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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?

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    If $BA = 4A$, doesn't that mean by ii) that the 1st, 2nd, and 3rd rows of $4A$ are the 3rd, 2nd, and 1st rows of $A$. That does not hold for all matrices $A \in \mathbb{R}^{3 \times 3}$. – 2017-02-26
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    For $A=I$ we have that $B=4I$, and the second condition does not hold. – 2017-02-26
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    The OP must mean those as separate questions with different answers. – 2017-02-26
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    They must refer to different B's. Should I post them as separate questions? – 2017-02-26
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    I suggest, as hinted to by ajotatxe, for each question, to take $A=I$ (since the conditions should work "for each $A$"), and see where it leads you. It is straightforward in both cases – 2017-02-26
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    Your title has $AB$ and the question has $BA$ – 2017-02-26

2 Answers 2

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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 $B=M$.

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    I think this is really slick, and much better than computing explicitly. – 2017-02-26
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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.

  1. $4A = \begin{pmatrix} 4a_{11} &4a_{12} &4a_{13} \\ 4a_{21} &4a_{22} &4a_{23} \\ 4a_{31} &4a_{32} &4a_{33} \end{pmatrix} = \begin{pmatrix} 4&0 &0 \\0&4 &0 \\0&0 &4 \\\end{pmatrix} \begin{pmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \end{pmatrix} = BA$ so that $B = \begin{pmatrix} 4&0 &0 \\0&4 &0 \\0&0 &4 \\\end{pmatrix}$.

  2. $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} $.

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    for someone just learning matrix multiplication, as I understand the OP is, it isn't obvious at all how you find $B$ in each case. – 2017-02-26