Mat1503 May/June 2013 Solutions

May/June 2013- Examination paper

Question 1

1.1 Describe the elementary row operations on a matrix.
solution:


• Multiply a row through by a nonzero constant


• Add a multiple of one row to another


• Interchange any two rows


1.2 Verify that

x = 19t − 35



y = 25 − 13t



z=t



is a solution of

2x + 3y + z = 5



5x + 7y − 4z = 0



solution:

Substituting for x, y and z in terms of t, we get

2(19t − 35) + 3(25 − 13t) + t
= 38t − 70 + 75 − 39t + t
=5

and

5(19t − 35) + 7(25 − 13t) − 4t
= 95t − 175 + 175 − 91t − 4t
= 4t − 4t

1

, =0

Thus we conclude that x = 19t − 35, y = 25 − 13t, z = t, t ∈ R is a solu-
tion of the system


1.3 a) Compute
   
3 2 1 3 0 −2
−5
5 1 0 1 −1 2
b) Find A in terms of B if 2A − B = 5(A + 2B)

solution:
   
3 2 1 3 0 −2
a) −5
5 1 0 1 −1 2
   
3 2 1 15 0 −10
= −
5 1 0 5 −5 10
−12 2 11
=
0 6 −10


1.3 b)
2A − B = 5(A + 2B)

2A − B = 5A + 10B

2A − 5A = B + 10B

−3A = 11B
11
∴A=− B
3

1.4 a) Given
 
3 −1
A=
0 −2


Compute A2 − A − 6I2

solution:
A2  
3 −1 3 −1
=
0 −2 0 −2
 
3.3 + −1.0 3. − 1 + −1. − 2
=
0.3 + −2.0 0. − 1 + −2. − 2
 
9 −1
=
0 4

2