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