Solve the following system:
$4x-12y+z=0\\ x-5y-z=0\\ -4x+12y+z=0$
So in matrix form it is $ \left(\begin {matrix} 4 & -12 & 1 \\ 1 & -5 & -1 \\ -4 & 12 & 1 \\ \end{matrix}\right) $. But if you do the row operation $R_3+R_1 \rightarrow R_3$ you get that $2z=0 \rightarrow z=0$, but the rest doesn't add up: The first row then gives $4x-12y=0$, and the second $x-5y=0$. What does this mean about the solutions?
Thanks!
It means that the only solutions are $(x,y,z)=(0,0,0)$. This is clear, because writing $A$ for the matrix, and $v=(x,y,z)^T$, we have the equation $$ Av=0 $$ for an invertible matrix $A$. Hence $v=A^{-1}(Av)=A^{-1}0=0$. The matrix is invertible because it has full rank (and nonzero determinant $-16$).