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)

 1 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