Question

Find the values of a and below `[(a + 3, b^2 + 2),(0, -6)] = [(2a + 1, 3b),(0, b^2 - 5b)]`

Answer 1

`[(a + 3, b^2 + 2),(0, -6)] = [(2a + 1, 3b),(0, b^2 - 5b)]`
comparing the corresponding elements
a + 3 = 2a + 1
⇒ 2a – a =3 – 1
⇒ a = 2
b² + 2 = 3b
⇒ b² – 3b + 2 = 0
⇒ b² – b – 2b + 2 = 0
⇒ b (b – 1) – 2 (b – 1) = 0
⇒ (b – 1) (b – 2) = 0.
Either b – 1 = 0,
then b = 1
or
b – 2 = 0,
then b = 2
Hence a = 2, b = 2 or 1.