Fsc Part 1 Mathematics (Complete Solution)

Q1
Evaluate the following determinants.

(i)
 
5 -2 -4
3 -1 -3
-2 1 2
 
SOLUTION:   
 
5 -2 -4
3 -1 -3
-2 1 2
 
 =  5
 
-1 -3
1 2
 
  - (-2) 
 
3 -3
-2 2
 
  + (-4)  
 
3 -1
-2 1
 

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

=5(1)+2(0)-4(1)

=5-4

=1


(ii)
 
5 2 -3
3 -1 1
-2 1 -2
 
SOLUTION:   
 
5 2 -3
3 -1 1
-2 1 -2
 
 =  5
 
-1 1
1 -2
 
  - 2 
 
3 1
-2 -2
 
  + (-3)
 
3 -1
-2 1
 

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

=5(1)-2(-4)-3(1)

=5+8-3

=10


(iii)
 
2 -3
-1 3 4
-2 5 6
 
SOLUTION:   
 
1 2 -3
-1 3 4
-2 5 6
 
 =  1
 
3 4
5 6
 
  - 2 
 
-1 4
-2 6
 
  +  (-3) 
 
-1 -3
-2 5
 

=1(18-20)-2(-6+8)-3(-5+6)

=1(-2)-2(2)-3(1)

=-2-4-3

=-9


(iv)
 
a+l  a-l a
a a+l a-l
a-l a a+l
 
SOLUTION:   
 
a+l a-l a
a a+l a-l
a-l a a+l
 
 =  3al 
 
1 a-l a
1 a+l a-l
1 a a+l
 
   
 =  
 
3a a-l a
3a a+l a-l
3a a a+l
 
  C1+(C2+C3
 =  3a
 
1 a-l a
0 2l -l
0 l l
 
  R2-R1
  R3-R1

Expand C1

 = 3a ( 1
 
2l -l
l l
 
  - 0 
 
a-l  a
l l
 
  + 0 
 
a-l a
2l -l
 
)

=3a [ 1(2l2+l2) +0+0]

=3a(3l2)

=9al2


(v)
 
1 2 -2
-1 1 -3
2 4 -1
 
SOLUTION:   
 
1 2 -2
-1 1 -3
2 4 -1
 
 =  1
 
1 -3
4 -1
 
  - 2 
 
-1 -3
2 -1
 
  + (-2)  
 
-1 1
2 4
 

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

=1(11)-2(7)-2(-6)

=11-14+12

=23-14

=9


(vi)
 
2a a a
b 2b b
c c 2c
 
SOLUTION:   
 
2a a a
b 2b b
c c 2c
 
 =  abc 
 
2 1 1
1 2 1
1 1 2
 
 Taking comman a from R1
 Taking comman b from R2
 Taking comman c from R3

Expand R1

 = abc ( 2
 
2 1
1 2
 
  - 1 
 
1 1
1 2
 
  + 1 
 
1 2
1 1
 
)

= abc[2(4-1)-1(2-1)+1(1-2)]

=abc[2(3)-1(1)+1(-1)]

=abc(6-1-1)

=abc(4)

=4abc

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