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I have the following system:

$$\left\{\begin{array}{cccccccc} 2x&+&3y&+&z&-&3v&=&2 \\ x&-&y&+&2z&+&v&=&0\\ 3x&+&2y&+&3z&-&2v&=&-2 \end{array}\right.$$

I have to show if the system does or doesn't have solutions using multidimensional vectors. I notice that it has more unknowns than equations so it is an undetermined system. If I form the matrix , I notice that the determinant is different from zero so this three vectors are linearly indipendent.Now what do I do to show if they have a solution or not?

Note: I have to use only determinants and linearly independent/dependent vector theory to show it.

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Since the appropriate matrix is $3\times 4$, the determinant is undefined. What do you mean by "determinant is different from zero"? – Dennis Gulko Dec 15 '12 at 11:49
No, I form the matrix with the vectors ( 2 1 3) (3 -1 2) and (1 2 3) ,and here the determinant is diff from zero...I know I have to form another matrix including 2 0 -2 and find the determinant ,but I don't know how to form it.. – Beyonce45 Dec 15 '12 at 11:55
@Beyonce45: So, what about the $v$'s?? – Babak S. Dec 15 '12 at 11:56
@Beyonce45: I edited your question just because I think it is easier to read this way. Feel free to revert my changes if you think otherwise. – Dennis Gulko Dec 15 '12 at 12:02

What you need to do is to row-reduce the extended matrix: $$\left[\left.\begin{array}{cccc}2&3&1&-3\\1&-1&2&1\\3&2&3&-2\end{array}\right|\begin{array}{c}2\\0\\-2\end{array}\right]\underset{R_3-R_2}{\overset{R_2-R_1}{\longrightarrow}}\left[\left.\begin{array}{cccc}2&3&1&-3\\1&-1&2&1\\0&0&0&0\end{array}\right|\begin{array}{c}2\\0\\-4\end{array}\right]$$ And to check whether there are any $0$-rows equal to non-zero or not. If there are rows of the form $[\begin{array}{c}0&0&0&0\end{array}|\begin{array}{c}a\end{array}]$ for some $a\neq 0$, then there are no solutions. Else, since the system is undetermined, there will be infinitely many solutions.
As we can see here, the last row is $[\begin{array}{c}0&0&0&0\end{array}|\begin{array}{c}-4\end{array}]$, hence there are no solutions.

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Exactly! Dennis. – Babak S. Dec 15 '12 at 11:57
Thanks, but In my textbook the answer is : No solutions :S – Beyonce45 Dec 15 '12 at 12:01
I didn't say there are. I just showed how to check whether there are any. – Dennis Gulko Dec 15 '12 at 12:03
ok,thanks you again – Beyonce45 Dec 15 '12 at 12:03
Can you take it from here? or do you need more details? – Dennis Gulko Dec 15 '12 at 12:04

I agree with Dennis, but she is saying that she has to prove it using linearly dependent and independent vectors. I don't think that this is possible in this case, because the vectors in the matrix form will never be linearly independent, because the determinant is always zero. ( To find the determinant in the matrix add zeros in the fourth row to convert it to a $4\times 4$ matrix)

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