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  • $A \in R^{2X3}$ with first row the $\pmatrix{1&2&3}$

  • $B \in R^{3x2}$ with first row the $\pmatrix{5&6}$

    • i) Does these matrices exist if $AB=I_2$ ?

    ii) Does these matrices exist if $BA=I_3$ ?

    Edit-Note:

    $I_{2}=\pmatrix{1&0\\0&1}$ and $I_{3}=\pmatrix{1&0&0\\0&1&0\\0&0&1}$

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      Have you tried writing down $AB$ and $BA$ and solve the $4$ or $9$ equations? – 2017-02-06
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      Hello. Yes I 've tried. Let $A=\pmatrix{1&2&3\\a&b&c}$ and $B=\pmatrix{5&6\\g&h\\i&j}$ If we multiply the two matrices there are many terms. – 2017-02-06
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      Then you could try to argue with dimensions that one of the cases is not possible, and find some necessary conditions for the other case, which should reduce the number of unknowns. – 2017-02-06
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      Consider the ranks of $A$, $B$, $BA$, and $I_3$. – 2017-02-06
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      @GerryMyerson thanks, could you read my answer, please? – 2017-02-06
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      @Hetebrij are dimensions useful in the exercise? – 2017-02-06
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      $AB=I_2$ works out to four equations in seven unknowns. You don't have to find *all* the solutions of the system, you only need *one*, and a little educated trial-and-error should get you there. You can even try assuming $a=0$, $b=0$, $c=1$, and see whether there are $g,h,i,j$ that work. – 2017-02-06
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      Any progress, MaketheDif? – 2017-02-08
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      Are you still there? – 2017-02-09
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      I'm voting to close this question as off-topic because OP has abandoned it. – 2017-02-12

    1 Answers 1

    2

    Thank you for all your help. I tried to do what you suggested.

    Is the following solution OK?

    $rank(A)=rank(\pmatrix{1&2&3\\a&b&c})=1$ or $ 2\implies$ $rank(\pmatrix{1&2&3\\a&b&c})\le2$

    $\implies rank(A)\le2$

    $rank(B)=rank(\pmatrix{5&6\\g&h\\k&m})=1 $ or $ 2 \implies rank(\pmatrix{5&6\\g&h\\k&m})\le2 $

    $\implies rank(B)\le2$

    i) $rankI_3=3$ |(1)|

    $rank(BA)\le min(rankA,rankB) \le 2 $, contradiction because of |(1)|, which implies that matrices $A,B$ don't exist.

    ii) $rankI_2=2$....

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      Since you want to find an upper bound on $\operatorname{rank}(BA)$ it is cleaner to write $\operatorname{rank}(A) \le 2$ and likewise for $B$. – 2017-02-06
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      You are right @Hetebrij Thanks! Any idea for ii) anybody? – 2017-02-06
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      For ii), try to find $A$ and $B$ such that their rank are $2$ as well as the rank of $AB$, – 2017-02-06
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      OK, I will try. It's not difficult, but it also it must be $AB=I_2$ – 2017-02-06