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=
 
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
 
 
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=
 
b
c d
 
 =  
 
-3 5/2
-10 8
 
 

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