matematika bab 3 “matriks”

MATEMATIKA BAB 3 “MATRIKS”

KRISNA MONICA IEA

1. Jika A = [2 −34 −1

] B = [2 −34 −1

] , tentukanlah:

a) A + B

b) A – B

c) AT + B

d) A – BT

a) A + B = [2 −34 −1

] + [2 −34 −1

]

= [4 −68 −2

]

b) A – B = [2 −34 −1

] − [2 −34 −1

]

= [0 00 0

]

c) AT + B = [

2 4−3 −1

] + [2 −34 −1

]

= [4 11 −2

]

d) A – BT

= [2 −34 −1

] − [2 4

−3 −1]

= [0 −77 0

]

2. Jika A = [1 −25 −1

] dan B = [−1 36 −2

]. Tentukan:

a). A x B = [1 −25 −1

] x [−1 36 −2

] = [−13 7−11 17

]

b). B x A = [−1 36 −2

] x [1 −25 −1

] = [14 −1−4 −10

]

c). AT x B = [

1 5−2 −1

] x [−1 36 −2

] = [−29 −7−4 −4

]

d). A x BT = [

1 −25 −1

] x [−1 63 −2

] = [−7 10−8 32

]


KRISNA MONICA IEA

3. Tentukanlah determinan dari matriks-matriks berikut:

a) K = [7 −86 −9

]

b) L = [10 −65 4

]

c) M = [3 2

−7 −1]

a). K = [7 −86 −9

]

|K|= |7 −86 −9

| = −63 − (−48) = −15

b) L = [10 −65 4

]

|L|= |10 −65 4

| = 40 − (−30) = 70

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 43 −2 1

]

b. Q = [5 1 1

−2 3 16 −2 1

]

a).P = [3 1 0

−2 1 43 −2 1

]

|P| = |3 1 0

−2 1 43 −2 1

|3 1

−2 13 −2


KRISNA MONICA IEA

= (3 + 12 + 0) − (0 − 24 − 2) = 41

b).Q = [5 1 1

−2 3 16 −2 1

]

|Q| = |5 1 1

−2 3 16 −2 1

|5 1

−2 36 −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

b) 2x − 5y = 3

4x − 9y = 5

a) 3x + 2y = 1

x − 5y = 6

|D|= |3 21 −5

| = −15 − 2 = −17

|Dx|= |1 26 −5

| = −5 − 12 = −17

|Dy|= |3 11 6

| = 18 − 1 = 17

x =Dx

D=

−17

−17= 1

y =Dy

D=

17

−17= −1

Jadi, x = 1

y = −1


KRISNA MONICA IEA

b) 2x − 5y = 3

4x − 9y = 5

|D|=|2 −54 −9

| = −18 + 20 = 2

|Dx|= |3 −55 −9

| = −27 + 25 = −2

|Dy|= |2 34 5

| = 10 − 12 = −2

x =Dx

D=

−2

2= −1

y =Dy

D=

−2

2= −1

Jadi, x = −1

y = −1

6. Dengan menggunakan determinan matriks selesaikan sistem persamaan linear berikut:

3x + 2y − 2z = 6

a). 4x + 3y − z = 10

−x − 2y + 5z = 1

2i1 + 5i2 − 3i3 = −6

b). i1 + 4i2 − 5i3 = −7

−i1 − 3i2 + 2i3 = 3

3x + 2y − 2z = 6

a). 4x + 3y − z = 10

−x − 2y + 5z = 1

|D| = |3 2 −24 3 −1

−1 −2 5|

3 24 3

−1 −2

= (45 + 2 + 16) − (6 + 6 + 40) = 11


KRISNA MONICA IEA

|Dx| = |6 2 −210 3 −11 −2 5

|6 210 31 −2

= (90 + (−2) + 40) − (−6 + 12 + 100)

= 128 − 106 = 22

|Dy| = |3 6 −24 10 −1

−1 1 5|

3 64 10

−1 1

= (150 + 6 + (−8) − (20 + (−3) + 120)

