matriks

1. BAB III MATRIKS
1. A = [
2 −3
4 −1
] B = [
2 −3
4 −1
]
a) A + B = [
2 −3
4 −1
] + [
2 −3
4 −1
]
= [
4 −6
8 −2
]
b)A – B = [
2 −3
4 −1
] − [
2 −3
4 −1
]
= [
0 0
0 0
]
c) AT
+ B = [
2 4
−3 −1
] + [
2 −3
4 −1
]
= [
4 1
1 −2
]
d) A – BT
= [
2 −3
4 −1
] − [
2 4
−3 −1
]
= [
0 −7
7 0
]
2. Diketahui : A = [
1 −2
5 −1
] dan B = [
−1 3
6 −2
]. Tentukan:
a). 𝐴 𝑥 𝐵 = [
1 −2
5 −1
] 𝑥 [
−1 3
6 −2
] = [
−13 7
−11 17
]
b). 𝐵 𝑥 𝐴 = [
−1 3
6 −2
] 𝑥 [
1 −2
5 −1
] = [
14 −1
−4 −10
]
c). AT
x B = [
1 5
−2 −1
] 𝑥 [
−1 3
6 −2
] = [
−29 −7
−4 −4
]
d). A x BT
= [
1 −2
5 −1
] 𝑥 [
−1 6
3 −2
] = [
−7 10
−8 32
]
3. Tentukanlah determinan dari matriks-matriks berikut:
a).K = [
7 −8
6 −9
]
|𝐾|= |
7 −8
6 −9
| = −63 − (−48) = −15
b) L = [
10 −6
5 4
]
|𝐿|= |
10 −6
5 4
| = 40 − (−30) = 70

2. BAB III MATRIKS
c)M = [
3 2
−7 −1
]
|𝑀|= |
3 2
−7 −1
| = −3 − (−14) = 11
4. Tentukan determinan dari matriks-matriks berikut:
a). |𝑃| = |
3 1 0
−2 1 4
3 −2 1
|
3 1
−2 1
3 −2
= (3 + 12 + 0) − (0 − 24 − 2) = 41
b). |𝑄| = |
5 1 1
−2 3 1
6 −2 1
|
5 1
−2 3
6 −2
= (15 + 6 + 4) − (18 − 10 − 2) = 19
5. Dengan menggunakan determinan matriks selesaikan sistem persamaan linear
berikut:
a) 3𝑥 + 2𝑦 = 1
𝑥 − 5𝑦 = 6
|𝐷|= |
3 2
1 −5
| = −15 − 2 = −17
|𝐷𝑥|= |
1 2
6 −5
| = −5 − 12 = −17
|𝐷𝑦|= |
3 1
1 6
| = 18 − 1 = 17
↔ 𝑥 =
𝐷𝑥
𝐷
=
−17
−17
= 1
↔ 𝑦 =
𝐷𝑦
𝐷
=
17
−17
= −1
b) 2𝑥 − 5𝑦 = 3
4𝑥 − 9𝑦 = 5
|𝐷|=|2 −5
4 −9
| = −18 + 20 = 2
|𝐷𝑥|= |
3 −5
5 −9
| = −27 + 25 = −2
|𝐷𝑦|= |
2 3
4 5
| = 10 − 12 = −2
↔ 𝑥 =
𝐷𝑥
𝐷
=
−2
2
= −1

3. BAB III MATRIKS
↔ 𝑦 =
𝐷𝑦
𝐷
=
−2
2
= −1
6.
3𝑥 + 2𝑦 − 2𝑧 = 6
a). 4𝑥 + 3𝑦 − 𝑧 = 10
−𝑥 − 2𝑦 + 5𝑧 = 1
|𝐷| = |
3 2 −2
4 3 −1
−1 −2 5
|
3 2
4 3
−1 −2
= (45 + 2 + 16) − (6 + 6 + 40) = 11
|𝐷𝑥| = |
6 2 −2
10 3 −1
1 −2 5
|
6 2
10 3
1 −2
= (90 + (−2) + 40) − (−6 + 12 + 100)
= 128 − 106 = 22
|𝐷𝑦| = |
3 6 −2
4 10 −1
−1 1 5
|
3 6
4 10
−1 1
= (150 + 6 + (−8) − (20 + (−3) + 120)
= 148 + 137 = 11
|𝐷𝑧| = |
3 2 6
4 3 10
−1 −2 1
|
3 2
4 3
−1 −2
= (9 − 20 − 48) − (−18 − 60 + 8)
= −59 − (−70) = 11
x =
Dx
D
=
22
11
= 2
y =
Dy
D
=
11
11
= 1
z =
Dz
D
=
11
11
= 1
2𝑖1 + 5𝑖2 − 3𝑖3 = −6
b). 𝑖1 + 4𝑖2 − 5𝑖3 = −7
−𝑖1 − 3𝑖2 + 2𝑖3 = 3

