Determinants and uniqueness of a solution

Question

I have a question here which says:

"If det$(A)\neq0,$then $Ax=x$ has a unique solution."

The $x$ on the right side isn't a typo. It's not supposed to be $b.$

I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.

I'm not really sure how to go about "proving" this.

I do notice that $Ax-x=0$ and thus $x(A-I)=0$. However, I can't really see where to go from there.

Answer 1

It's clearly false. Take

$A = I; \tag 1$

then

$\det(A) = 1 \ne 0, \tag 2$

but

$\forall x, \; Ax = Ix = x; \tag 3$

the solution to

$Ax = x \tag 4$

is certainly not unique!

This example shows that the proposed hypothetical is not true, but why? I think the most concise reason I can give is that $\det(A) \ne 0$ does not preclude $\det (A - I) = 0$; if

$\det(A - I) \ne 0, \tag 5$

then

$(A - I)x = 0 \tag 6$

has the unique solution $0$; but when $\det(A - I)$ vanishes, in general we will have

$\ker (A - I) \ne \{0\}; \tag 7$

indeed, $\dim(\ker(A - I))$ may be as great as $\text{size}(A)$, as the above example indicates.