if $det(A) \neq 0$ for a square matrix $A$, then a unique solution always exists for the system $Ax = b$ for any vector $b$.
proof: let, $A_{n \times n}$ be a square matrix, having $det(A) \neq 0$.
now, let $b$ be any $n$-dimensional vector.
we want to prove that there exists a vector $x$ such that $Ax = b$.
since, $det(A) \neq 0$, we know that $A$ is invertible.
let, $A^{-1}$ be the inverse of $A$.
consider the vector $x = A^{-1}b$.
we will show that this $x$ is a solution to $Ax = b$
$x = A^{-1}b$ ...............................(1)
multiplying $A$ both sides,
$Ax = A(A^{-1}b)$
$Ax = (AA^{-1})b)$ (by associativity of matrix multiplication)
$Ax = Ib$ (since $AA^{-1} = I$, where $I$ is the identity matrix)
$Ax = b$ (since $Ib = b$ for any vector $b$)
therefore, we have found an $x$ (namely, $A^{-1}b$) such that $Ax = b$.
this $x$ exists for any arbitrary vector $b$ because $A^{-1}$ exists (due to $det(A) \neq 0$) and can be applied to any vector $b$.
moreover, this solution is unique.
If there were another solution $y$ such that $Ay = b$, then:
$Ay = b$
multiplying $A^{-1}$ both sides,
$A^{-1}Ay = A^{-1}b$
$Iy = A^{-1}b$
$y = A^{-1}b$ ..............................(2)
from equation (1) and (2),
$x = y$
hence, the solution $x$ is unique for any $b$ in $Ax = b$.
Thus, we have proved that when $det(A) \neq 0$, for any vector $b$, there always exists a unique solution $x$ to the equation $Ax = b$, and this solution is given by $x = A^{-1}b$.
