# Rules on replacing a solution to one system by another

Suppose a linear system $$Ax=b \tag 1$$ is given, $A\in\mathbb{R}^{n\times n}$, $b\in\mathbb{R}^n$, and a solution exists. Now, suppose the system is multiplied from the left by some $C\in\mathbb{R}^{n\times n}$, resulting in $$CAx=Cb. \tag 2$$ A solution to $(1)$ is also a solution to $(2)$ (solving $(2)$ can be replaced by solving $(1)$), but the system $(2)$ might have more solutions than $(1)$, so solving $(1)$ cannot be replaced by solving $(2)$. Under which conditions on $C$ can solving $(1)$ be replaced by solving $(2)$?

What is with a solution to the system obtained by adding $(1)$ and $(2)$? Does it hold that a solution to the system obtained by summing $Ax=b$ and $Cx=d$, for some $A, C$ and $b, d$ (solutions exist, and it is known that the systems have at least one common solution), would yield a common solution (system (3) $(A+C)x=b+d$. In other words, can it happen that (3) has some solution that does not satisfy both $Ax=b$ and $Cx=d$?

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C has to be regular –  user127.0.0.1 Nov 29 '12 at 14:12

The condition you're looking for is that $C$ must have an inverse (meaning its determinant must not be equal to $0$). Such matrices are called invertible, nondegenerate, nonsingular, and, rarely, regular.
As per your second question, the answer is positive - it can happen that the solution to the combined system has more solutions, beyond the one common to both $(1)$ and $(2)$. For example, soppose $A$ is an invertible matrix and $C = -A$. Therefore, $Ax=b$ and $Cx=d$ have exactly one solution each(and they're the same solution, as per your condition). Note that $b+d$ is necessarily $0$. The equation $(A+C)x=b+d$, therefore, has infinitely many solutions.