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I know that the identity matrix would look something like $\begin{pmatrix} 1 & 0 & 0 \\0 &1 & 0 \\ 0 & 0 &1 \end{pmatrix}$, which I know is non singular, but I am not sure how to go about proving that a square matrix is nonsingular if and only if it is row-equivalent to an identity matrix.

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  • $\begingroup$ Maybe you can start by considering the system $Ax = b$. Matrix being nonsingular $\Longleftrightarrow$ the system has unique solution $\Longleftrightarrow$ row-equivalent to an identity matrix $\endgroup$ – 3x89g2 Sep 16 '16 at 16:48
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Row reduction is the same as multiplying by elementary matrices (which are invertible).

So for any $A$ there is $B$, invertible so that $BA$ is reduced row echelon. If $BA$ is the identity then $BA=I$ and $B$ is the inverse for $A$. If $BA$ is not the identity then it has a zero row and thus $BA$ cannot be invertible and since $B$ is invertible $A$ cannot be invertible.

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