Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab




Solution Manual




Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

,Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab




Solutions to Exercises in Chapter 1

Section 1.2
1.1 A matrix is an orthogonal matrix if
XTX = I
Is the following matrix an orthogonal matrix?
 −1 −1 
1
X=  1 −1 
2 −1 1 
 1 1 

Solution:

x={{-1.,-1},{1,-1},{-1,1},{1,1}}/2;
Transpose[x].x//MatrixForm

yields
 1 0 
 0 1 
 

Therefore, X is an orthogonal matrix.

1.2 If
 1 −1   1 1 
A=  B=
 2 −1   4 −1 

does (A + B)2 = A 2 + B 2?
Solution:

a={{1,-1},{2,-1}};
b={{1,1},{4,-1}};
((a+b).(a+b)-a.a-b.b)//MatrixForm

yields
 0 0 
 0 0 
 

Therefore, the expressions are equal.



Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

,Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab




1.3 Given the two matrices
 1 4 −3   4 1 
A=  and B= 2 6 
 2 5 4   0 3 

Find the matrix products AB and BA.

Solution:

 1 4 −3  4 1   12 16 
AB =   = 
 2 6
 2 5 4   0 3   18 44 
 
 4 1   6 21 −8 
   1 4 −3  
BA = 2 6 
 =  14 38 18
 0 3   2 5 4   6 15 12 


Aa={{1,4,-3},{2,5,4}};
Bb={{4,1},{2,6},{0,3}};
Aa.Bb//MatrixForm
Bb.Aa//MatrixForm

1.4 Given the following matrices and their respective orders: A (nm), B (pm), and C (ns).
Show one way in which these three matrices can be multiplied. What is the order of the resulting
matrix?

Solution:
CT ABT → (n  s)T (n  m)( p  m)T → (s  n)(n  m)(m  p) → (s  p)


1.5 Given
 ab b2 
A= 
 −a −ab 
2


Determine A2.

Solution: From Eq. (1.13)




Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab

, Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab




a12   a11 + a12a21 a12 (a11 + a22 )
 a11 a   a 2 
AA = 
a22  =  a (a + a ) a a + a2 
12 11
 a21 a22   a21 
21 11 22 21 12 22

 a2b2 − a2b2 b2 (ab − ab) 
=  =0
 −a (ab − ab ) −a b + a b 
2 2 2 2 2




Aa={{a b, b^2},{-a^2,-a b}};
Aa.Aa//MatrixForm

1.6 Given the matrix
 −4 −3 −1 
A= 2 1 1 
 
 4 −2 4 

Determine the value of 4I − 4A − A2 + A3.

Solution:
 6 11 −3 
A2 =  −2 −7 3 
 −4 −22 10 

 −14 −1 −7 
3 
A = 6 −7 7 
 12 −30 22 

Then,
 1 0 0   −4 −3 −1 
  − 4 
4I − 4 A − A + A = 4 0 1
2 3 0 2 1 1
 
 0 0 1   4 −2 4 
 6 11 −3   −14 −1 −7 

− −2 −7 3  + 6 −7 7 
 −4 −22 10   12 −30 22 

 0 0 0 
= 0 0 0 
 0 0 0 
Mathematica verification
Aa={{-4,-3,-1},{2,1,1},{4,-2,4}};


Solution Manual for Advanced Engineering Mathematics with Mathematica 1st edition by Edward Magrab