1. Soalan Penyelesaian dan Skema Markah Markah
1. a) (1)6 - k(2) = 0
– 2k = -6
K = 3
b) (i) Matrikssongsang= 







 12
26
)2(2)6(1
1
= 







12
26
2
1
=










2
11
13
ii) 





62
21






y
x
= 





4
1






y
x
= 







 12
26
)2(2)6(1
1






4
1
= 







 )4(1)1(2
4)2()1(6
46
1
= 





2
2
2
1
= 





1
1
Maka, x = -1,y = 1
P1
P1
P1
K1
N1 N1
1
1
1
1
2
6
Markah
2. a) K(3) – (-4)(1) = 0 , (ad-bc= 0)
3k + 4 = 0
K = 3
4

b) Q-1 = 







 31
51
)1(5)1(3
1
= 







31
51
2
1
= 









2
3
2
1
2
5
2
1
c) Q 





y
x
= 





2
14








11
53






y
x
= 





2
14






y
x
= 







 31
51
)1(5)1(3
1






2
14
P1
K1
N1
P1
1
1
1
1

2. = 







 2)3()14(1
2)5()14(1
53
1
= 





8
4
2
1
= 





4
2
Maka x = 2 , y = 4
3.
a) 




 
n3
21 -1 = 





 13
2
)3)(2()(1
1 n
n
= 





 13
2
6
1 n
n
= 










6
1
62
3
6
2
6
n
nn
n
.
Diberi 





 m3
25
= 










6
1
62
3
6
2
6
n
nn
n
5 =
𝒏
−𝒏+𝟔
-5n + 30 = n
6n = 30
n = 5
Masukkan n = 5 dalam m =
−𝟏
−𝒏+𝟔
m =
−𝟏
−𝟓+𝟔
= -1
Maka n = 5, m = -1
4.
a) 







13
24 -1 = 







 43
21
3)2()1(4
1
= 







 43
21
64
1
=












2
2
3
1
2
1

3. 








nm
1
2
1
=












2
2
3
1
2
1
Maka m =  2
3
, n = 2)
b) 4x  2y = 10
3x  y = 6








13
24






y
x
= 





6
10






y
x
= 







 43
21
)3)(2()1(4
1






6
10
= 







 )6(4)10(3
)6(2)10(1
64
1
= 





6
2
2
1
= 





 3
1
Maka x = , y= -3
5.
a) 







65
43 -1 = 







 35
46
)5)(4()6(3
1
= 







35
46
2
1
Diberi 







35
6 p
m = 







35
46
2
1
Maka m =
1
2
, p = 4
b) 3x  4y = 1
5x  6y = 2








65
43






y
x
= 





2
1






y
x
= 







 35
46
)5)(4()6(3
1






2
1

4. = 







 )2(3)1(5
)2(4)1(6
2018
1
= 





11
14
2
1
= 





5.5
7
Maka x = 7 , y = 5.5
6.
a) 





1
2
n
m -1 = 







 2
1
)()1(2
1
n
m
nm
= 







 2
1
2
1
n
m
mn
Diberi matriks songsang 







2
31
8
1
n
= = 







 2
1
2
1
n
m
mn
m = 3
Masukkan m =3 dalam(8 = 2-mn)
8 = 2 – 3n
n = -2
Maka m = 3 , n = -2
b) 












8
16
k
h
M






 12
32






k
h
= 





 8
16






k
h
= 




 
 22
31
)2(3)1(2
1






 8
16
= 







 )8(2)10(3
)8)(3()16(1
62
1
= 





16
40
8
1
= 





2
5
Maka h = 5 , k = 2

5. 7.
a) M-1 =








 65
32
5)3()2(6
1
Diberi 





 65
3
15
1 a
ab
= 







 65
32
5)3()2(6
1
a = -2
masukkana = -2 dalamab = -12
-2b = -12
b = 6
Maka a= -2 , b = 6
b) 











2
0
f
e
M








25
36












2
0
f
e






f
e
= 







 65
32
5)3()2(6
1






2
0
= 







 )2(6)0(5
)2(3)0(2
1512
1
= 





12
6
3
1
= 





4
2
Maka e = 2 , f = 4
8. a) 2p + 7 q = -2
3p + 8q = 3






83
72






q
p
= 





3
2
Maka matriks F = 





83
72
b) (i) Diberi 




















3
2
3
7
5
1
n
m
q
p
F-1
= 







 23
78
)5(7)8(2
1
= 







 n
m
3
7
5
1
Maka m = 8 , n = 2

6. (ii) 




















3
2
23
78
5
1
q
p
= 







 )3(2)2(3
)3)(7()2(8
5
1
= 





 12
37
5
1
=













5
12
5
37
Maka : p = 5
37
, q = 5
12 )
9.
a) Diberi 







23
47
Q = 





10
01
Q = 







23
47 -1
= 







 73
42
)3(4)2(7
1
= 








73
42
2
1
=








2
7
2
3
21
b)
7p + 4q = 13
3p + 2q = 7








23
47






q
p
= 





7
13






q
p
= 







 73
42
)3(4)2(7
1






7
13
= 








)7)(7()13(3
)7)(4()13(2
2
1
= 








10
2
2
1
=






5
1
Maka p = 1 , q = 5

7. 10.
Diberi P =






32
53
a) Jika PQ = QP = I






32
53
Q = 





10
01
Q = 





32
53 -1
= 







 32
53
)2(5)3(3
1
= 








32
53
1
1
= 







32
53
b) 3s + 5t = 8
2s + 3t = 6






32
53






t
s
= 





6
8






t
s
= 







 32
53
)2(5)3(3
1






6
8
= 








)6(3)8(2
)6)(5()8(3
1
1
= 






2
6
1
1
= 





 2
6
Maka s = -6 , t = 2
11.
a) 





