If the rank of xxxxx2xxxx+1is 1, then
x=0(or)x=1
x=0
x=1
x=±2
xxxxx2xxxx+1R2−R1,R3−R1
~xxx0x2-x0001
If rank =1, Thenx2−x=0
⇒x(x−1)=0
⇒x=0 ∵if x=1, then rank=2
Is the possible answer.