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I saw the statement and the reasoning in some pdf Can anybody please explain it to me.

statement - If (A| B) is row equivalent to (A' | B' ), then the two systems AX = B and A' X = B' have the same solutions.

Reasoning: Since ρ(P Q) = ρ(P)Q we have AX = B if and only ρ(A)X = ρ(B) for any composition ρ of finitely many elementary row operations.

Thank you.

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If $\left[A | B\right]$ is row equivalent to $\left[A' | B'\right]$. It means there exists some invertible matrices $P$ such that

$P\left[A | B\right]=\left[A' | B'\right] $

which is equivalent to $PA = A'$ and $PB = B'$.

Hence if $AX=B$, we can multiply $P$ to the equation from the left and obtain $PAX=PB$ which means $A'X=B'$

Conversely, if $A'X=B'$, again, we can multiply $P^{-1}$ from the left and obtain $AX=B$.

Hence the two systems are equivalent.

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    If [A|B][A|B] is row equivalent to [A′|B′][A′|B′]. It means there exists some invertible matrices PP such that P[A|B]=[A′|B′]. Could you please give a refrence for this property. All I know is that there exists set of elementary operations on rows using which we can obain one from another. I have no clue about the existence of that invertible matrix. Thankyou – 2017-01-07
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    Each elementary row operations correspond to an elementary matrix multiplication on the left which is known to be invertible. Product of multiple elementary matrices produces the invertible matrix. – 2017-01-07