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Taking $M$ to be of the form \begin{pmatrix} a &b &c \\ d & e & f\\ g& h & i \end{pmatrix} we get (from the $2$ given conditions) $6$ equations whereas total number of variables are $9$. So i think $(D)$ is the correct option.Am i going in the right direction? Please help.Thanks in advance for your time.

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up vote 1 down vote accepted

Hint: Is the vector $(6,0,0)^T$ a linear combination of the vectors $(1,2,3)^T$ and $(4,5,0)^T$? (For ease of typing we are using $(x,y,z)^T$ for the column vector with entries $x$, $y$, and $z$.)

Remark: It is true that there are $9$ variables and only $6$ equations. But if the question had used, say, $(6,8,3)^T$ instead of $(6,0,0)^T$, the result would be determined. So simple counting of equations and unknowns is not enough.

Your suggested approach can be made to work, for example by exhibiting two solutions to the system of equations that give different results on $(6,0,0)^T$. This is equivalent to showing that the first column of the matrix $M$ is not determined.

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Sir,I can not find the vector $(5,6,0)^T$ in the equations.Did you mean $(4,5,0)^T$? – user53386 Jan 20 '13 at 6:19
@user53386: Thanks for noticing the typo. Corrected. – André Nicolas Jan 20 '13 at 6:21

Yours way is correct. My way of thnking goes like this : Since $$\pmatrix{6 \\ 0 \\ 0}=a \pmatrix{1 \\ 2 \\3}+b \pmatrix{4 \\ 5 \\0}$$ where such $a$ and $b$ do not exist, hence we cannot determine the matrix $M$

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Your last $6$ was meant to be a zero? – Gerry Myerson Jan 20 '13 at 6:14
@GerryMyerson: yeah , edited – Idonknow Jan 20 '13 at 6:16
@Idonknow thanks a lot.I have got it.+1 from me. – user53386 Jan 20 '13 at 6:24

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