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 Further Mathematics Paper 2, Nov/Dec. 2013
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength

Question 11

(a)        Given that A = ;
(i)         find A2.
(ii)        determine the scalars and such that I +  +  A² = 0, where I = 2 × 2 unit matrix               and 0 is a 2 × 2 null matrix.
(b)        Solve the triangle ABC

A

b
9 cm

20o                  C
B                                    15 cm
(NOT DRAWN TO SCALE)

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Observation

This question was reportedly answered by majority of the candidates and they performed very well in
it.  However, it was also reported that candidates performed better in part(b) than in part(a).

In part(a), candidates were expected to multiply the matrix A = by itself to obtain  = .  The equation I +  + A² = 0 became  +  =  i.e.  =   This implied that                           1 +  - 3 = 0 .... (eqn.1), 2 + 4 = 0 ... (eqn. 2), -2 - 4 = 0 ... (eqn.3).  From (eqn.2), 2 + 4 = 0.  Therefore  = -2  Substituting  for  in (equation 1) gave 1 - 2 - 3 = 0.  This implied that 1 = 5 or  = .  Hence,  = -2 = .

In part(b), two methods were popular among the candidates.  They were the sine rule and the cosine rule.  Using the sine rule,  = .  Therefore SinA =   = 0.5700.  Hence, A = 34.8o.  Also, B = 180o – 20o – 34.8o = 125.2o.  Using the Sine rule,   =  i.e. b =             = 21.5 cm.  Therefore, the angles of the triangle were 20o, 34.8o and 125.2o while the side b was 21.5 cm long.