Mat1503 May /June 2014 Solutions

May-June 2014 Examination paper

Question
 1 
1 2 0
a)(i)
0 1 1
(ii) x = −2y ∴ x = −2

b)
 
1 1 −1 4
2 1 3 0
0 1 −5 8
−2R1 + R2 → R2
 
1 1 −1 4
0 −1 5 −8
0 1 −5 8
−R2 → R2
 
1 1 −1 4
0 1 −5 8
0 1 −5 8
−R2 + R3 → R3
 
1 1 −1 4
0 1 −5 8
0 0 0 0




1

, Question
1.
c)     
1 −2 2 1 2 −4 1 0 0
i) AB = 2 1 1 −1 −1 3 = 0
     1 0 = I3
1 0 1 −1 −2 5 0 0 1
    
1 2 −4 1 −2 2 1 0 0
BA = −1 −1 3  2 1 1 = 0 1 0  = I3
−1 −2 5 1 0 1 0 0 1
∴ B = A−1

ii) AX = Y

A−1 (AX) = A−1 Y

IX = A−1 Y

X = A−1
    
1 2 −4 3 11
X = −1 −1 3   0  =  −9  ∴ x1 = 11, x2 = −9, x3 = −13
−1 −2 5 −2 −13




1

, Question
1 d)
Since C is an inverse of B,

we have
CB = I.



Multiplying both sides on the right by D gives

(CB)D = ID = D.



But we also have by the associative property, that

(CB)D = C(BD) = CI = C



since D is an inverse and so we get that C = D




1