Page 112 - Modul Aljabar

= √90

= 2√10
c. d(A, C) = ǁA − C ǁ

= < A − C, A − C > 1/2

2
2
2
2
= √(1 − 6) + (2 − 3) +(3 − 2) + (3 − 1)
2
2
2
2
= √(−5) + (−1) +(1) + (2)
= √25 + 1 + 1 + 4
= √31
d. d(C, B) = ǁC − B ǁ
= < C − B, C − B > 1/2

2
2
2
2
= √(6 + 1) + (3 − 7) +(2 − 6) + (1 − 2)
= √7 + (−4) +(−4) + (−1)
2
2
2
2
= √49 + 16 + 16 + 1
= √82
4. Penyelesaian :
+ 1 5 1 3 5 1
= [ 4 2 − 1 −3] = [4 + 1 3 ]
2 4 2 4 − 2
maka diperoleh:

+ 1 = 3, = 2
2 − 1 − + 1, = 2

2 − 2, 2, ℎ = 4

5. Penyelesaian :
Kita perlu mengeco hasil kali dalam antara p dan q.

Perhatikan bahwa
( . ) = 3. 2. 1 + 2. (−1). 3

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