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This is from Artin Algebra, proposition 1.2.20:

Proposition: Let $A$ be a square matrix that has either a left inverse or a right inverse, a matrix $B$ such that either $BA=I$ or $AB=I$. Then $A$ is invertible and $B$ is its inverse.

Proof: Suppose $AB=I$. We perform row reduction on $A$. Say $A'=PA$, where $P=E_k\cdots E_1$ is the product of corresponding elementary matrices, and $A'$ is a row echelon matrix. Then $A'B=PAB=P$. Because $P$ is invertible, its bottom row is not zero. Then bottom row of $A'$ can't be zero either. $\dots$

I can't understand how bottom row of $A'$ can't be zero.

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    $\begingroup$ What will be the determinant value of the matrix if a row/column is zero at any point of time? $\endgroup$ – Aditya Nov 3 '17 at 6:52
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Suppose the bottom row of $A'$ is zero. Then, carrying out the matrix multiplication, you will see that the bottom row of $A'B$ is zero too.

If you're comfortable with block matrix partitioning, suppose

$$ A' = \begin{bmatrix} A'_1 \\ 0\end{bmatrix}, B = \begin{bmatrix} B_1 & B_2 \end{bmatrix} $$

Then

$$ A'B = \begin{bmatrix} A'_1 \\ 0\end{bmatrix}\begin{bmatrix} B_1 & B_2 \end{bmatrix} = \begin{bmatrix} A_1' B_1 & A_1' B_2 \\ 0 & 0 \end{bmatrix} $$

Which has a row of zeros.

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  • $\begingroup$ Oh thank you. Now I feel silly. $\endgroup$ – Silent Nov 3 '17 at 6:52
  • $\begingroup$ @Silent Don't feel silly at all. Block partitioning is a very powerful (but under-taught in undergraduate classes in my experience) tool in matrix algebra/analysis. $\endgroup$ – eepperly16 Nov 3 '17 at 6:55
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    $\begingroup$ There's an ocw course on linear Algebra.. ocw.mit.edu/courses/mathematics/… $\endgroup$ – Aditya Nov 3 '17 at 6:55
  • $\begingroup$ @eepperly16, no I meant that this should have clicked to me! $\endgroup$ – Silent Nov 3 '17 at 6:58

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