= 148 + 137 = 11

|Dz| = |3 2 64 3 10

−1 −2 1|

3 24 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

Jadi, x = 2

y = 1

z = 1

2i1 + 5i2 − 3i3 = −6

b). i1 + 4i2 − 5i3 = −7

−i1 − 3i2 + 2i3 = 3

|D| = |2 5 −31 4 −5

−1 −3 2|

2 51 4

−1 −3

= (16 + 25 + 9) − (12 + 30 + 10) = −2


KRISNA MONICA IEA

|Di1| = |−6 5 −3−7 4 −53 −3 2

|−6 5−7 43 −3

= (−48 − 75 − 63) − (−36 − 90 − 70)

= (−186) + 196 = 0

|Di2| = |2 −6 −31 −7 −5

−1 3 2|

2 −61 −7

−1 3

= (−28 − 30 − 9) − (−21 − 30 − 12)

= (−67) + 63 = −4

|Di3| = |2 5 −61 4 −7

−1 −3 3|

2 51 4

−1 −3

= (24 + 35 + 18) − (24 + 42 + 15)

= 77 − 81 = −4

i1 =Di1D

=10

−2= −5

i2 =Di2D

=−4

−2= 2

i3 =Di3D

=−4

−2= 2

Jadi, i1 = 5

i2 = 2

i3 = 2

7. Tentukan invers dari matriks-matriks berikut:

a) A = [3 −82 −9

]

b) B = [3 −65 2

]

c) C = [3 5

−1 −1]


KRISNA MONICA IEA

a) A = [3 −82 −9

]

|A| = |3 −82 −9

| = −27 + 16 = −11

A−1 = −1

11[−9 8−2 3

]

= [

9

11−

8

112

11−

3

11

]

b) B = [3 −65 2

]

|B| = |3 −65 2

| = 6 + 11 = 17

B−1 =1

17[

2 6−5 3

]

= [

2

17

6

17

−5

17

3

17

]

c) C = [3 5

−1 −1]

|B| = |3 5

−1 −1| = −3 − (−5) = 2

B−1 =1

2[−1 −51 3

]

= [−

1

2−

5

21

2

3

2

]

8. Tentukan invers dari matriks-matriks berikut:

a) 𝐷 = [4 2 0

−2 1 11 −2 1

]

b) 𝐸 = [3 1 2

−2 5 14 −2 1

]


KRISNA MONICA IEA

a. 𝐷 = [4 2 0

−2 1 11 −2 1

]

|𝐷| = |4 2 0

−2 1 11 −2 1

|4 2

−2 11 −2

|𝐷| = (4 + 2 + 0) − (0 − 8 − 9)

= 6 + 12 = 18

𝐷11 = + |1 1

−2 1| = 3

𝐷12 = − |−2 11 1

| = 3

𝐷13 = + |−2 11 −2

| = 3

𝐷21 = − |2 0

−2 1| = −2

𝐷22 = + |4 0

−2 1| = 4

𝐷23 = − |4 21 −2

| = 10

𝐷31 = + |2 01 1

| = 2

𝐷32 = − |4 0

−2 1| = −4

𝐷33 = + |4 2

−2 1| = 8

𝐴𝑑𝑗 𝐷 = 𝐷𝑇 = [3 −2 23 4 −43 10 8

]

𝐷−1 =1

det 𝐷𝑥 𝐴𝑑𝑗 𝐷

𝐷−1 =1

18𝑥 [

3 −2 23 4 −43 10 8

]

𝐷−1 =

[ 1

6⁄ −19⁄

19⁄

16⁄

29⁄ −2

9⁄

16⁄

59⁄

49⁄ ]

b. 𝐸 = [3 1 2

−2 5 14 −2 1

]

|𝐸| = |3 1 2

−2 5 14 −2 1

|3 1

−2 54 −2


KRISNA MONICA IEA

|𝐸| = (14 + 4 + 8) − (40 − 6 − 2)

= 27 − 32 = −5

𝐸11 = +|5 1

−2 1| = 3

𝐸12 = −|−2 14 1

| = 6

𝐸13 = +|−2 54 −2

| = −5

𝐸21 = −|1 2

−2 1| = −5

𝐸22 = +|3 24 1

| = −3

𝐸23 = −|3 14 −2

| = 10

𝐸31 = +|1 25 1

| = −6

𝐸32 = −|3 2

−2 1| = −7

𝐸33 = +|3 1

−2 5| = 17

𝐴𝑑𝑗 𝐸 = 𝐸𝑇 = [3 −5 −66 −3 −7

−5 10 17]

𝐸−1 =1

det 𝐸𝑥 𝐴𝑑𝑗 𝐸

𝐸−1 = −1

5𝑥 [

3 −5 −66 −3 −7

−5 10 17]

𝐸−1 =

[ 7

5⁄ −1 −65⁄

65⁄ −3

5⁄ −75⁄

−1 2 175⁄ ]

9. Dengan menggunakan invers matriks selesaikan sistem persamaam linear berikut:

a. 5𝑥 + 4𝑦 = 20

2𝑥 − 3𝑦 = 3

b. 3𝑥 − 𝑦 = 7

7𝑥 − 2𝑦 = 16

a). 5𝑥 + 4𝑦 = 20

2𝑥 − 3𝑦 = 3

[5 42 −3

] [𝑥𝑦] = [

203

]


KRISNA MONICA IEA

A x = B

|𝐴| = |5 42 −3

| = −15 − 8 = −23

x = 𝐴−1 B

[𝑥𝑦] = −

1

23[−3 −4−2 5

] [203

]

[𝑥𝑦] = −

1

23[−72−25

]

[𝑥𝑦] = [

7223⁄

2523⁄

]

Jadi, x = 7223⁄

y = 2523⁄

b). 3𝑥 − 𝑦 = 7

7𝑥 − 2𝑦 = 16

[3 −17 −2

] [𝑥𝑦] = [

716

]

A x = B

|𝐴| = |3 −17 −2

| = −6 − (−7) = 1

x = 𝐴−1 B

[𝑥𝑦] =

1

1[−2 1−7 3

] [716

]

[𝑥𝑦] = [

2−1

]

[𝑥𝑦] = [

2−1

]

Jadi, x = 2


KRISNA MONICA IEA

y = −1

10. Dengan menggunakan invers matriks selesaikan persamaan linear berikut:

2𝑥 + 𝑦 − 2𝑧 = 4

a). 3𝑥 + 2𝑦 − 4𝑧 = 6

−5𝑥 − 3𝑦 + 6𝑧 = −10

3i1 + 4i2 − 2i3 = 3

b). 6i1 + 43 − 3i3 = 2

−3i1 − 2i2 + 25 = 2

a) [2 1 −23 2 −4

−5 −3 6] [

𝑥𝑦𝑧] = [

46

−10]

A x = B

|𝐴| = |2 1 −23 2 −4

−5 −3 6|

2 13 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 −22 −4

|


KRISNA MONICA IEA

𝐷32 = − |2 −23 −4

|

𝐷33 = |2 13 2

|

𝑎𝑑𝑗𝐴 = [𝐴𝑇] = [0 0 02 2 21 1 1

]

x = 𝐴−1 B

x = 1

det𝐴𝑥 𝐴𝑑𝑗 𝐴 𝑥 𝐵

[𝑥𝑦𝑧] =

1

0[0 0 02 2 21 1 1

] [46

−10]

[𝑥𝑦𝑧] = 𝑡𝑖𝑑𝑎𝑘 𝑡𝑒𝑟𝑑𝑒𝑓𝑖𝑛𝑖𝑠𝑖

b) [3 4 −26 3 −3

−3 −2 5] [

𝑖1𝑖2𝑖3

] = [322]

|𝐴| = |3 4 −26 3 −3

−3 −2 5|

3 46 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 −23 −3

| = −12 − (−6) = −6

𝐷32 = − |3 −26 −3

| = −9 − (−12) = −3

𝐷33 = |3 46 3

| = 9 − 24 = −15

𝑎𝑑𝑗𝐴 = [𝐴𝑇] = [9 −16 −6

−21 9 −3−3 −6 −15

]


KRISNA MONICA IEA

x = 𝐴−1 B

x = 1

det𝐴𝑥 𝐴𝑑𝑗 𝐴 𝑥 𝐵

[𝑖1𝑖2𝑖3

] = −1

51[

9 −16 −6−21 9 −3−3 −6 −15

] [322]

[𝑖1𝑖2𝑖3

] = [

1751⁄

11

]

Jadi, i1 = 1751⁄

i2 = 1

i3 = 1

matematika bab 3 “matriks”

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