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Given that: $$AB= \left[ {\matrix{ 3 & 1 \cr 2 & 1 \cr } } \right]$$ and $$BA= \left[ {\matrix{ 5 & 3 \cr -2 & -1 \cr } } \right]$$ find $A$ and $B$.

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It would be nice if you could ask your question not like an order, don't you think so? –  a.r. Feb 13 '11 at 7:24
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2 Answers

up vote 13 down vote accepted

First note that both the products are of full rank and hence $A$ and $B$ are also of full rank. So we would expect one set of solutions with one degree of freedom since if $A$ and $B$ satisfy our equations, so will $kA$ and $\frac{1}{k}B$.

Let $$A = \left[ {\matrix{ a_1 & a_2 \cr a_3 & a_4 \cr } } \right],B = \left[ {\matrix{ b_1 & b_2 \cr b_3 & b_4 \cr } } \right]$$

$$AB= \left[ {\matrix{ 3 & 1 \cr 2 & 1 \cr } } \right], BA= \left[ {\matrix{ 5 & 3 \cr -2 & -1 \cr } } \right]$$

Hence, $$B \left[ {\matrix{ 3 & 1 \cr 2 & 1 \cr } } \right] = B(AB) = (BA)B= \left[ {\matrix{ 5 & 3 \cr -2 & -1 \cr } } \right] B$$

\begin{align*} 3b_1 + 2b_2 & = 5b_1 + 3b_3\\ b_1 + b_2 & = 5b_2 + 3b_4\\ 3b_3 + 2b_4 & = -2b_1 - b_3\\ b_3 + b_4 & = -2b_2 - b_4 \end{align*}

Rearranging, we get,

\begin{align*} 2b_2 & = 2b_1 + 3b_3\\ b_1 & = 4b_2 + 3b_4\\ 2b_1 + 4b_3 + 2b_4 & = 0\\ 2b_2 + b_3 + 2b_4 & = 0 \end{align*}

Set $b_2 = k$, and $b_4 = m$, we get $b_1 = 4k+3m$ and $b_3 = -2(k+m)$

Hence, $B = \left[ {\matrix{ 4k+3m & k \cr -2(k+m) & m \cr } } \right]$ where $k,m \neq 0$, $B^{-1} = \frac{1}{3m^2+6mk+2k^2} \left[ {\matrix{ m & -k \cr 2(k+m) & 4k+3m \cr } } \right]$.

Hence, $A = B^{-1} \times \left[ {\matrix{ 5 & 3 \cr -2 & -1 \cr } } \right]$

$A = \frac{1}{3m^2+6mk+2k^2} \left[ {\matrix{ 5m+2k & 3m+k \cr 2k+4m & 2k+3m \cr } } \right]$

There are two degrees of freedom.

EDIT

Thanks to Qiang Li for pointing out the other degree of freedom

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you missed another set of solutions for $B$, which is $$B = m \left[ {\matrix{ -4 & -1 \cr 2 & 0 \cr } } \right]$$ –  Qiang Li Feb 14 '11 at 23:37
    
@Qiang: Thanks. I have updated accordingly. –  user17762 Feb 15 '11 at 0:07
    
Thank you in advance for your solution! –  Piseth Feb 18 '11 at 5:15
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Since $AB$ and $BA$ are both defined, we know that $A$ and $B$ are both $2x2$ matrices.

Let:
$A= \left[ {\matrix{ x_1 & x_2 \cr x_3 & x_4 \cr } } \right]$

$B= \left[ {\matrix{ y_1 & y_2 \cr y_3 & y_4 \cr } } \right]$

Then we get the following equations :

From $AB$:
$x_1y_1+x_2y_3=3$
$x_1y_2+x_2y_4=1$
$x_3y_1+x_4y_3=2$
$x_3y_3+x_4y_4=1$

From $BA$:
$y_1x_1+y_2x_3=5$
$y_1x_2+y_2x_4=3$
$y_3x_1+y_4x_3=-2$
$y_3x_2+y_3x_4=-1$

Then we have eight equations in 8 unknowns, and it should be possible to find A and B.

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I am not sure if this will be of help to solve the problem in an elegant way. –  user17762 Feb 13 '11 at 7:57
    
You are right, this is a pure "brute force" approach. It shouldn't be too hard after doing this work to input the equations into something like wolfram alpha to get results. –  fdart17 Feb 13 '11 at 8:04
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