 12
31
R = 





10
01
.
R = 





 12
31 -1
= 




 
 12
31
)2(3)1(1
1
= 




 
12
31
5
1

8. Diberi 




 
1
311
km
= 




 
12
31
5
1
Maka m = 5 , k = 2
b)



















 5
5
12
31
y
x






y
x
= 




 
 12
31
)2(3)1(1
1






 5
5
= 







)5(1)5(2
)5)(3()5(1
5
1
= 





5
10
5
1
= 





1
2
Maka x = 2 , y = 1
12. a) Diberi






















10
01
62
7
42
761 s
r








42
761
r
= 







42
76 -1








42
761
r
= 







 ss 2
76
)2)(7()6(
1








42
761
r
= 







 ss 2
76
146
1
S = 4
Masukkans =4 dalam r = -6s + 14
r = -6 (4) + 14
r = -10
Maka s = 4 , r = -10
b )




















6
3
42
76
y
x






y
x
= 







 62
74
)2(7)4(6
1






6
3
= 







 )6)(6()3(2
)6)(7()3(4
1424
1

9. = 








30
30
10
1
= 





3
3
Maka x = 3 , y = 3
13.
a) 





 24
35 -1 = 




 
 54
32
)4(3)5(2
1
= 




 
54
32
22
1
Diberi
m 




 
5
32
n
= 




 
54
32
22
1
Maka m = 1/22 n = 4
b)
5x + 3y = 26
-4x + 2y = -12






 24
35






y
x
= 





12
26






y
x
= 




 
 54
32
)4(3)5(2
1






12
26
= 







)12(5)26(4
)12)(3()26(2
22
1
= 





44
88
22
1
= 





2
4
Maka x = 4 , y = 2

10. 14.
a)








32
23 -1 =








 32
23
)2)(2()3(3
1
= 








32
23
5
1
Diberi :






3
231
mR
= 








32
23
5
1
Maka R = 5 , m = -2
b)
3x - 2y = 8
2x - 3y = 7




















7
8
32
23
y
x






y
x
= 







 32
23
)2)(2()3(3
1






7
8
= 








)7(3)8(2
)7)(2()8(3
5
1
= - 





5
10
5
1
= 





1
2
Maka x = 2 , y = -1
15.
a) M 







23
47
= 





10
01
M = 







23
47 -1
= 







 73
42
)3)(4()2(7
1
= - 







73
42
2
1
=














2
7
2
3
21

11. b)
7x - 4y = -2
3x - 2y = 4








23
47






y
x
= 





4
2






y
x
= 







 73
42
)3)(4()2(7
1






4
2
= 








)4)(7()2(3
)4(4)2(2
2
1
= 






34
20
2
1
= 







17
10
Maka x = -10 , y = -17
16.
a) 5h + 3s = 240
4h + 2s = 176






24
35






s
h
= 





176
240
b)






24
35






s
h
= 





176
240






s
h
= 







 54
32
)4)(3()2(5
1






176
240
= 








)176)(5()240(4
)176)(3()240(2
2
1
= 








80
48
2
1
= 





40
24
Maka h = 24 , s = 40
17. a) 8x + 40y = 440
6x + 25y = 290






256
408






y
x
= 





290
440

12. b)






256
408






y
x
= 





290
440






y
x
= 







 86
4025
)6)(40()25(8
1






290
440
= 








)290)(8()440(6
)290)(40()440(25
2
1
= 








320
600
40
1
= 





8
15
Maka x = 15 , y = 8
18.
a) i) N = 





34
1t
ad – bc = 0
3t - 1(4)= 0
t =
4
3
ii)
Diberi t = 2
N = 





34
12
N-1 = 







 24
13
)4)(1()3(2
1
= 







24
13
2
1
=














12
2
1
2
3
b)
2x + y = 1
4x + 3y = 5






34
12






y
x
= 





5
1






y
x
= 







 24
13
)4)(1()3(2
1






5
1
= 







)5)(2()1(4
)5)(1()1(3
2
1

13. = 





6
2
2
1
= 





3
1
Maka x = -1 , y = 3
19.
Diberi P= 





40
03
, Q = 





y
x
0
0
PQ = P + Q






40
03






y
x
0
0
= 





40
03
+ 





y
x
0
0








)(4)0)(0()0(4)(0
)(0)0)(3()0(0)(3
yx
yx
= 







y
x
400
003






y
x
40
03
= 







y
x
40
03
3x =3+x
2X =3
X =
3
2
4y = 4 + y
3y = 4
y =
4
3
Maka X =
3
2
, y =
4
3
20.






 n
m
5
3
16
1





 
35
2n
= 





10
01








)3()2(5)5()(5
)3()2(3)5()(3
16
1
nnn
mmn
= 





10
01
-6 + 3m = 0