Higher Engineering Mathematics, Sixth Edition

The solution of simultaneous equations by matrices and determinants 247

8. Kirchhoff’s laws are used to determine the
   current equations in an electrical network and
   show that
   i 1 + 8 i 2 + 3 i 3 =− 31
   3 i 1 − 2 i 2 +i 3 =− 5
   2 i 1 − 3 i 2 + 2 i 3 = 6
   Use determinants to find the values ofi 1 ,i 2
   andi 3. [i 1 =− 5 ,i 2 =− 4 ,i 3 =2]


9. The forces in three members of a framework
   areF 1 ,F 2 andF 3. They are related by the
   simultaneous equations shown below.
   
   
   1. 4 F 1 + 2. 8 F 2 + 2. 8 F 3 = 5. 6
   
   
   2. 2 F 1 − 1. 4 F 2 + 5. 6 F 3 = 35. 0
   
   
   3. 2 F 1 + 2. 8 F 2 − 1. 4 F 3 =− 5. 6
      Find the values of F 1 , F 2 and F 3 using
      determinants.
      [F 1 = 2 ,F 2 =− 3 ,F 3 =4]
   
   
   
   


10. Mesh-current analysis produces the following
    three equations:
    20 ∠ 0 ◦=( 5 + 3 −j 4 )I 1 −( 3 −j 4 )I 2
    10 ∠ 90 ◦=( 3 −j 4 + 2 )I 2 −( 3 −j 4 )I 1 − 2 I 3
    − 15 ∠ 0 ◦− 10 ∠ 90 ◦=( 12 + 2 )I 3 − 2 I 2
    Solve the equations for the loop currentsI 1 ,I 2
    andI 3.
    ⎡
    ⎣

I 1 = 3. 317 ∠ 22. 57 ◦A

I 2 = 1. 963 ∠ 40. 97 ◦A

I 3 = 1. 010 ∠− 148. 32 ◦A

⎤

⎦

23.3 Solution of simultaneous

equations using Cramers rule

Cramers rule states that if

a 11 x+a 12 y+a 13 z=b 1

a 21 x+a 22 y+a 23 z=b 2

a 31 x+a 32 y+a 33 z=b 3

then x=

Dx

D

,y=

Dy

D

andz=

Dz

D

where D=

∣ ∣ ∣ ∣ ∣ ∣

a 11 a 12 a 13

a 21 a 22 a 23

a 31 a 32 a 33

∣ ∣ ∣ ∣ ∣ ∣

Dx=

∣ ∣ ∣ ∣ ∣ ∣

b 1 a 12 a 13

b 2 a 22 a 23

b 3 a 32 a 33

∣ ∣ ∣ ∣ ∣ ∣

i.e. thex-column has been replaced by the R.H.S.

bcolumn,

Dy=

∣ ∣ ∣ ∣ ∣ ∣

a 11 b 1 a 13

a 21 b 2 a 23

a 31 b 3 a 33

∣ ∣ ∣ ∣ ∣ ∣

i.e. they-column has been replaced by the R.H.S.

bcolumn,

Dz=

∣ ∣ ∣ ∣ ∣ ∣

a 11 a 12 b 1

a 21 a 22 b 2

a 31 a 32 b 3

∣ ∣ ∣ ∣ ∣ ∣

i.e. thez-column has been replaced by the R.H.S.

bcolumn.

Problem 7. Solve the following simultaneous

equations using Cramers rule.

x+y+z= 4

2 x− 3 y+ 4 z= 33

3 x− 2 y− 2 z= 2

(This is the same as Problem 2 and a comparison of

methods may be made). Following the above method:

D=

∣ ∣ ∣ ∣ ∣ ∣

111

2 − 34

3 − 2 − 2

∣ ∣ ∣ ∣ ∣ ∣

= 1 ( 6 −(− 8 ))− 1 ((− 4 )− 12 )

+ 1 ((− 4 )−(− 9 ))= 14 + 16 + 5 = 35

Dx=

∣ ∣ ∣ ∣ ∣ ∣

411

33 − 34

2 − 2 − 2

∣ ∣ ∣ ∣ ∣ ∣

= 4 ( 6 −(− 8 ))− 1 ((− 66 )− 8 )

+ 1 ((− 66 )−(− 6 ))= 56 + 74 − 60 = 70

Dy=

∣ ∣ ∣ ∣ ∣ ∣

14 1

233 4

32 − 2

∣ ∣ ∣ ∣ ∣ ∣

= 1 ((− 66 )− 8 )− 4 ((− 4 )− 12 )+ 1 ( 4 − 99 )

=− 74 + 64 − 95 =− 105

Dz=

∣ ∣ ∣ ∣ ∣ ∣

114

2 − 333

3 − 22

∣ ∣ ∣ ∣ ∣ ∣

= 1 ((− 6 )−(− 66 ))− 1 ( 4 − 99 )

+ 4 ((− 4 )−(− 9 ))= 60 + 95 + 20 = 175