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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

1 Answer 1

up vote 2 down vote accepted

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.

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And the second quest? –  user506901 Nov 29 '12 at 14:17
    
@user506901: I don't understand it. Can you please ask specifically? What conditions should C satisfy so that what? –  Armen Tsirunyan Nov 29 '12 at 14:18
    
please see the edit. –  user506901 Nov 29 '12 at 15:38
    
@user506901: See my edit –  Armen Tsirunyan Nov 29 '12 at 16:41
    
Thanks. It would be good if you update your answer to clearly state that it can happen that the solution to the combined system has more solutions, beyond the one common to both 1) and 2). –  user506901 Nov 29 '12 at 17:35

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