Introduction to Optimum Design 4th Edition (2016) - Jasbir Singh Arora - Solutions Manual (PDF)

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SOLUTIONS MANUAL

, APPENDIX
A
Vector and Matrix Algebra


A.1
Evaluate the following determinant:
Solution:
2 1 3 0 3 1 3 1
1 2 1 = 1 2 1 = (–1) 1+2
(1)  = (–1)[( 3  2 ) – ( 5 1 )] = 1
5 2
3 1 5 0 5 2
A.2
Evaluate the following determinant:
Solution:

0 2 3 2 0 3 20
0 0 2 1 3
0 4 5 4 0 1 0 = (–1)2+3 (–1) 1 3 1 = (–1)1+3 (2)
0 = 2(–2 + 9) = 14
=
1 2 2 1 1 3 2 1 3 2 1 3 2
3 1 2 1 3 2 2 1
A.3
Evaluate the following determinant:
Solution:

0 0 0 2 2 3 3 2
1
0 0 5 3 = 0 1 1 = 2 x 1 x 5 x (–2) = –20
0 1 1 1 0 0 5 3
2 3 3 2 0 0 0 2
A.4
Calculate values of the scalar 𝜆𝜆 for which the determinant vanishes:
Solution:
2 1 0 2 1
1 3 0 = (2 –  ) 1 3   = (2 –  )[(2 –  )(3 –  ) – 1] = 0;
0 3 2  




or ( 2   )(  2 - 5  5)  0; ;  1 = (5 – 5 )/2,  2 = 2,  3 = (5 + 5 )/2



Arora, Introduction to Optimum Design, 3e A-1

, Appendix A Vector and Matrix Algebra


A.5
Calculate values of the scalar 𝜆𝜆 for which the determinant vanishes:
Solution:

2 2 0
2 2
1 2 0 = (2 –  ) = (2 –  )[(2 –  )2 – 2] = 0;
1 2  
0 0 2  



or ( 2   )(  2 - 4   2)  0; 1 = 2 – 2 , 2  2 , 3  2  2
A.6
Determine rank of the following matrix:  
Solution:
 3 0 1 3  2 0 3 2 1 0 3/ 2 1 1 0 3/ 21
 0 1 0 1
2 0 3 2 2 1 2 1 1 5 1 5
  ~   ~   ~   ~
 0 2 8 1  3 0 1 3 0 0 7/ 2 0 0 0 7/ 2 0
   1 0 2 8 1 0 0 2 3
 2 1 2 1  0 2 8     
1 0 3/ 2 1 1 0 0 0
0
1 5 1 0 1 0 0
  ~   ; Rank is 4.
0 0 1 0 0 0 1 0
0  0 0 1
 0 0 1   0 
A.7
Determine rank of the following matrix:
Solution:
1 2 2 2 4 1 0
0 0 0 0 1 0 0 0 0 1 0 0 00
 3 4 1 2 1 0 1 4 2 1 0 1 0 00
1 6 3 0  ~   ~   ~   ~
2
1 23 23 53 12 0 0 2 1 1 6 0 1 2 1 6 0 0 2 3 7
1 0 3 0 0 1 3 3 0 0 1 3 3
   3     
1 0 0 0 0  1 0 0 0 0
0   0 0
 1 0 0 0  0 1 0  ; Rank is 4.
~
0 0 1 0 0  0 0 1 
0 0
0 2 0 1 0
 0 0 13 /   0 0 




Arora, Introduction to Optimum Design, 4e A-2

, Appendix A Vector and Matrix Algebra

A.8
Determine rank of the following matrix:
Solution:
1
0 20 30 41  1
0
0 0 0  1 0 0 0  1 0 0 0  1 0 0 0 
1  0 0 0 1  0 0 0 1  0 1 0 0 
   0 0       
3 2 3 0  ฀ 0 4 6 12 ฀ 0 1 5  4  ฀ 0 1 0 0  ฀ 0 0 1 0  ;
 0
2 3 1 4  0  4  0 4 6 12 0 18 4  0 0 0 1
   1 5       
2 0 6 0 0 4 0  8  0 4 0  8  0 0 0 8 0 0 0 0
1         
 2 1 4  0 0 2 0  0 0 2 0  0 0 2 0  0 0 0 0
Rank is 4.
A.9
Obtain solution of the following equations using the Gaussian elimination procedure:
Solution:
2x1+2x2+x3=5
x1−2x2+2x3=1
x2+2x3=3

2 2 1 5 1 1 1/ 2  1 1  1 1 
A = 1 2 2 1 ~ 0 3 3/ 2 3/ 2 ~ 0 1 1/ 2 1/ 2  ~ 0 1 1/ 2 1/ 2  ;
   
       
0 1 2 3 0 1 2 3  0 0 5/ 2 5/ 2 0 0 1 1 
1 1 1/ 2   x1  5/ 2
0 1 1/ 2  x2  = 1/ 2  ; x = 1; x = (1/2) + (1/2)x = 1; x = (5/2) – x – (1/2)x = 1
     3 2 3 1 2 3
 0 0 1  x3   1 
A.10
Obtain solution of the following equations using the Gaussian elimination procedure:
Solution:
x2−x3=0
x1+x2+x3=3
x1−3x2=−2

0 1 1 0 1 11 3 1 1 1 3 1 1 1 3


A = 1 1 1 3 ~ 0 1 0 ~ 0
1   1 1 0 ~ 0 1 1 0  ~
     
1 3 0 2
  1 3 0 2 0 4 1 5 0 0 5 5
1
0 1 1 0 ; or 0 1
1 1 3 1 1 1  x1   3
1  x2  =  0 ; x = 1; x = x = 1; x = 3 – x – x = 1
       3 2 3 1 1 2
 1
0 0 1 1 0 0 1 x3 




Arora, Introduction to Optimum Design, 4e A-3