ASSIGNMENT 3 SOLUTIONS

MAT2691 – SOLUTIONS – ASSIGNMENT 3 – SEMESTERS 1+2
MPHO DIKO: TUTOR
TOTAL MARKS = 43



QUESTION 1 [marks = 3]



 3 1 1
Given A    and B    . Determine AB and BA.
4 7   5

SOLUTION


For AB:
Putting the orders of A and B in juxtaposition (i.e. side by side): 2  2 2 1.
Therefore, the product AB exists (i.e. it is defined or it is possible to compute it).


 3 1  1 
AB    
 4 7   5 
 31   1 5  
 
  4 1   7  5  
 8 
 
 31


For BA:

The orders of B and A: 2 1 2 2

Therefore, the product BA does not exist!




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, QUESTION 2 [marks = 10]


1 2   3 1 
Given X   and Y    , show that  X  Y  X  Y   X  Y
2 2
 
 0 3   2 1

SOLUTION


1 2   3 1   2 1
XY     
0 3   2 1  2 2 


1 2   3 1  1 2   3 1  4 3
XY       
0 3   2 1 0 3   2 1   2 4 


 2 1  4 3  2  4    1 2   2  3   1 4   6 2
 X  Y  X  Y     

 2 2   2 4    2  4    2  2   2  3   2  4    4 2


1 2 1 2 11   2  0  1 2    2  3 1 8
X2 =   

0 3  0 3    0 1   3 0   0  2    3 3  0 9 


 3 1   3 1    3 3  1 2   31  1 1  11 4 
Y2    

 2 1  2 1  2  3   1 2   2 1   1 1 8 3 


1 8  11 4  1 8  11 4   10 4 
X2  Y2      
0 9   8 3  0 9   8 3  8 6 


Therefore,


 6 2   10 4 
 4 6   8 6 
  




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