Fsc Part 1 Mathematics (Complete Solution)

Q1
Solve the following systems of linear equations by cramer's rule.

(i)  
2x+2y+z=3 
3x-2y-2z=1
5x+y-3z=2
}

SOLUTION:

lAl= 
 
2 1
3 -2 -2
5 1 -3
 

=2(6+2)-2(-9+10)+1(3+10)

=16-2+13

=27

lA1l= 
 
2 1
1 -2 -2
2 1 -3
 

=3(6+2)-2(-3+4)+1(1+4)

=24-2+5

=27

x=
lA1l
lAl
=
27
27

=1

lA2l= 
 
2 3 1
3 1 -2
5 2 -3
 

=2(-3+4)-2(-9+10)+1(6-5)

=2-3+1

=0

y=
lA2l
lAl
=
0
27

=0

lA3l= 
 
2 2 3
3 -2 1
5 1 2
 

=2(-4-1)-2(6-5)+3(3+10)

=-10-2+39

=27

z=
lA3l
lAl
=
27
27

=1

So,

x=1   ,y=0  and  z=1

(ii)  
      2x1-x2+x3=5 
4x1+2x2+3x3=8
     3x1-4x2-x3=3
}

SOLUTION:

lAl= 
 
-1 1
4 2 3
3 -4 -1
 

=2(-2+12)+1(-4-9)+1(-16-1)

=20-13-22

=-15

lA1l= 
 
-1 1
8 -2 3
3 -4 -1
 

=5(-2+12)+1(-8-9)+1(32-6)

=50-17-38

=-5

x1=
lA1l
lAl
=
-5
-15
=
1
lA2l= 
 
2 5 1
4 8 3
3 3 -1
 

=2(-8-9)-5(-4-9)+1(12-24)

=-34+65-12

=19

x2=
lA2l
lAl
=
19 
-15
lA3l= 
 
2 5 1
8 3
3 3 -1
 

=2(6+32)+1(12-24)+5(-18-6)

=76-12-110

=-46

x3=
lA3l
lAl
=
-46
-15
=
46
15

So,

x1=1/3   ,x2=-19/15  and  x3=46/15


(iii)  
      2x1-x2+x3=8 
     x1+2x2+2x3=6
     x1-2x2-x3=1
}

SOLUTION:

lAl= 
 
-1 1
1 2 2
1 -2 -1
 

=2(-2+4)+1(-1-2)+1(-2-2)

=4-3-4

=-3

lA1l= 
 
8 -1 1
6 2 2
1 -2 -1
 

=8(-2+4)+1(-6-2)+1(-12-2)

=16-8-14

=-6

x1=
lA1l
lAl
=
-6
-3
= 2
lA2l= 
 
2 8 1
1 6 2
1 1 -1
 

=2(-6-2)-8(-1-2)+1(1-6)

=-16+24-5

=3

x2=
lA2l
lAl
=
3
-3

=-1

lA3l= 
 
2 -1 8
1 2 6
1 -2 1
 

=2(2+12)+1(1-6)+8(-2-2)

=28-5-32

=-9

x3=
lA3l
lAl
=
-9
-3
= 3

So,

x1=2   ,x2=-1  and  x3=3

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