280 MATRICES AND DETERMINANTS
Following 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
43then y=−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=21 m/s
whent= 3 .5 s andv=33 m/s whent= 6 .1s,
use determinants to find 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
DwhereDuis 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)=2.6Thusu
12. 6=−a
− 12=1
26i.e. u=12. 6
2. 6= 4 .846 m/sand a=12
2. 6= 4 .615 m/s^2 ,each correct to 4 significant
figuresProblem 5. Applying Kirchhoff’s laws to an
electric circuit results in the following equations:(9+j12)I 1 −(6+j8)I 2 = 5
−(6+j8)I 1 +(8+j3)I 2 =(2+j4)Solve the equations forI 1 andI 2Following the procedure:(i) (9+j12)I 1 −(6+j8)I 2 − 5 = 0−(6+j8)I 1 +(8+j3)I 2 −(2+j4)= 0(ii)I 1
∣
∣
∣
∣−(6+j8) − 5
(8+j3) −(2+j4)∣
∣
∣
∣=−I 2
∣
∣
∣
∣(9+j12) − 5
−(6+j8) −(2+j4)∣
∣
∣
∣=1
∣
∣
∣
∣(9+j12) −(6+j8)
−(6+j8) (8+j3)∣
∣
∣
∣