The document contains examples of matrix operations including addition, subtraction, multiplication, transpose, and inverse of matrices. It also contains examples of using determinants and inverse matrices to solve systems of linear equations. Some key examples are: 1) Finding the sum, difference, product, and transpose of various 2x2 matrices 2) Computing the determinants of several 2x2 and 3x3 matrices 3) Using determinants and inverse matrices to solve 3 systems of 2 linear equations

1. Latihan3.3
1. A.𝐴 = [2 −3
4 −1
] ; 𝐵 = [2 −3
4 −1
]
𝐴 + 𝐵 = [4 −6
8 2
]
B.𝐴 = [2 −3
4 −1
] ; 𝐵 = [2 −3
4 −1
]
𝐴 − 𝐵 = [0 0
0 0
]
C. 𝐴 = [ 2 4
−3 −1
] ; 𝐵 = [2 −3
4 −1
]
𝐴 𝑇
+ 𝐵 = [4 1
1 −2
]
D. 𝐴 = [2 −3
4 −1
] ; 𝐵 = [ 2 4
−3 −1
]
𝐴 − 𝐵 𝑇
= [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. 𝐴 𝑻
𝑥 𝐵 = [ 1 5
−2 −1
] 𝑥 [−1 3
6 −2
] = [29 −7
−4 −4
]
D. 𝐴 𝑥 𝐵 𝑇
= [1 −2
5 −1
] 𝑥 [−1 6
3 −2
] = [−7 10
−8 32
]

2. Latihan3.3
3. Carilah Determinandari :
A. 𝐾 = [7 −8
6 −9
] , 𝑑𝑒𝑡 𝐾 = (7 𝑥 (−9)) − (−8 𝑥 6) = −63 + 48 = −15
B. 𝐿 = [10 −6
5 4
] , det 𝐿 = (10 𝑥 4)— 6 𝑥 5 = 40 + 30 = 70
C. 𝑀 = [ 3 2
−7 −1
] , 𝑑𝑒𝑡 𝑀 = (3 𝑥 (−1)) − (2 𝑥 (−7)) = −3 + 14 = 11
4. Tentukan Determinandari matriks-matriks berikut :
A. 𝑃 = [
3 1 0
−2 1 4
3 −2 1
]
3 1
−2 1
3 −2
= ((3 . 1 .1) + (1 . 4 . 3) + 0) − (0 + (3.4. (−2)) + (1.(−2). 1))
= 15 + 26 = 41
B. 𝑄 = [
5 1 1
−2 3 1
6 −2 1
]
5 1
−2 3
6 −2
= ((5 . 3 .1) + (1 . 1 . 6) + (1. (−2). (−2)) − ((1.3.6) + (5.1. (−2)) + (1. (−2). 1))
= 25 − 6 = 19
5. Dengan menggunakanDeterminan matriks, Selesaikansystempersamaan linierberikut:
A. 3𝑥 + 2𝑦 = 1
𝑥 − 5𝑦 = 6
𝐷 = |3 2
1 5
| = −17 | 𝐷𝑥 = |1 2
6 −5
| = −17 | 𝐷𝑦 = |3 1
1 6
| = 17
𝑥 =
−17
−17
= 1 𝑦 =
17
−17
= −1
B. 2𝑥 − 5𝑦 = 3
4𝑥 − 9𝑦 = 5
𝐷 = |2 −5
4 −9
| = 2 | 𝐷𝑥 = |3 −5
5 −9
| = −2 | 𝐷𝑦 = |2 3
4 5
| = −2
𝑥 =
−2
2
= −1 𝑦 =
−2
2
= −1