4. BAB III MATRIKS
|𝐷| = |
2 5 −3
1 4 −5
−1 −3 2
|
2 5
1 4
−1 −3
= (16 + 25 + 9) − (12 + 30 + 10) = −2
|𝐷 𝑖1| = |
−6 5 −3
−7 4 −5
3 −3 2
|
−6 5
−7 4
3 −3
= (−48 − 75 − 63) − (−36 − 90 − 70)
= −186 + 196 = 10
|𝐷 𝑖2| = |
2 −6 −3
1 −7 −5
−1 3 2
|
2 −6
1 −7
−1 3
= (−28 − 30 − 9) − (−21 − 30 − 12)
= −67 + 63 = −4
|𝐷 𝑖3| = |
2 5 −6
1 4 −7
−1 −3 3
|
2 5
1 4
−1 −3
= (24 + 35 + 18) − (24 + 42 + 15)
= 77 − 81 = −4
𝑖1 =
𝐷 𝑖1
D
=
10
−2
= −5
𝑖2 =
𝐷 𝑖2
D
=
−4
−2
= 2
𝑖3 =
𝐷 𝑖3
D
=
−4
−2
= 2
7.
a). 𝐴 = [
3 −8
2 −9
] |𝐴| = −27 + 16 = −11
𝐴−1
= −
1
11
[
−9 8
−2 3
]
= [
9
11⁄ − 8
11⁄
2
11⁄ − 3
11⁄
]
a).𝐵 = [
3 −6
5 2
] |𝐵| = 6 + 11 = 17
𝐵−1
=
1
17
[
2 6
−5 3
]

5. BAB III MATRIKS
= [
2
17⁄ 6
17⁄
− 5
17⁄ 3
17⁄
]
c). 𝐶 = [
3 5
−1 −1
]
|𝐵| = 3 − (−5) = 2
𝐵−1
=
1
2
[
−1 −5
1 3
]
= [
− 1
2⁄ − 5
2⁄
1
2⁄ 3
2⁄
]
8. Tentukan invers dari matriks-matriks berikut:
a) 𝐷 = [
4 2 0
−2 1 1
1 −2 1
]
|𝐷| = (4 + 2 + 0) − (0 − 8 − 9) = 18
𝐷11 = |
1 1
−2 1
| = 3
𝐷12 = − |
−2 1
1 1
| = 3
𝐷13 = |
−2 1
1 −2
| = 3
𝐷21 = − |
2 0
−2 1
| = −2
𝐷22 = |
4 0
−2 1
| = 4
𝐷23 = − |
4 2
1 −2
| = 10
𝐷31 = |
2 0
1 1
| = 2
𝐷32 = − |
4 0
−2 1
| = −4
𝐷33 = |
4 2
−2 1
| = 8
𝐴𝑑𝑗 𝐷 = 𝐷 𝑇
= [
3 −2 2
3 4 −4
3 10 8
]
𝐷−1
=
1
18
𝑥 [
3 −2 2
3 4 −4
3 10 8
]

6. BAB III MATRIKS
𝐷−1
=
[
1
6⁄ − 1
9⁄ 1
9⁄
1
6⁄ 2
9⁄ − 2
9⁄
1
6⁄ 5
9⁄ 4
9⁄ ]
b) 𝐸 = [
3 1 2
−2 5 1
4 −2 1
]
|𝐷| = |
3 1 2
−2 5 1
4 −2 1
|
3 1
−2 5
4 −2
= (14 + 4 + 8) − (40 − 6 − 2) = −5
𝐸11 = |
5 1
−2 1
| = 3
𝐸12 = − |
−2 1
4 1
| = 6
𝐸13 = |
−2 5
4 −2
| = −5
𝐸21 = − |
1 2
−2 1
| = −5
𝐸22 = |
3 2
4 1
| = −3
𝐸23 = − |
3 1
4 −2
| = 10
𝐸31 = |
1 2
5 1
| = −6
𝐸32 = − |
3 2
−2 1
| = −7
𝐸33 = |
3 1
−2 5
| = 17
𝐴𝑑𝑗 𝐸 = 𝐸 𝑇
= [
3 −5 −6
6 −3 −7
−5 10 17
]
𝐸−1
= −
1
5
𝑥 [
3 −5 −6
6 −3 −7
−5 10 17
]
𝐸−1
=
[
7
5⁄ −1 − 6
5⁄
6
5⁄ − 3
5⁄ − 7
5⁄
−1 2 17
5⁄ ]

