Minggu, 26 Februari 2012

Matbis (tentang matriks)


( 07 3252 0564 ) Fransiskus Asisi Verry Kristanto
1
B =
1
A =
-1
4 0
C = 
-1 -6 3
D = 
1 6 2
2 -3 3 -4 2 -1 0 3 0
0 1 3x2 5 4x1 3x2 4 1 2 3x3
2 2 -4 0 9 0 3 11 -4 3
0 6 -2 + 0 3 0 = 0 9 -2
-4 0 10 3 9 9 -1 9 19
3 1 0 3 4 + 2 3 -2 0 = 3 3 1 4
2 -1 2 -5 -4 5 -3 2 -2 4 -1 -3
-3 0 4 -1 -3 6 -4 3 -6 6 0 2
5 -2 3 2 0 1 -5 1 5 -1 -2 3
4
1 4 6 -1 -5 5
-2 7 - 7 2 = -9 5
6 9 1 0 7 9
5 1 2 3 4 4 0 2 1 -3 2 1 3
2 0 4 5 - 3 2 5 0 = -1 -2 -1 5
1 3 0 2 2 4 4 2 -1 -1 -4 4
6 A = 2 1 B = -6 -5 C =  -2 -1
3 -3 2 -3 -3 3
a. 2 ( A - 2B ) =
2
2 1 -
4
-6 -5
=
4 2
-
-24 -20
=
28 22
3 -3 2 -3 6 -6 8 -12 -2 6
b. 3 ( A - C ) + B = 3 2 1 - 3 -2 -1 + -6 -5
3 -3 -3 3 2 -3
= 6 3 - -6 -3 + -6 -5 = 6 -5
9 -9 -9 9 2 -3 2 -21
c. 2B - 3A + 2C =
2
-6 -5 - 3 2 1 + 2 -2 -1
2 -3 3 -3 -3 3
=
-12 -10 - 6 3 + -4 -2
=
-22 -15
4 -6 9 -9 -6 6 -11 9
7. 2 -4 4 0 = 12 -12
3 2 -1 3 10 6
8. 2 0 3 1 = 23
-1 4 5 4 50
7
9. 2 0 3 11 -4 3 = 19 19 63
-1 4 5 0 9 -2 -16 85 84
-1 9 19
10. A = 1 -1 0 B =  0 0 -1 C =  1 0
0 1 1 2 -1 0 2 -1
0 0 2 0 1
a. A2 = 1 -1 0 1 -1 0 = ( tidak terdefinisi karena jumlah kolom pada matriks pertama tidak sama dengan jumlah baris pada matriks kedua )
0 1 1 0 1 1
b. B3
=
0 0 -1
0 0 -1 0 0 -1
2 -1 0 2 -1 0 2 -1 0
0 0 2 0 0 2 0 0 2
=
0
0
-2 0 0 -1 = 0 0 -4
-2 1 -2 2 -1 0 2 -1 -2
0 0 4 0 0 2 0 0 8
c. ( AC )2 = 1 -1 0 1 0 1 -1 0 1 0
0 1 1 2 -1 0 1 1 2 -1
0 1 0 1
=
-1 1
-1 1 = 3 -1
2 0 2 0 -2 2
11. A =
1
2
B =
1 3
C =
1 0 D = 1 2 -1
0 -1 4 -1 1 2 2 0 2
7 0
a. 3A' + D  = 3 1 0 7 + 1 2 -1
2 -1 0 2 0 2
= 3 0 21 + 1 2 -1 = 4 2 20
6 -3 0 2 0 2 8 -3 2
b. 2B' - 3C' = 
2
1 4 -
3
1 1
3 -1 0 2
=
2 8
-
3 3
=
-1 5
6 -2 0 6 6 -8
c. C' - D
=
1 1 - 1 2 -1 = ( tidak terdefinisi karena kedua matriks tidak berordo sama )
0 2 2 0 2
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