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.

  • 2
    $\begingroup$ 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 $A^{-1}0=z\ne 0$. $\endgroup$
    – Joe
    Sep 20, 2021 at 12:48
  • $\begingroup$ 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 : $x=A^{-1}b$ $\endgroup$
    – Peter
    Sep 20, 2021 at 12:54
  • $\begingroup$ Generally, a function $f$ has a left inverse $f^{-1}_{\text{left}} \circ f(z)=z$ if and only if it's injective. $\endgroup$
    – William
    Sep 20, 2021 at 13:33

1 Answer 1


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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .