Mat1503 Oct /Nov 2014 Solutions

October/November 2014
Question 1
a)  
1 −3
i)coefficient matrix A =
0 2
ii) Using back substitution, we get

2x2 = 6
=⇒ x2 = 6/2 = 3

x1 − 3x2 = 2
x1 = 2 + 3x2 = 2 + 3(3) = 11

b)      
1 2 2 1 1 0
A= ,B= ,C=
3 4 −3 2 2 1
A(BC)
    
1 2 2 1 1 0
=
3 4 −3 2 2 1
  
1 2 4 1
=
3 4 1 2
6 5
=
16 11


(AB)C 
   
1 2 2 1 1 0
=
3 4 −3 2 2 1
  
−4 5 1 0
=
−6 11 2 1
 
6 5
=
16 11
∴ A(BC) = (AB)C
     
1 2 2 1 1 0
ii) A(B + C) = +
3 4 −3 2 2 1
  
1 2 3 1
=
3 4 −1 3


1

,  
1 7
=
5 15
     
1 2 2 1 1 2 1 0
AB + AC = +
3 4 −3 2 3 4 2 1
   
−4 5 5 2
= +
−6 11 11 4
 
1 7
=
5 15
∴ A(B + C) = AB + AC

c) Let D and E be nonsingular matrices then

(DE)(DE)−1

= (DE)(E −1 D −1 )

= D(EE −1 )D −1

= D(I)D −1

= DD −1

=I
 
−1 1
d) If F =
1 0
   
0 −1 0 1
F = −1
−1
=
−1 −1 1 1
       
−1 1 0 1 1 0 0 1 −1 1
FF =−1
= = = F −1 F
1 0 1 1 0 1 1 1 1 0




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