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I understand that the invertibility theorem tells us that Ax=b has at least one solution for every b in R^n . I'm also aware that Ax=0 will have ONLY the trivial solution.

What is an example of an invertible matrix, A where there is more than one solution for a particular b?

Thanks.

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2 Answers

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What you've written isn't actually quite right. If $A$ is invertible, then $A^{-1}$ exists, and thus $$ Ax = b \iff A^{-1}b = A^{-1}Ax = x.$$ Thus this is the only solution, and it always exists. (Note that $b = 0$ implies $x=0$.)

If $A$ is not invertible, then $Ax = b$ may have no solutions, an exact solution ($x^*$) or an uncountably large family of solutions given by $$x = x^* + \lambda_1 v_1 + \cdots \lambda_r v_r,$$ for vectors $v_1,...,v_r$. (So the solution space then has dimension $r$, if it exists.)

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I think you got confused. If $A$ is invertible, then $Ax = b \implies x = A^{-1}b$, and so $x$ is unique and determined by the expression $A^{-1}b$. Now consider: $$\begin{cases} x-y = 0 \\ 2x-2y = 0\end{cases}$$Your matrix $A = \begin{bmatrix} 1 & -1 \\ 2 & -2\end{bmatrix}$ is not invertible and the system has the solution set $\{ (x,x) \in \Bbb R^2 \mid x \in \Bbb R \}$.

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