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Let $A$ be $n\times n$ matrix and there is at least one non trivial solution for the system $Ax=0$.

For any real column vector $b$ with n components,the equation $Ax=b$ has:

$1.$Unique solution.

$2.$Finite many solutions.

$3.$Infinitely many solutions.

$4.$No solution.

I think first and second options are false.

I have problem about the third and fourth options and I think both could be true but I am not sure.

What you will suggest the answer?

Thanks

user163993
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  • Take $b=0$, you know $Ax=0$ has at least one non-trivial solution (thus it has infinitely many...), hence (4) is excluded... – Milly Oct 22 '14 at 21:03

1 Answers1

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You're correct: $Ax = b$ will either have infinitely many solutions or no solutions. For example, take $n = 2$ and consider: $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$ Notice that not all columns of $A$ contain a pivot, so $Ax = 0$ has (infinitely) many (nontrivial) solutions. Yet if: $$ b = \begin{bmatrix} 7 \\ 1 \end{bmatrix} $$ then $Ax = b$ has no solution. On the other hand, if: $$ b = \begin{bmatrix} 7 \\ 0 \end{bmatrix} $$ then $Ax = b$ has infinitely many solutions.

Adriano
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