This document provides questions and solutions for a mock test on engineering concepts. Q1 asks about matrix properties related to singularity. Q2 solves a linear system using Gaussian elimination and Cramer's rule. Q3 finds the eigenvalues and eigenvectors of a given matrix. The solutions show the step-by-step working and reasoning for each question.

1. School of Engineering – UCSI UNIVERSITY EE107 (Jan-April 2013)
Mock Test1 (XY Feb 2013): Questions & Solutions
a b
Q1 Consider the matrix A   .
c d 
(a) Find the value of , as a function of a, b, c, d and , that makes A
nonsingular.
(0.5 marks)
(b) Find the value of  that makes A nonsingular.
(0.5 marks)
(c) If  = , find the value of a that makes A singular.
(0.5 marks)
 bc
Q1. (a).  
ad
(0.5 mark)
 ad
(b).  
bc
(0.5 mark)
bc
(c). a 
d
(0.5 mark)
Q2. Solve the linear system by (a) Gauss elimination and (b) Cramer Rule:
AX  B
1 3 1   x   1 
 2 1 0   y    2
    
1 1 2   z   3 
    
Solution: Using Gaussian Elimination Method:
1 3 1 1 
 
 Augmented  Matrix  2 1 0 2
1 1 2 3 
 
 R3  2R3  R2  R2  R2  2R1  R3  5R3  R2
1 3 1 1  1 3 1 1 1 3 1 1
     
 2 1 0 2 0  5  2 0  0  5  2 0 
0 1 4 4
  0 1
 4 4  0 0 18 20
 
1

2. School of Engineering – UCSI UNIVERSITY EE107 (Jan-April 2013)
 18z  20  z  10 / 9
 5 y  2z  0
 5 y  2(10 / 9)  0  y  4 / 9
x  3y  z  1
x  3(4 / 9)  (10 / 9)  1  x  11/ 9
(2 marks)
(b) Using Cramer’s Rule:
1 3 1
 A1  2 1 0
A
x 1  
A
3
 1 2

2 1 1 3
 A1  1 02  1  10  11
3 1 2 1
 From  (a)  knows  that : A  9
x  11/ 9
1 1 1
 A2  2 2 0
A
 y 2  
A
1
 3 2
2 2 1 1
 A2  1 02  400 4
1 3 1 2
y  4 / 9
1 3 1
 A3  2 1 2
A
z 3  
A
1
 1 3
1 2 2 2 2 1
 A3  1 3 1  1  12  1  10
1 3 1 3 1 1
z  10 / 9
(2marks)
 5 2 
Q3. Given A    ; find the eigenvalues and corresponding eigenvectors.
 2  2
 The eigenvalues are determined via characteristic equation:
A  I  0
2

3. School of Engineering – UCSI UNIVERSITY EE107 (Jan-April 2013)
 5 2  1 0
     0 1   0
 2  2  
  5 2   0 
    0
 2  2  0  
 5   2 
  0
 2  2   
  5    2     (2)(2)  0
 10  5  2  2  4  0
 2  7  6  0
   6   1  0
 1  6 2  1
 The corresponding eigenvectors are determined via:
 A  I X  0
 when   1  6 :
 A  (6) I X  0
  5 2  1 0   x 
  2 2   (6) 0 1   y   0

     
1 2   x 
    0
 2 4  y 
 x  2y  0
1
 Lets  x  k  y   k
2
 1 
 Hence,the  eigenvecto r  is :  
  1 / 2
 when   2  1 :
 A  (1) I X  0
  5 2  1 0   x 
  
 2  2  (1) 0 1   y   0
     
 4 2   x 
    0
 2  1  y 
 4 x  2 y  0
 Lets  x  k  y  2k
1
 Hence,the  eigenvecto r  is :  
 2
(3 marks)
3