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So I have the system of equations: $$ \begin{align*} 2a+3b+c-6d&= 1 \\ a-b+c+2d&=0\\ 3a+2b+3c-4d&=-1 \end{align*} $$

I have to prove that this system has no solutions. So, first I prove that all of them are linearly independent, this happens when the determinant is different from zero. I form the matrix $$ \begin{bmatrix} 2 & 3 & 1 \\ 1 & -1 & 1 \\ 3 & 2 & 3 \end{bmatrix} $$ and the determinant is indeed different from zero (It would take long to write it here). The system shouldn't have any solutions, so some of the vectors should be linearly independent as well. Can you form for me, the matrix that proves that they are linearly independent?

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So, what was happened to $d$'s and their coefficients? – Babak S. Dec 15 '12 at 8:32
What is that ?? – Engmajor Dec 15 '12 at 8:34
You have 3 equations but 4 parameters. Which of them are your unknown variables? – Kaster Dec 15 '12 at 8:38
All of them? Why dont you just form the matrice for me,it is kinda urgent...please. – Engmajor Dec 15 '12 at 8:40
Why is it urgent? Is it an assignment? Or an exam problem? – Christopher A. Wong Dec 15 '12 at 8:41

You will not be able to show that there are no solutions. Check out for example $a=\dfrac{9}{5}$, $b=-\dfrac{1}{5}$, $c=-2$, $d=0$.

I suggest going through the row reduction process.

Remark: If you really want to use a determinant, it can be done, though I don't advise it. But you could bring the $d$ stuff to the right hand side, treating $d$ as a parameter, and see what the determinant of the resulting $3\times 3$ tells you.

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Sorry but this is the question on my textbook :/ – Engmajor Dec 15 '12 at 8:53

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