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In Hoffman and Kuntze's Linear Algebra, it is stated that given $2$ systems of equations $A$ and $B$ where each equation of $B$ is a linear combination of the equations of $A$, then "then every solution of $A$ is a solution of $B$. Of course it may happen that some solutions of $B$ are not solutions of $A$. This clearly does not happen if each equation in $A$ is a linear combination of the equations of $B$."

Now, my question: Is the only reason that $B$ might contain more solutions than $A$ because each equation in $B$ might have a $0$ for some specific equation of $A$? And thus could we equally well say that $A$ and $B$ have have exactly the same solution set if $B$ is a set of equations all of which are linear combinations of $A$ and such that every equation in $A$ is multiplied by a non-zero scalar in at least one of the equations of $B$?

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  • $\begingroup$ What if $A$ contains several equations and $B$ is simply the sum of those equations? Then the one equation in $B$ is a linear combination of those in $A$, and every equation in $A$ appears in this linear combination with a nonzero coefficient. But with a few examples, you'll quickly see that $B$ can have a larger solution set. $\endgroup$ – cws Jul 21 '14 at 1:35
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Think of it this way: Even if A and B have the same number of equations each, you might "lose" equations when you go from A to B -- for instance A has two equations, B also has two equations, one being the sum of the equations in A and the other being twice that same sum. The two equations in B are really equivalent, so you've "lost" one equation! And with fewer equations you have more solutions.

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