Page 20 - Top Class F5 - Mathematics (Chapter 2)

Mathematics Form 5 Chapter 2 Matrices

 7 –5  1 1 3 

(b) It is given that m n 1 is the inverse matrix of (c) It is given that 8 –1 m is the inverse matrix of
1 5
 –1 7  .  m –3  .
n
1
 7 –5   1 5  1 1 3 m –3

Diberi m n 1 ialah matriks songsang bagi –1 7 . Diberi 8 –1 m  ialah matriks songsang bagi  1 n  .
 1 5  Inverse matrix of / Matriks songsang bagi  m –3 
Inverse matrix of / Matriks songsang bagi –1 7 1 n
1 7 –5 1 n 3
=   =  
1(7) – 5(–1) 1 1 mn – (–3)(1) –1 m
1 7 –5 1  n 3 
=   =
12 1 1 mn + 3 –1 m

1 7 –5 7 –5 1  n 3  1 1 3 
  = m   mn + 3 –1 m = 8 –1 m
12 1 1 n 1
1 n = 1, mn + 3 = 8
Hence / Maka, m = , n = 1
12 m + 3 = 8
m = 5
Hence / Maka, m = 5, n = 1

26. Write the following simultaneous linear equations in matrix form. PL 3
Tuliskan persamaan linear serentak berikut dalam bentuk matriks.

Example (a) 3x + 6y = 15 (b) m – 4n = –7
x = 5y – 1 2x + 3y = 8 –2m + 3n = 4
x + 2y = 6
15
   
3 6 x
      1 –4 m = –7
=
x = 5y – 1 ⇒ x – 5y = –1 2 3 y 8 –2 3 n 4
x + 2y = 6
   
 1 –5 x = –1
1 2 y
6
(c) p + 2q – 5 = 0 (d) x = 3y + 6 (e) 2n = –3m
2p – 3 = 0 4x + 5y = 7 m = 5n – 8
p + 2q – 5 = 0 ⇒ p + 2q = 5 x = 3y + 6 ⇒ x – 3y = 6 2n = –3m ⇒ 3m + 2n = 0
2p – 3 = 0 ⇒ 2p = 3 4x + 5y = 7 m = 5n – 8 ⇒ m – 5n = –8
   
   
      1 –3 x 6  3 2 m 0
5
1 2 p
2 0 q = 3 4 5 y = 7 1 –5 n = –8
27. Solve the following equations. PL 4
Selesaikan persamaan berikut.
Example 4 6 x 10
(a)     
=
9
3 5 y
     x 1 5 –6 10
3 4 x
7
=
1 2 y
–3

 
  = 3(2) – 4(1) –1 3 –3   = 4(5) – 6(3) –3 4  

x
2 –4 7
1
9
y
y

= 1 50 – 54 
1 14 + 12 2 –30 + 36
=  
 
2 –7 – 9 = 1 –4
1 26 2 6
=  
–2
 
x
2 –16   = ad – bc  d –b p q =  
1
y
  –c a 3
13
= –8 Hence / Maka, x = –2, y = 3
Hence / Maka, x = 13, y = –8

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