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?