Page 21 - Top Class F5 - Mathematics (Chapter 2)

Mathematics Form 5 Chapter 2 Matrices

   
 2 –5 m –8  1 –1 m 5
   
(b) 3 –6 n = 6 (c) 4 8 n = 12
 
 
m
  1  –6 5 –8   1  8 1 5
m
n = 2(–6) – (–5)(3) –3 2 6 n = 1(8) – (–1)(4) –4 1 12
1 48 + 30 1 40 + 12
=   =  
3 24 + 12 12 –20 + 12
1 78 1 52
=   =  
3 36 12 –8
26
  1
= 12 4 2
= 3
Hence / Maka, m = 26, n = 12  –
3
1 2
Hence / Maka, m = 4 , n = –
3 3

28. Solve the following simultaneous linear equations using matrix method. PL 4
Selesaikan persamaan linear serentak berikut dengan kaedah matriks.

Example (a) 3p + 2q = –2
5p + 4q = 2 4p + 3q = –5
6p + 7q = 9
–2
3 2 p
=
    
     4 3 q –5
2
5 4 p
6 7 q = 9   = 1  3 –2 –2
 
p
  1  7 –4 2 q 3(3) – 2(4) –4 3 –5
 
p
q =

5(7) – 4(6) –6 5 9
= 1 –6 + 10 
1 14 – 36 1 8 – 15
=  
4
11 –12 + 45 =  
1 –22 –7
=  
11 33 Hence / Maka, p = 4, q = –7
–2
 
= 3
Hence / Maka, p = –2, q = 3
(b) 2x = 6 – 7y ⇒ 2x + 7y = 6 (c) 2m = 3n – 9 ⇒ 2m – 3n = –9
–2x + 3y = 4 2n = 3m – 4 ⇒ 3m – 2n = 4
 –2 3 y = 6  2 –3 m = –9
   
2 7 x
   
3 –2 n
4
4
  = 2(3) – 7(–2) 2 2 4   = 2(–2) – (–3)(3) –3 2  
–2 3 –9

m
3 –7 6

x
 
1
1
y
n
4
1 18 – 28 1 18 + 12
=   =  
20 12 + 8 5 27 + 8
1 –10 1 30
=   =  
20 20 5 35
  =  
6
1
–
= 2 7
1 Hence / Maka, m = 6, n = 7
1
Hence / Maka, x = – , y = 1
2
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