invertible matrix that has one unique solution Let $A$ be such a 4x4 matrix that the equation $Ax = (1, 0, 0, 0)$ has a unique solution $x$. Is $A$ necessarily invertible?
I know that if A is invertible, the equation $Ax = y$ has for all $y \in \mathbb R^n$ a unique solution $x = A^{-1} y \in \mathbb R^n$. But there's only one unique solution, could someone talk me through this?
 A: Have you studied the algorithm to compute the general solution of a linear system? It should come quite early in a linear algebra course (for instance, in my course it came before the definition of a matrix inverse). It is an easy consequence of that algorithm that if a square matrix $A$ is not invertible then for each $b$ the equation $Ax=b$ has either no solutions or an infinite number of them (in $\mathbb{R}^n$).
A: Theorem: Let $A$ is a $4\times 4$ real matrix.  If $Ax=(1,0,0,0)$ has a unique solution then $A$ is invertible.
Proof: We'll do a proof by contrapositive.  In this case the contrapositive states: let $A$ is a $4\times 4$ real matrix.  If $A$ is not invertible then $Ax=(1,0,0,0)$ does not have a unique solution.
Let $A$ be a non-invertible $4\times 4$ real matrix.  If $(1,0,0,0)\not\in\operatorname{col}(A)$, then it's trivially true that $Ax=(1,0,0,0)$ does not have a unique solution.  So WLOG, let $A$ such that $(1,0,0,0) \in \operatorname{col}(A)$.  Then, by the matrix invertibility theorem, $A$ must have a non-trivial kernel.  That means that there must be at least one nonzero vector, let's call it $w$, such that $Aw=0$.  Because $(1,0,0,0)\in\operatorname{col}(A)$, there exists a vector $y$ such that $Ay=(1,0,0,0)$.  Then $y+w$ must also be a solution to $Ax=(1,0,0,0)$ because $$A(y+w) = Ay+Aw = (1,0,0,0) + (0,0,0,0) = (1,0,0,0)$$  Therefore $Ax=(1,0,0,0)$ does not have a unique solution.$\ \ \ \ \color{red}{\square}$
A: Yes it is.
Since $A$ is a 4 by 4 matrix, then A can be seen as a linear function $A: E \rightarrow F$ where $E,F$ are vector spaces with $dim(E)=dim(F)=4$
Because the Vector Spaces in consideration are finite dimensional and in fact have the same dimension one can reason as follows:
If A is not invertible (bijection) then A is not injective.
Because since $E$ and $F$ have the same finite dimension, one has that $A$ is invertible iff A is injective (or surjective)). So $dim$ $Ker(A) \geq 1$. 
Take $v \in ker(A)$ with $v \neq 0$, now since $x$ is a solution to $Ax = (1,0,0,0)$ then $A(x+v) = Ax + Av = Ax = (1,0,0,0)$ would also be a solution to your equation. But $x+v \neq x$ sice we took $v \neq 0$, so we have produced a solution to our original equation that is not $x$. This violates our assumption that $x$ is unique, so this means it cannot happen that $A$ is not invertible.
This proves that A must be invertible (bijection).
Hope that helps.
