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Q: If $A$ is a $3 \times 3$ matrix and $y$ is a vector in $\mathbb{R}^3$ such that $Ax = y$ does not have a solution, then there exists no vector $z$ in $\mathbb{R}^3$ such that the equation $Ax = z$ has a unique solution.


This is true right? If $A$ doesn't span $\mathbb{R}^3$ then the only possibility is either no solution or infinite solutions on an $\mathbb{R}^2$ plane?

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You're right. Do you need to prove your argument? –  Johannes Kloos Sep 19 '12 at 7:48
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up vote 1 down vote accepted

You have the right idea: it is indeed true, but the case of infinite solutions might be a line or all of $\Bbb R^3$ rather than a plane. It all depends on the rank of $A$. If there is a vector $y$ such that $Ax=y$ has no solution, then the rank of $A$ is less than $3$.

  • If it’s $2$, the null space is a straight line through the origin, the range is a plane through the origin, and $Ax=y$ has a straight line’s worth of solutions if $y$ is in the range and none if it is not.

  • If it’s $1$, the null space is a plane through the origin, the range is a line through the origin, and $Ax=y$ has a plane’s worth of solutions if $y$ is in the range and none if it is not.

  • If it’s $0$, $A$ is the zero matrix, the null space is all of $\Bbb R^3$, and the only $y$ for which $Ax=y$ has a solution is $y=0$, in which case every $x\in\Bbb R^3$ is a solution.

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Thanks for the clarification Brian. –  Dirk Calloway Sep 19 '12 at 13:12
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