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Let $AX=Y$ be a non-homogeneous linear system where $A$ is an $n\times n$ matrix. If this system has a unique solution, can we deduce that $\det A\neq 0$ directly from cramer's rule?

Using the relation $x_i\det A = \det A_i$, if $\det A = 0$ then each $A_i=0$. Then I have no idea of what to do next.

I know that the existence of a non-trivial solution to the homogeneous system $AX=0$ implies that $\det A = 0$. But is the result even true in the non-homogeneous case?

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It is true. Not sure it is a good idea to use Cramer's rule because if you use it then you assume the determinant is non zero. Anyway, let's say $u$ is the only solution to the system $Ax=y$. Suppose $\det(A)=0$. Then the system $Ax=0$ has a non trivial solution, let's call it $z$. Now let $v=u-z$ which implies $z=u-v$. Since $z\ne 0$ we know that $u\ne v$. But now look what we get:

$0=Az=A(u-v)=Au-Av$

So $Av=Au=y$. Since $v\ne u$ we get that the system $Ax=y$ has two different solutions and this is a contradiction.

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