Find matrix $B$ such that $BA=4A$ Find the unique $B \in \mathbb{R}^{3 \times 3}$ such that for every $A \in \mathbb{R}^{3 \times 3}$ we have 
i) $BA=4A$
ii) The 1, 2 and 3 rows of $BA$ are the 3,2 and 1 rows of A. 
This problem is in the section where they define matrix multiplication. My only idea was to set up a giant system of equations, but well...it was too giant. Is there a smart way to solve it?

Since each conditions works for every matrix $A\in \mathcal{M}_3(\mathbb{R})$, take $A=I$, then
i) $BA=4A$ becomes $BI=4I$, hence $B=4I$.
ii) Let $M$ be the matrix whose rows are the rows 3, 2, 1 of $I$. Then, $BA=BI=M$, therefore $B=M$.
