Fsc Part 1 Mathematics (Complete Solution)

# Q9 Solve the following matrix equations for X

(i)

 4 3 2 2

A -

 2 3 -1 -2

=

 -1 -4 3 6

SOLUTION:

 4 3 2 2

A -

 2 3 -1 -2

=
<
 -1 -4 3 6

 4 3 2 2

A =

 2 3 -1 -2

+
<
 -1 -4 3 6

 4 3 2 2

A =

 2-1 3-4 -1+3 -2+6

 4 3 2 2

A =

 1 -1 2 4

Let  A=

 a b c d

then

 4 3 2 2

 a b c d

=

 1 -1 2 4

 4a+3c 4b+3d 2a+2c 2b+2d

=

 1 -1 2 4

Comparing the corresponding elements, we get

4a+3c=1 ...(1),

4b+3d=-1 ....(2),

2a+2c=2  .........(3),

2b+2d=4 ....(4)

eq (1) and (3) gives 4a+3c=1 ....(1) and a+c=1 ......(3)

to find a and c , put c=1-a in (3) ,

then 4a+3(1-a)=1

=> a=-2

then

c=1-a

c=1-(-2)

c=1+2

=>  c=3

eq (2) and (4) give

4b+3d=-1 ...(2)  and b+d=2 ......(4)

To find b and d , put d=2-b from (4) in (2) , then

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

4b-3b=-1-6

=>  b=-7

then

d=2-b

d=2-(-7)

d=2+7

d=9

Hence  A=

 a b c d

=

 -2 -7 3 9

(ii) A

 3 1 4 2

-

 -1 2 3 1

=
<
 2 0 -1 5

SOLUTION:

A

 3 1 4 2

-

 -1 2 3 1

=

 2 0 -1 5

 3 1 4 2

=

 -1 2 3 1

+

 2 0 -1 5

A

 3 1 4 2

=

 -1+2 2+0 3-1 1+5

 3 1 4 2

=

 1 2 2 6

Let  A=

 a b c d

then

 a b c d

 3 1 4 2

=

 1 2 2 6

 3a+4b a+2b 3c+4d c+2d

=

 1 2 2 6

Comparing the corresponding elements, we get

3a+4b=1  .....(1),

a+2b=2   ....(2),

3c+4d=2 .....(3),

c+2d=6  .......(4)

To find a and b , put a=2-2b from (2) in (1), then

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

6-6b+4b=1

=> b=5/2

then

a=2-2b

a=2-5

=> a=-3

To Find c and d , put  c=6-2d  from (4) in (3) , then

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

18-6d+4d=2

-2d=-16

d=8

then

c=6-2d

c=6-2(8)

c=6-16

c=-10

Hence  A=

 a b c d

=

 -3 5/2 -10 8