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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"?2012-12-15
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    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..2012-12-15
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    @Beyonce45: So, what about the $v$'s??2012-12-15
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    @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.2012-12-15

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Edit:
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.2012-12-15
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    Thanks, but In my textbook the answer is : No solutions :S2012-12-15
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    I didn't say there are. I just showed how to *check* whether there are any.2012-12-15
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    ok,thanks you again2012-12-15
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    Can you take it from here? or do you need more details?2012-12-15
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    I can actually see from inspection that row 3 is a linear combination of row 1 + row 2. From that, you can see that the last row will be the case of 0 = 02012-12-15
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    @user43956 You mean $0=4$.2012-12-15
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    @Beyonce45: I edited my answer to be complete. My notation $R_i-R_j$ means that the result goes into the first one.2012-12-15
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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)