3. Latihan3.3
6. Dengan menggunakandeterminanmatriks, Selesaikansystempersamaan linierberikut:
A. 3𝑥 + 2𝑦 − 2𝑧 = 6
4𝑥 + 3𝑦 − 𝑧 = 10
−𝑥 − 2𝑦 + 5𝑧 = 1
𝐷 = [
3 2 −2
4 3 −1
−1 −2 5
] = 3[ 3 −1
−2 5
] − 2 [ 4 −1
−1 5
] − 2 [ 4 3
−1 −2
]
= 3(15 − 2) − 2(20 − 1) − 2(−8 + 3)
= 39 − 38 + 10 = 11
𝐷𝑥 = |
6 2 −2
10 3 −1
1 −2 5
| = 6 [ 3 −1
−2 5
] − 2[10 −1
1 5
] − 2 [10 3
1 −2
]
= 6(15 − 2) − 2(50 + 1) − 2(−20 − 3)
= 78 − 102 + 46 = 22
𝐷𝑦 = |
3 6 −2
4 10 −1
−1 1 5
| = 3[10 −1
1 5
] − 6[ 4 −1
−1 5
] − 2[ 4 10
−1 1
]
= 3(50 + 1) − 6(20 − 1) − 2(4 + 10)
= 153 − 114 − 28 = 11
𝐷𝑧 = |
3 2 6
4 3 10
−1 −2 1
| = 3[ 3 10
−2 1
] − 2[ 4 10
−1 1
] + 6 [ 4 3
−1 −2
]
= 3(3 + 20) − 2(4 + 10) + 6(−8 + 3)
= 69 − 28 − 30 = 11
𝒙 =
𝑫 𝒙
𝑫
=
𝟐𝟐
𝟏𝟏
= 𝟐 𝒚 =
𝑫 𝒚
𝑫
=
𝟏𝟏
𝟏𝟏
= 𝟏 𝒛 =
𝑫 𝒛
𝑫
=
𝟏𝟏
𝟏𝟏
= 𝟏
B. 2𝑖1 + 5𝑖2 − 3𝑖3 = −6
𝑖1 + 4𝑖2 − 5𝑖3 = −7
−𝑖1 − 3𝑖2 + 2𝑖3 = 3
𝐷 = |
2 5 −3
1 4 −5
−1 −3 2
| = 2 [ 4 −5
−3 2
] − 5 [ 1 −5
−1 2
] − 3 [ 1 4
−1 −3
]
= 2(8 − 15) − 5(2 − 5) − 3(−3 + 4)
= −14 + 15 − 3 = −2
𝐷𝑖1 = |
−6 5 −3
−7 4 −5
3 −3 2
| − 6 [ 4 −5
−3 2
] − 5[−7 −5
3 2
] − 3[−7 4
3 −3
]
= −6(8 − 15) − 5(−14 + 15) − 3(21 − 12)
= 42 − 5 − 27 = 10

4. Latihan3.3
𝐷𝑖2 = |
2 −6 −3
1 −7 −5
−1 3 2
| = 2 [−7 −5
3 2
] + 6 [ 1 −5
−1 2
] − 3[ 1 −7
−1 3
]
= 2(−14 + 15) + 6(2 − 5) − 3(3 − 7)
= 2 − 18 + 12 = −4
𝐷𝑖3 = |
2 5 −6
1 4 −7
−1 −3 3
| = 2 [ 4 −7
−3 3
] − 5 [ 1 −7
−1 3
] − 6[ 1 4
−1 −3
]
= 2(12 − 21) − 5(3 − 7) − 6(−3 + 4)
= −18 + 20 − 6 = −4
𝒊 𝟏 =
𝑫𝒊 𝟏
𝑫
=
𝟏𝟎
−𝟐
= −𝟓 𝒊 𝟐 =
𝑫𝒊 𝟏
𝑫
=
−𝟒
−𝟐
= 𝟐 𝒊 𝟑 =
𝑫𝒊 𝟏
𝑫
=
−𝟒
−𝟐
= 𝟐
7. Tentukan Inversdari matriks-matriks berikut:
a. 𝐴 = [3 −8
2 −9
] = −27 + 16 = −11
𝑨−𝟏
=
𝟏
−𝟏𝟏
[−𝟗 𝟖
−𝟐 𝟑
] = [
𝟗
𝟏𝟏
−
𝟖
𝟏𝟏
𝟐
𝟏𝟏
−
𝟑
𝟏𝟏
]
b. 𝐵 = [3 −6
5 2
] = 6 + 30 = 36
𝑩−𝟏
=
𝟏
𝟑𝟔
[ 𝟐 𝟔
−𝟓 𝟑
] = [
𝟏
𝟏𝟖
𝟏
𝟔
−
𝟓
𝟑𝟔
𝟏
𝟏𝟐
]
c. 𝐶 = [ 3 5
−1 −1
] = −3 + 5 = 2
𝑨−𝟏
=
𝟏
𝟐
[−𝟏 −𝟓
𝟏 𝟑
] = [
−
𝟏
𝟐
−
𝟓
𝟐
𝟏
𝟐
𝟑
𝟐
]
8. Tentukan Inversdari matriks-matriks berikut:
a. 𝐷 = [
4 2 0
−2 1 1
1 −2 1
] det 𝐷 = ( 4 + 2 + 0) − (0 − 8 − 4) = 18
𝐷11 = [ 1 1
−2 1
] = 3 𝐷21 = − [ 2 0
−2 1
] = −2 𝐷31 = [2 0
1 1
] = 2
𝐷12 = − [−2 1
1 1
] = 3 𝐷22 = [4 0
1 1
] = 4 𝐷32 = − [ 4 0
−2 1
] = −4
𝐷13 = [−2 1
1 −2
] = 3 𝐷23 = − [4 2
1 −2
] = 10 𝐷33 = [ 4 2
−2 1
] = 8
𝑎𝑑𝑗 𝐷 = [
3 3 2
3 4 −4
3 10 8
] → 𝐷−1
=
1
18
[
3 3 2
3 4 −4
3 10 8
] = [
1/6 1/6 1/9
1/6 2/9 −2/9
1/6 5/9 4/9
]

