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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.

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up vote 2 down vote accepted

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 $(0)$.

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Many Many thanks! – Edwardo Aug 24 '12 at 18:24
by the way, Did you know a good website in which I can learn linear algebra stuff? @Brian – Edwardo Aug 24 '12 at 18:26
@Edwardo: I’ve never really looked for one. You might try this site; I’ve not looked at these notes, but on the basis of the author’s calculus notes I’d expect them to be decent though not great. – Brian M. Scott Aug 24 '12 at 18:32

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.

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