j$, so the $(2,1)$ entry is the entry in row $2$, column $1$. Starting with the matrix
$$\left[\begin{array}{rr|rr}
2&1&3&-1\\
4&4&5&-1
\end{array}\right]\;,$$
subtract twice the first row, which is $\begin{bmatrix}4&2&6&-2\end{bmatrix}$, from the second row to get
$$\left[\begin{array}{rr|rr}
2&1&3&-1\\
0&2&-1&0
\end{array}\right]\;.\tag{0}$$
This actually corresponds to two matrices, one for each of the linear systems:
$$\left[\begin{array}{rr|rr}
2&1&3\\
0&2&-1
\end{array}\right]\tag{1}$$
for the first system, and
$$\left[\begin{array}{rr|rr}
2&1&-1\\
0&2&0
\end{array}\right]\tag{2}$$
for the second. From the second row of $(1)$ I see that $2y=-1$, so $y=-\frac12$. The first row of $(1)$ corresponds to the equation $2x+y=3$; I now know that $y=-\frac12$, so this is $2x-\frac12=3$, $2x=\frac72$, and $x=\frac74$.
Similarly, the second row of $(2)$ corresponds to the equation $2y=0$, so $y=0$. The first row corresponds to $2x+y=-1$, and $y=0$, so $2x=-1$, and $x=-\frac12$. I now have the solutions to both of the original systems.
With a little practice you won’t have to split $(0)$ into $(1)$ and $(2)$; you’ll be able to do the same work directly from $(0)$.
A: Hint: Second row, first column (there is a $4$ there). To "eliminate" it, add $-2$ times the first (full) row to the second full row.   