5. Latihan3.3
b. 𝐸 = [
3 1 2
−2 5 1
4 −2 1
] det 𝐸 = ( 15 + 4 + 8) − (40 − 6 − 2) = −5
𝐸11 = [ 5 1
−2 1
] = 7 𝐸21 = − [ 1 2
−2 1
] = −5 𝐸31 = [1 2
5 1
] = −9
𝐸12 = − [−2 1
4 1
] = 6 𝐸22 = [3 2
4 1
] = −5 𝐸32 = − [ 3 2
−2 1
] = −7
𝐸13 = [−2 5
4 −2
] = −16 𝐸23 = − [3 1
4 −2
] = 10 𝐸33 = [ 3 1
−2 5
] = 17
𝑎𝑑𝑗 3 = [
7 −5 −9
6 −5 −7
−16 10 17
] → 3−1
=
1
−5
[
7 −5 −9
6 −5 −7
−16 10 17
] = [
−7/5 1 9/5
−6/5 1 7/5
16/5 −2 −17/5
]
9. Dengan menggunakaninvers matriks, selesaikansystempersamaan linierberikut:
A. 5𝑥 + 4𝑦 = 20
2𝑥 − 3𝑦 = 3
[5 4
2 −3
] [
𝑥
𝑦
] = [20
3
] → det 𝐴 = [5 4
2 −3
] = −15 − 8 = −23
𝐴−1
=
1
−23
[−3 −4
−2 5
] → [
𝑥
𝑦
] =
1
−23
[−3 −4
−2 5
] [20
3
]
=
1
−23
[−60 − 12
−40 + 15
] =
1
−23
[−72
−25
] = [
72/23
25/23
]
B. 3𝑥 − 𝑦 = 7
7𝑥 − 2𝑦 = 16
[3 −1
7 −2
] [
𝑥
𝑦
] = [ 7
16
] → det 𝐵 = [3 −1
7 −2
] = −6 + 7 = 1
𝐴−1
=
1
1
[−2 1
−7 3
] → [
𝑥
𝑦
] = [−2 1
−7 3
] [ 7
16
]
= [−14 + 16
−49 + 48
] = [ 2
−1
]
10. Dengan menggunakanInvers matriks, selesaikansystem persamaan linierberikut:
a. 2𝑥 + 𝑦 − 2𝑧 = 4
3𝑥 + 2𝑦 − 4𝑧 = 6
−5𝑥 − 3𝑦 + 6𝑧 = −10
det 𝐴 = [
2 1 −2
3 2 −4
−5 −3 6
]
2 1
3 2
−5 −3
= (24 + 20 + 18) − (20 + 24 + 18)
= 0
Karena determinan A = 0, maka x, y, dan z benilai ≠∞

6. Latihan3.3
b. 3𝑖1 + 4𝑖2 − 2𝑖3 = 3
6𝑖1 + 3𝑖2 − 3𝑖3 = 2
−3𝑖1 − 2𝑖2 + 5𝑖3 = 2
det 𝐵 = [
3 4 −2
6 3 −3
−3 −2 5
]
3 4
6 3
−3 −2
= (45 + 36 + 24) − (18 + 18 + 120) = 105 − 156 = −51
𝑀11 = | 3 −3
−2 5
| = 9 𝑀21 = | 4 −2
−2 5
| = −16 𝑀31 = |4 −2
3 −3
| = 6
𝑀12 = | 6 −3
−3 5
| = −21 𝑀22 = | 3 −2
−3 5
| = 9 𝑀32 = |3 −2
6 −3
| = −3
𝑀13 = | 6 3
−3 −2
| = −3 𝑀23 = | 3 4
−3 −2
| = −6 𝑀33 = |3 4
6 3
| = −15
[
𝑥
𝑦
𝑧
] =
1
−51
[
9 −16 6
−21 9 −3
−3 −6 −15
] [
3
2
2
]
=
1
−51
[
27 − 32 + 12
−63 + 18 − 6
−9 − 12 − 30
]
=
1
−51
[
−7
−51
−51
]
[
𝑥
𝑦
𝑧
] = [
7/51
1
1
]