(a) |A|=2≠0 ∴ρ(A)=3
(b) Apply R2−R1,R3−R1 ∴|A|=−4≠0
∴ρ(A)=3
(c) R3−R2,R2−R1 makes two rows identical.
∴|A|=0
minor of order 2 is not zero
ρ(A)=2
(d) R2+R3 gives |A|=0
∴ρ(A)=2 as above
(e) ρ(A)=2 exactly as in part (c)
(f) R3+2R1 amkes R3 a row of zeros.
∴|A|=0
∴ρ(A)=2 as above.