244 Higher Engineering Mathematics
Dy=∣
∣
∣
∣
∣a 1 c 1
a 2 c 2∣
∣
∣
∣
∣i.e. the determinant of the coefficients left
when they-column is covered up,and D=∣
∣
∣
∣
∣a 1 b 1
a 2 b 2∣
∣
∣
∣
∣i.e. the determinant of the coefficients left
when the constants-column is covered up.Problem 3. Solve the following simultaneous
equations using determinants:3 x− 4 y= 127 x+ 5 y= 6. 5Following the above procedure:
(i) 3x− 4 y− 12 = 0
7 x+ 5 y− 6. 5 = 0(ii)x
∣
∣
∣
∣− 4 − 12
5 − 6. 5∣
∣
∣
∣=−y
∣
∣
∣
∣3 − 12
7 − 6. 5∣
∣
∣
∣=1
∣
∣
∣
∣3 − 4
75∣
∣
∣
∣i.e.x
(− 4 )(− 6. 5 )−(− 12 )( 5 )=−y
( 3 )(− 6. 5 )−(− 12 )( 7 )=1
( 3 )( 5 )−(− 4 )( 7 )i.e.x
26 + 60=−y
− 19. 5 + 84=1
15 + 28i.e.x
86=−y
64. 5=1
43Sincex
86=1
43thenx=86
43= 2and since−y
64. 5=1
43theny=−64. 5
43=− 1. 5Problem 4. The velocity of a car, accelerating at
uniform accelerationabetween two points, is given
byv=u+at,whereuis its velocity when passing
the first point andtis the time taken to pass
between the two points. Ifv=21m/s whent= 3 .5s
andv=33m/s whent= 6 .1s, use determinants tofind the values ofuanda, each correct to 4
significant figures.Substituting the given values inv=u+atgives:21 =u+ 3. 5 a (1)
33 =u+ 6. 1 a (2)(i) The equations are written in the form
a 1 x+b 1 y+c 1 = 0 ,
i.e. u+ 3. 5 a− 21 = 0
and u+ 6. 1 a− 33 = 0(ii) The solution is given by
u
Du=−a
Da=1
D
whereDuis the determinant of coefficients left
when theucolumn is covered up,i.e. Du=∣
∣
∣
∣
∣3. 5 − 216. 1 − 33∣
∣
∣
∣
∣=( 3. 5 )(− 33 )−(− 21 )( 6. 1 )
=12.6Similarly, Da=∣
∣
∣
∣1 − 21
1 − 33∣
∣
∣
∣
=(1)(−33)−(−21)(1)
=− 12and D=∣
∣
∣
∣13. 5
16. 1∣
∣
∣
∣
=(1)(6.1)−(3.5)(1)=2.6Thus
u
12. 6=
−a
− 12=
1
26i.e. u=12. 6
2. 6= 4 .846m/sand a=12
2. 6= 4 .615m/s^2 ,each correct to 4 significant
figures.Problem 5. Applying Kirchhoff’s laws to an
electric circuit results in the following equations:( 9 +j 12 )I 1 −( 6 +j 8 )I 2 = 5
−( 6 +j 8 )I 1 +( 8 +j 3 )I 2 =( 2 +j 4 )Solve the equations forI 1 andI 2