Fsc Part 1 Mathematics (Complete Solution)

# Q5 Solve That

If

 ι 2ι 1 -ι

,B =

 -ι 1 2ι ι

and C=

 2ι -1 -ι ι

(i) (AB)C=A(BC)

SOLUTION:

L.H.S=(AB)C

= (

 ι 2ι 1 -ι

 -ι 1 2ι ι

)

 2ι -1 -ι ι

=

 -ι2+4ι2 ι+2ι2 ι-2ι2 1-ι2

 2ι -1 -ι ι

=

 1-4 ι-2 ι+2 1+1

 2ι -1 -ι ι

=

 -3 ι-2 ι+2 2

 2ι -1 -ι ι

=

 -6ι-ι2+2ι 3+ι2-2ι -2ι2+4ι-2ι ι-2+2ι

=

 -4ι+1 -2ι+2 2ι+2 3ι-2

R.H.S=A(BC)

=

 ι 2ι 1 -ι

(

 -ι 1 2ι ι

 2ι -1 -ι ι

)
=

 ι 2ι 1 -ι

 -2ι2+ι ι+ι 4ι2-ι2 -2ι+ι2

=

 ι 2ι 1 -ι

 2-ι 2ι -4+1 -2ι-1

=

 ι 2ι 1 -ι

 2-ι 2ι -3 -2ι_-1

=

 2ι-ι2-6ι 2ι2-4ι2-2ι 2-ι-6ι 2ι+2ι2+ι

=

 -4ι+1 -2ι+2 2ι+2 3ι-2

L.H.S=R.H.S

Hence (AB)C=A(BC)

(ii) (A+B)C=AC+BC

SOLUTION:

L.H.S=(A+B)C

= (

 ι 2ι 1 -ι

+

 -ι 1 2ι ι

)

 2ι -1 -ι ι

=

 ι-ι 2ι+1 1+2ι -ι+ι

 2ι -1 -ι ι

=

 0 2ι+1 1+2ι 0

 2ι -1 -ι ι

=

 0_2ι2-ι 2ι2+ι 4ι2+2ι -2ι-1

=

 -2(-1)-ι 2(-1)+ι 4(-1)+2ι -2ι-1

=

 2-ι -2+ι -4+2ι -2ι-1

R.H.S=AC+BC

AC=

 ι 2ι 1 -ι

 2ι -1 -ι ι

=

 0 -ι-2 2ι-1 0

BC=

 -ι 1 2ι ι

 2ι -1 -ι ι

=

 2-ι 2ι -5 -2ι-1

AC+BC=

 0 -ι-2 2ι-1 0

+

 2-ι 2ι -5 -2ι-1

=

 0+2-ι -ι-2+2ι 2ι+1-5 0-2ι-1

=

 2-ι -2+ι -4+2ι -2ι-1

Hence L.H.S=R.H.S

So proved