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 2 1 3 -2 -2 5 1 -3

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

=16-2+13

=27

lA1l=

 3 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=

 2 -1 1 4 2 3 3 -4 -1

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

=20-13-22

=-15

lA1l=

 5 -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 3
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 4 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=

 2 -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