Page 257 - MATHEMATICS COURSE FOR SECONDARY SCHOOLS BOOK 2

EXAMPLE 4
(-5 7) 04
(b) 13 P - 12 Q = 13 = (
(6 -9 )- 12 (2 0 ) 312 48
(a) Given that
(b) Given that B =
(a) kA = 2 =
=
(b) kB = -6 =
-18 ==
(
)(
)
A =
and k = 2, find kA.
=
14
24
( 8 -2 )and k = -6, find kB. 3 0
= ( 2 - 1 -3 + 0 ) 1- 2 4- 4
(
-5 7 04
)
) )
=(1 -3) -1 0
(2 x [-5] 2 x0
Alternatively:
(-10 14 08
(2 x 6 2x [-9] ) ( 2 x 3 2 x 12 +
2 x 7 2 x4
(a) 2P – Q =
=
2 (6 -9 3 12
– )(
2 0 48
)
(8-2) =
(6 x 8 -6 x [-2]) = -6 x 3 -6 x 0
( 10 -18)
(b) 13 P - 12 Q = 13 (6 -9 )- 12 (2
(-48
12)
30
( 12 + [-2] 6 + [-4]
-18 + 0 ) 24 + [-8]
0
)( 2 -3 - 1 0
)
1 x 6 1 x [-9] 33
1 x 2 1 x 0
11
3 x 3 3 x12
2
2
) 2 16
(12 -18)+(-2 0 ) 6 24 -4 -8
-
22
1 x4 1 x 8
-2 0 - 4 -8
EXAMPLE 5
3 12 4
0 ) 8
If P =
(a) 2P – Q
(a) 2P – Q
11
2 0 3 x 3 3 x12 2 2
( 6 -9 3 12
1 x4 1 x 8 )()()
and Q = , evaluate 48
)(
(b) 13 P - 12 Q
6 -9 2 0
14
- 2 -4
= 2 ( ) – ( ) )
= ( 1 -3 ) -1 0
3 12
48
( )) -1 0
( 2 x 6 2 x [-9] (
= – 2 0
)
2 - 1 -3 + 0 = 1- 2 4- 4
2 x 3 2 x12
48
= 1 -3
(12-18 2 0
(12 - 2 -18 - 0 ) 6 - 4 24 - 8
(10 -18) 2 16
=–
=
=
6 24
)() 48
251
( 1 x 6 1 x [-9] ) ( 1 x 2 1 x 0 ) =33+22
= 2 -3 + -1 0
=
( 2 + [-1] 1 + [-2]
-3 + 0 ) 4 +[-4]