Math question using matrices 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.
 A: 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.
A: 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)
