How to prove that multiplying the equation $Ax = B$ from the left with invertible matrices doesn't change set of solutions of the given system ?
In the book of Linear Algebra in Action by Harry Dym, at page 21, it is given the above statement, and I don't know which actions would change the set of solution of the given system, so I basically stuck.
We discussed a proof already in the comments, so I'll post an answer. Most people would just write something like this: Suppose $M$ is invertible. Then $$ Ax = b \implies MAx = Mb \implies M^{-1} MAx = M^{-1} M b \implies Ax = b. \text{QED}. $$
But I'll also write out a proof in more detail. We are given an $m \times n$ matrix $A$ and an $m \times 1$ column vector $b$. We want to show that if $M$ is an invertible $m \times m$ matrix, then the set of all solutions to $Ax = b$ is equal to the set of all solutions to $MAx = Mb$.
So, let $S_1 = \{ x \mid Ax = b \}$ and let $S_2 = \{ x \mid MAx = Mb \}$. We'll show that $S_1 \subset S_2$ and also that $S_2 \subset S_1$, which will prove that $S_1 = S_2$.
First, let's show that $S_1 \subset S_2$. If $x \in S_1$, then $Ax = b$, so multiplying by $M$ we find that $MAx = Mb$. This shows that $x \in S_2$. So, $S_1 \subset S_2$.
Next, we'll show that $S_2 \subset S_1$. If $x \in S_2$, then $MA x = Mb$. Multiplying by $M^{-1}$, we find that $M^{-1} MAx = M^{-1} Mb$, which implies that $Ax = b$. This shows that $x \in S_1$. So, $S_2 \subset S_1$. This completes the proof that $S_1 = S_2$.