# Why do we need to ensure that equation $Ax=b$ has at most one solution for each value of $b$ for the matrix $A$ to have an inverse?

For every point to have a solution in $$Ax=b$$ the matrix must contain at least one set of $$m$$ linearly independent columns.

But I wonder why for the matrix to have an inverse, we additionally need to ensure that equation $$Ax=b$$ has at most one solution for each value of $$b$$. To do so, we need to make certain that the matrix has at most $$m$$ columns. Otherwise there is more than one way of parameterizing each solution.

I am neither very smart nor very quick to understand in math, don't hesitate to explain it to me with dumb examples as if I was a teenager, I would be very glad.

• If $Ax=Ay=b$ for $x \ne y$, then $A(x-y)=Az=0$ for $z=x-y \ne 0$, so $A$ cannot have an inverse, since it's not possible for $A^{-1}0=z\ne 0$.
– Joe
Sep 20, 2021 at 12:48
• First of all, is $A$ a $m\times n$-matrix ? If yes, note that it only can have an inverse if $m=n$ (otherwise $A$ has only a pseudoinverse). The first step only ensures the existence of a solution, but if $A$ is invertible, the solution is unique : $x=A^{-1}b$ Sep 20, 2021 at 12:54
• Generally, a function $f$ has a left inverse $f^{-1}_{\text{left}} \circ f(z)=z$ if and only if it's injective. Sep 20, 2021 at 13:33

Let's say the equation has two solutions for some $$b$$. That means there exist two different vectors $$x_1, x_2$$ such that $$Ax_1=b$$ and $$Ax_2=b$$.

Now, suppose the matrix $$A$$ has an inverse $$A^{-1}$$. The property of the inverse matrix must be that if $$A$$ maps the vector $$x$$ to $$y$$, then $$A^{-1}$$ maps the vector $$y$$ to $$x$$.

Well, we know that $$A$$ maps $$x_1$$ to $$b$$, so we know that $$A^{-1}$$ must map $$b$$ to $$x_1$$, right?

And we also know that $$A$$ maps $$x_2$$ to $$b$$, so we know that $$A^{-1}$$ must map $$b$$ to $$x_2$$, right?

Can you see the problem above?

A more rigorous proof would be to assume that $$x_1, x_2$$ are any two solutions of the equation $$Ax=b$$, and then note that if $$A$$ is invertible, then $$Ax_1=b$$ can be multiplied by the inverse to get $$A^{-1}Ax_2=A^{-1}b$$ which simplifies to $$A^{-1}b=x_2$$.

Similarly, $$Ax_2=b$$ means that $$A^{-1}b=x_2$$ which means that $$x_1,x_2$$ are identical.

$$x_1=A^{-1}b=x_2$$ so the two