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Given two $n times n$ matrices $A$ and $B$, form a new block matrix

$$P := beginbmatrixI_n&B\-A&0endbmatrix$$

Then by using only elementary row operations, show that $P$ can be transformed into

$$P' := beginbmatrixI_n&B\0&ABendbmatrix $$

The solution to this problem is:

$$P = beginbmatrixI_n&B\-A&0endbmatrix sim beginbmatrixI_n&B\-A + AI_n &0 + ABendbmatrix sim beginbmatrixI_n&B\0&ABendbmatrix$$

I don't understand this solution. Why can $A$ be multiplied from the left on the first half of the matrix and then be added to the second half of the matrix to form a sequence of elementary row operations?

Let
$$
beginbmatrixR_1 \ R_2endbmatrix = beginbmatrixI_n&B\-A&0endbmatrix
$$

Then
$$
beginbmatrixR_1 \ R_2endbmatrix undersettextRowOpmapsto beginbmatrixR_1 \ R_2 + A R_1endbmatrix
implies
beginbmatrixI_n&B\-A&0endbmatrix
undersettextRowOpmapsto
beginbmatrixI_n&B\-A+AI_n& 0+ABendbmatrix
=
beginbmatrixI_n&B\0& ABendbmatrix
$$

Does this make it any clear?

I guess you're trying to prove Sylvester rank inequality.

This works just like the elementary row operations. We can do this with block matrices:

$$
M = beginpmatrixA&B\C&Dendpmatrix sim beginpmatrixA+K,C&B+K,D\C&Dendpmatrix
$$

What we are doing here behind the scenes is multiplying the matrix by an elementary matrix $T$:

$$
T = beginpmatrixI&K\0&Iendpmatrix
$$

So we get:

$$
TM =
beginpmatrixI&K\0&Iendpmatrix beginpmatrixA&B\C&Dendpmatrix = beginpmatrixA+K,C&B+K,D\C&Dendpmatrix
$$

As you can see, $T$ has full rank and therefore invertible, which means it doesn't change rank of the matrix:

$$operatornamerkTM = operatornamerkM$$