# True or false? The linear system of equations $Ax=b$ is uniquely solvable for every $b \in \mathbb{R}^{n}$ if and only if det(A) $\neq$ 0 [duplicate]

Let $n \in \mathbb{N}$, let $A \in \mathbb{R}^{n \times n}$

True or false? The linear system of equations for $Ax=b$ is uniquely solvable for every $b \in \mathbb{R}^{n}$ if and only if $\text{det(A)} \neq$ 0

I think the statement is true. Making $\text{det(A)} \neq 0$ should avoid us getting non-unique results such as $0=0$.

Remebering this and knowing $\text{det(A)} \neq 0$ gives linearly independent vectors, we can only get unique solution.

Edit: We don't know anything about $x$. If $x \neq 0$, then the statement will be true, right?

Please correct me if I'm wrong / unclear, these tasks will be asked in my exam and I want pass it :)

• Diagonalize your matrix . You know that $detA \ne 0$, so there are non-zero elements on diagonal . – openspace Mar 11 '17 at 11:17

If $det(A)\neq 0$ then A is invert-able, meaning there is $A^{-1}$. Using this we conclude:
$$Ax = b \Leftrightarrow x= A^{-1} b$$ Which gives us a unique solution.
On the other hand, if $Ax = b$ has a unique solution for every $b\in \mathbb{R}$, this means this is true for $b=\vec{0}$, meaning only one vector solves the equation
$$Ax = 0$$ This means that $dim(N(A))=0$ and $dim(C(A))=n$, making $A$ invert-able and $det(A)\neq 0$.
True. Because if you have $\det A \neq 0$, you can invert $A$. So $x= A^{-1}b$ and every equation $Ax=b$ has solution. The solution is unique because $A$ is a bijetion.