How to know if equation adds more info to a set of equations? I'm looking for a formalized way to be able to know whether adding an equation to a set of linear equations adds new information or not, does anyone know of a way?
For a really simple case, let's say I have these equations:
$A = 3\\
B+C = 4\\
D = 8
$
Obviously, the equation below doesn't add any new information:
$-A = -3$
But it gets less obvious when adding an equation like this:
$B + D = 6$
Or:
$C = 2$
When the list of equations is longer, it gets a lot less obvious to me as well:
$
A=3\\
B+C=4\\
D=8\\
C=4\\
F=3\\
E=12\\
F+G=4\\
H=1
$ 
If I want to add a new equation:
$D+E=7$
I can see that $D$ and $E$ are already fully specified, so this doesn't really add any new information, but a more complex expression would be harder for me to rule out.
Is there a formalized way to see whether or not adding a new equation will add more information?  Possibly by looking at the matrix form of the equations?
Here is the last group of equations:
$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}
* 
\begin{bmatrix}
A \\
B \\
C \\
D \\
E \\
F \\
G \\
H \\
\end{bmatrix}
=
\begin{bmatrix}
3 \\
4 \\
8 \\
4 \\
3 \\
12 \\
4 \\
1 \\
\end{bmatrix}
$
And here is the equation I wanted to add to the set:
$
\begin{bmatrix}
0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
\end{bmatrix}
* 
\begin{bmatrix}
A \\
B \\
C \\
D \\
E \\
F \\
G \\
H \\
\end{bmatrix}
=
7
$
Thanks for any help you can provide!
 A: Let's say you have a set of $n$ equations in $m$ variables.  You already know how to represent this set of equations as an $n \times m$ matrix $A$.
Now you want to augment that set of equations with one more.  This new, larger set of equations can be represented by an $(n+1) \times m$ matrix, $B$.
The question "Does my last equation add any information to the collection?" is equivalent to "Is the rank of $B$ larger than the rank of $A$?".  To answer this question, you need to:


*

*Perform row reduction on $A$ and count the number of pivots;

*Perform row reduction on $B$ and count the number of pivots;

*Compare one number to the other.

A: Given a set of equations,
$$
\begin{cases}
    A+B = 5\\
    2D = 3\\
    A + B + C + 3D = 10,
\end{cases}
$$
You can identify it with a matrix like this,
$$
\begin{bmatrix}
1 & 1 & 0 & 0 &  5 \\ 
0 & 0 & 0 & 2 &  3 \\ 
1 & 1 & 1 & 3 &  10 \\ 
\end{bmatrix}
$$
Notice that each column of the matrix corresponds to each variable A, B, C and D in order, the last column corresponds to the result of the equation, and each number in the rows (except for the results of the equations) correspond to the integer coefficient that is next to each variable. For example the first row corresponds to the equation $A+B = 5$ that is the same as $A + B + 0C + 0D = 5$ so we should write in the first row 1 1 0 0 5.
This association is very useful because the matrix has the same properties as the system of equations when you want to add equations is the same as adding rows, so you already know that when two equations have the same "information" then the can "cancel each other out". For example,
\begin{cases}
    Z+Y = 1\\
    -5Z-5Y=-5,\\
\end{cases}
Divide the second equation by 5,
\begin{cases}
    Z+Y = 1\\
    -Z-Y=-1\\
\end{cases}
Sum the first one to the last one,
\begin{cases}
    Z+Y = 1\\
    -Z-Y + Z+ Y=-1 + 1\\
\end{cases}
We get,
\begin{cases}
    Z+Y = 1\\
    0=0\\
\end{cases}
And we already know that so we end with, 
\begin{cases}
    Z+Y = 1\\
\end{cases}
But this method also works with matrices,
We associate a matrix to the system of equations and do the same "operations",
\begin{cases}
    Z+Y = 1\\
    -5Z-5Y=-5,\\
\end{cases}
$\hspace{87 mm} \equiv$
\begin{bmatrix}
1 & 1 & 1 \\ 
-5 & -5 & -5 \\ 
\end{bmatrix}
We divide the last row by 5,
\begin{bmatrix}
1 & 1 & 1 \\ 
-1 & -1 & -1 \\ 
\end{bmatrix}
the we sum the first row to the last row,
\begin{bmatrix}
1 & 1 & 1 \\ 
-1 + 1 & -1 + 1& -1 + 1\\ 
\end{bmatrix}
is the same as,
\begin{bmatrix}
1 & 1 & 1 \\ 
0 & 0& 0\\ 
\end{bmatrix}
then we just have that the associated system of equation is the 1x1 matrix:
\begin{bmatrix}
1 & 1 & 1 \\  
\end{bmatrix}
We showed that the last equation wasn't adding "new information" because we ended up with
\begin{cases}
    Z+Y = 1\\
\end{cases}
Finally, what we have done with the associated matrix is to check if its rows are linearly independent, then two equations have "different information" if and only if their "correspondent rows" are linearly independent in their associated matrix.
