MAT2691 ASSIGNMENT 3 SEMESTER 2 2023

MAT2691
ASSIGNMENT 3
SEMESTER 2
2023

, QUESTION 1




Solution:



1.1).


4 −2 5
A=[ ] and B = [ ]
−2 5 1


4 −2 5
AB = [ ][ ]
−2 5 1
20 − 2
=[ ]
−10 + 5
18
=[ ]
−5




BA cannot be computed because the number of columns of B are not equal to the rows of
A.




1.2).


−6 2 −3 4
X=[ ] and Y=[ ]
−5 −11 −1 9

, −6 2 −3 4 −6 2 −3 4
(X − Y)(X + Y) = ([ ]−[ ]) ([ ]+[ ])
−5 −11 −1 9 −5 −11 −1 9
−6 − (−3) 2−4 −6 + (−3) 2+4
=[ ][ ]
−5 − (−1) −11 − 9 −5 + (−1) −11 + 9
−3 −2 −9 6
=[ ][ ]
−4 −20 −6 −3
27 + 12 −18 + 6
=[ ]
36 + 120 −24 + 60
39 −12
=[ ]
156 36




−6 2 −6 2
X2 = [ ][ ]
−5 −11 −5 −11
36 − 10 −12 − 22
= [ ]
30 + 55 −10 + 121
26 −34
= [ ]
85 111


−3 4 −3 4
Y2 = [ ][ ]
−1 9 −1 9
9 − 4 −12 + 36
=[ ]
3 − 9 −4 + 81
5 24
=[ ]
−6 77


26 −34 5 24
X2 − Y2 = [ ]−[ ]
85 111 −6 77
26 − 5 −34 − 24
=[ ]
85 − (−6) 111 − 77
21 −58
=[ ]
91 34



(X − Y)(X + Y) = [ 39 −12] ≠ [21 −58] = X 2 − Y 2
156 36 91 34
(X − Y)(X + Y) ≠ X 2 − Y 2 Shown