Given $$M = \left(\matrix{ 1 & 2 & 3 & { - 1} & 1 \cr 3 & 2 & 1 & { - 1} & 1 \cr 2 & 3 & 1 & 1 & 1 \cr 5 & 5 & 2 & 0 & {2x + 1}\cr } \right) $$ solve for $x$ such that $\mbox{rank}(M)=2$
What I have tried here is to convert this to row echelon form, which is: $$\left( {\matrix{ 1 & 2 & 3 & { - 1} & 1 \cr 0 & { - 4} & { - 8} & 2 & { - 2} \cr 0 & 0 & { - 3} & {{5 \over 2}} & { - {1 \over 2}} \cr 0 & 0 & 0 & 0 & {2x - 1} \cr } } \right)$$ It seems that this matrix cannot have the rank of 2, but can I prove it without converting the matrix to row echelon form? Thank you.
You can compute the determinant of $$\pmatrix{1 &2 &3 \cr 3 & 2 & 1\cr 2 & 3 & 1}.$$ It is non-zero, so there is no linear relation between the first three rows and the matrix must have rank 3 at least.