7. BAB III MATRIKS
9.
a). 5𝑥 + 4𝑦 = 20
2𝑥 − 3𝑦 = 3
[
5 4
2 −3
] [
𝑥
𝑦] = [
20
3
]
A x = B
|𝐴| = |
5 4
2 −3
| = −15 − 8 = −23
x = 𝐴−1
B
x =
1
det 𝐴
𝐵
[
𝑥
𝑦] = −
1
23
[
−3 −4
−2 5
] [
20
3
]
[
𝑥
𝑦] = −
1
23
[
−72
−25
]
[
𝑥
𝑦] = [
72
23⁄
25
23⁄
]
Jadi, x = 72
23⁄
y = 25
23⁄
b). 3𝑥 − 𝑦 = 7
7𝑥 − 2𝑦 = 16
[
3 −1
7 −2
] [
𝑥
𝑦] = [
7
16
]
A x = B

8. BAB III MATRIKS
|𝐴| = |
3 −1
7 −2
| = −6 − (−7) = 1
x = 𝐴−1
B
x =
1
det 𝐴
𝐵
[
𝑥
𝑦] =
1
1
[
−2 1
−7 3
] [
7
16
]
[
𝑥
𝑦] = [
2
−1
]
[
𝑥
𝑦] = [
2
−1
]
Jadi, x = 2
y = −1
10).
a) [
2 1 −2
3 2 −4
−5 −3 6
] [
𝑥
𝑦
𝑧
] = [
4
6
−10
]
A x = B
|𝐴| = |
2 1 −2
3 2 −4
−5 −3 6
|
2 1
3 2
−5 −3
𝐷11 = |
2 −4
−3 6
|
𝐷12 = − |
3 −4
−5 6
|
𝐷13 = |
3 2
−5 −3
|
𝐷21 = − |
1 −2
−3 6
|
𝐷22 = |
2 −2
−5 6
|
𝐷23 = − |
2 1
−5 −3
|
𝐷31 = |
1 −2
2 −4
|

9. BAB III MATRIKS
𝐷32 = − |
2 −2
3 −4
|
𝐷33 = |
2 1
3 2
|
𝑎𝑑𝑗𝐴 = [𝐴 𝑇] = [
0 0 0
2 2 2
1 1 1
]
x = 𝐴−1
B
x =
1
det 𝐴
𝑥 𝐴𝑑𝑗 𝐴 𝑥 𝐵
[
𝑥
𝑦
𝑧
] =
1
0
[
0 0 0
2 2 2
1 1 1
] [
4
6
−10
]
[
𝑥
𝑦
𝑧
] = 𝑡𝑖𝑑𝑎𝑘 𝑡𝑒𝑟𝑑𝑒𝑓𝑖𝑛𝑖𝑠𝑖
b) [
3 4 −2
6 3 −3
−3 −2 5
] [
𝑖1
𝑖2
𝑖3
] = [
3
2
2
]
|𝐴| = |
3 4 −2
6 3 −3
−3 −2 5
|
3 4
6 3
−3 −2
= (45 + 36 + 24) − (18 + 18 + 120)
= (105 − 156) = −51
𝐷11 = |
3 −3
−2 5
| = 15 − 6 = 9
𝐷12 = − |
6 −3
−3 5
| = 30 − 9 = −21
𝐷13 = |
6 3
−3 −2
| = −12 − (−9) = −3
𝐷21 = − |
4 −2
−2 5
| = 20 − 4 = −16
𝐷22 = |
3 −2
−3 5
| = 15 − 6 = 9
𝐷23 = − |
3 4
−3 −2
| = −6 − (−12) = −6
𝐷31 = |
4 −2
3 −3
| = −12 − (−6) = −6
𝐷32 = − |
3 −2
6 −3
| = −9 − (−12) = −3
𝐷33 = |
3 4
6 3
| = 9 − 24 = −15
𝑎𝑑𝑗𝐴 = [𝐴 𝑇] = [
9 −16 −6
−21 9 −3
−3 −6 −15
]

10. BAB III MATRIKS
x = 𝐴−1
B
x =
1
det 𝐴
𝑥 𝐴𝑑𝑗 𝐴 𝑥 𝐵
[
𝑖1
𝑖2
𝑖3
] = −
1
51
[
9 −16 −6
−21 9 −3
−3 −6 −15
] [
3
2
2
]
[
𝑖1
𝑖2
𝑖3
] = [
17
51⁄
1
1
]