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This is not an assignment or anything from my book, just a pure interest. If I wanted to show this, wouldn't I have to say that the square matrix can be reduced into the identity square matrix and then say that the determinant is not zero. After I say that and use the fact that the only element in the kernel is the zero vector, could I conclude that the matrix is invertible?

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I would do it using the rank-nullity theorem and properties of linear maps. If $Ker(A) = {0}$ Then the linear map which the matrix represents is injective and has nullity zero. Hence the rank of the matrix is $n$ (if the matrix is $n\times n$,) which means that it is a surjective map. Since the map is linear and a bijection, it is an isomorphism which means it has an inverse. Hence the matrix must be invertible.

(Of course, to say this rigourously you'd have to explicitly state where the map is from and to and all the rest of it, but if you know a little bit about it it's not hard to do.)

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You don't need determinants for this.

Imagine doing Gauss-Jordan elimination on the matrix $A$. The result is a matrix $GA$ in reduced row echelon form, where $G$ is invertible. If $GA=I$, then $A$ is invertible (and $G$ is its inverse). Otherwise there is a non-pivot column, and you can use that to construct a nonzero column vector $X$ such that $(GA)X=0$. But since $G$ is invertible, $G(AX)=0$ implies $AX=0$ so $X$ is in the kernel of $A$, and the kernel is therefore nontrivial.

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