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?
 A: 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 $B=M$.
A: Taking them to be separate questions, 
You should start by looking at what $4A$ is equal to as an arbitrary matrix. Then, seeing what matrix can a multiply by in order to get this this product? In the second case, the same reasoning follows. See below. Let $A$ be the matrix with a standard representation. 


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*$4A = \begin{pmatrix} 4a_{11} &4a_{12} &4a_{13} \\ 4a_{21} &4a_{22} &4a_{23} \\ 4a_{31} &4a_{32} &4a_{33} \end{pmatrix} = \begin{pmatrix} 4&0 &0 \\0&4 &0 \\0&0 &4 \\\end{pmatrix} \begin{pmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \end{pmatrix} = BA$ so that $B =   \begin{pmatrix} 4&0 &0 \\0&4 &0 \\0&0 &4 \\\end{pmatrix}$. 

*$BA = \begin{pmatrix} a_{31} &a_{32} &a_{33} \\ a_{21} &a_{22} &a_{23} \\ a_{11} &a_{12} &a_{13} \\ \end{pmatrix}= \begin{pmatrix} 0&0&1\\ 0&1&0\\ 1&0&0 \end{pmatrix} \begin{pmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \end{pmatrix}\ $ so that $B = \begin{pmatrix} 0&0&1\\ 0&1&0\\ 1&0&0 \end{pmatrix} $. 
