NAME:
20.1100
INTRODUCTION TO ENGINEERING ANALYSIS
STUDIO/LAPTOP VERSION
TEST #2-Solution
Spring 1999
February 26
The exam is open book/open notes/open laptop. No internet connection is allowed,
however. You have one hour to complete the exam. You may use Maple to solve any
problem that does not specifically say “solve by hand,” but you should still clearly show
your problem setup and explain your solution method. For those problems that do
require solution by hand, you may use Maple to check your answers. There are six
problems with a total of 110 points (10 bonus points).
1. (25%) Two balls are stacked as below. Ball A has radius 0.15m and mass 5Kg. Ball
B has radius 0.35m and mass 10Kg.
(a) Draw the Free Body Diagram. (10%)
(b) Verify that the force balance equation is of the form (NA = magnitude of the
reaction force from the wall to ball A at contact A, NB = magnitude of the reaction
force from the wall to ball B at contact B, NC = magnitude of the reaction force
from the well to ball B at contact C, ND = magnitude of the reaction force
between balls A and B): (5%)
NA - 0.8 ND = 0
0.6 ND = 49 N
-NB + 0.8 ND = 0
-0.6 ND + NC = 98 N
(c) Write the force balance equation in A x = b form. Is the set of simultaneous
equations consistent? (i.e., does at least one solution exist?) If so, is the solution
unique? You don't have to solve the equations. (10%)
A
A
D
B
C
0.9m
B
NAME:
Solution:
a.)
y
ND


NA
x
98.1 N
49.1 N
ND
NC
where  = arcsin(.4/.5) = 53.13
b.) Writing out the equilibrium equations for the two FBDs, we get:
1. FX = NA - sin ND = 0
FY = cos ND - 49 N = 0
2. FX = -NB + sin ND = 0
FY = - cos ND + NC - 98 N = 0
Substituting for cos and sin, we get the same equations as given above.
c.)
1
0
0

0
0
0
1
0
0 .8N A  0 
0 .6 N B  49 

0 .8 NC  0 
1 .6

ND 
 
98 

The null space of this equation is:
1 0 0 .8
0 0 0 .6
0 1 0 .8
0 0 1 .6

0
0
0
0

NB
NAME:
Right away, we can see that all of the unknowns must be zero to satisfy this matrix
equation. Since the N(A)={0}, there must be one unique solution to this system of
equations.
NAME:
2. (20%) Do the following problems by hand
c.) (5%) Find A+B
 1 0 4
2
A  
B


1 2 3 

 4
T
(b) (5%) Find A
(c) (10%) Find det(C)
1 0
C  1 2

6 0
2 3
1 5

4 
3 
10 

Solution:
c.) Add like elements:
 1  2 0  2 4  3 1 2 1
A  B  

1  4 2  1 3  5 
 
 3 1 8 

c.) Interchange rows and columns:
1 1 
A  0 2 

4 3

T
c.) Use the method of cofactors:
1 0 4 
2
11
det 1 2
3  1(1)
0

6 0 10

det(C)  1(1)(20)  4(1)(12)
det(C)  20  48  28
3
10
 4(1)
13
1 2
6
0
NAME:
(b) (15%)
(b) Find the inverse of A by hand with Gauss-Jordan Elimination (8%)
1 4
A  
2 6 

(c) Find the inverse of A by hand using Cramer's Rule (7%)
Solution:
a.)
To find the inverse with Gauss Jordan Elimination, first augment A with the identity
matrix, the use ERO's:
1 4

2 6
1 0 
1 4
 2R1  R2 

0 1 

0 2
 4R2  R1 
1 0

0 1
1 0 
1 4
 R2 /(2) 

2 1 

0 1
3
2 
1 1/ 2

3 2 
1
A  
1 1/ 2

b.)
Using Cramer's rule, first find the matrix of cofactors:
C11  (1)
11
(6)
C12  (1)1 2 (2)
C21  (1)21 (4)
C22  (1)2 2 (1)
6 2 
C  
4 1 

The adjoint will be the transpose of this matrix:
6
adjoint( A)  C T  
2
4
1

0 
1 1/ 2 

1
NAME:
Next, find the determinant of A (this is easy for a 2x2 matrix):
det( A)  6  (8)  2
And finally, the inverse is the adjoint divided by the determinant (i.e., divide each
element of the adjoint by det(A)):
3 2 
adjoint A / det( A)  
1 1/ 2 

Which is the same answer we got in part a.
NAME:
4. (15%)
(a) Write the equations corresponding to the following equation (where a is a
constant) (5%)
1 0
1 2

2 0
4x1  2 
3 x 2   0 
8

a 

x3 
 
(b) (10%) For what value(s) of a does there exist one and only one solution?
infinitely many solutions? no solution?
Solution:
a.)
The equations are:
X1 - 4 X3 = 2
- X1 + 2 X2 + 3 X3 = 0
2 X1 - 8 X3 = a
b.)
Augment A with b and use ERO's:
1 0 4 2 
1 0 4
1 2 3 0  2R1  R3  1 2 3


2 0 8 a 

0 0 0
2 
0 
4  a

The last row says that 0 = -4 + a
If -4 + a does not equal zero, there will be no solution.
For a solution to exist, -4 + a = 0, and a = 4.
This solution can never be unique, since we will be left with two equations and three
unknowns.
Summary:
No solution: a  4
Infinite solution: a = 4
Unique solution: No a exists to satisfy this condition.
NAME:
NAME:
5. (25%)
(a) (10%) Consider the vectorial equation A x = b:
 1 1
 1 
x1 
 0 2    0 
x 2   

4 3 

1
i.
Show that the equation has no solution. (5%)
ii.
Find the vector x such that ||Ax-b|| is minimized. (5%)
(b) (15%) Consider another vectorial equation A x = b:
1 0

1 2
i.
ii.
iii.
x 
4 1
1 
x 2  
3 
  
1

x3 
Show that the equation has infinitely many solutions. (5%)
Parameterize all the solutions by x2. (5%)
Find the vector x that is of the minimum norm among all the solutions.
(5%)
Solution:
a.) i. Using Gauss Jordan Elimination:
1 1 1 
1 1 1
1 1 1
0 2
0  4R1 R3  0 2 0 2R3  R2  0 0 6



4 3 1

0 1 3

0 1 3

The second equation says that 0=6, which can never be true. Therefore, this system of
equations has no solution.
ii. Since this is a "tall" matrix, we want to find
x = (ATA)-1ATb
Using Maple, we find that:

1 0 4


1 2 3


1
 1 1
1 
1 0 4 
 6 / 23 
 0 2  
 0  

 1 2 3
1/ 23





4 3 
1
NAME:
b.) i.
Using Gauss Jordan Elimination:
1 0 4 1 
1 0 4 1 
 1 0 4

R1

R2


R2
/
2




1 2 3 1

0 2 1 0 

0 1 1/ 2
1
0

These equations can be written out as:
X1 - 4X3 = 1
X2 -X3/2 = 0
These two equations have three unknowns; therefore, an infinite number of solutions
exist.
ii. We can write this solution parametrically (in terms of X2) as:
1  8x 2  1  8x 2 
x   x2  0   x2 

0 
 
 2x 
 
2x 

2
2
iii. Since this is a "fat" matrix, we want to find
x = AT(AAT)-1b
Using Maple, we find that:
 1 1
1 0 4
 0 2 
1 2 3



4 3 
1
 1 1
 5/ 69 
 1 
 0 2      8 / 69 
 1


4 3 

16 / 69

NAME:
6. (10%)
Find the null space and range space of the following matrix
1 0
A  1 2

2 0
4
3 
8

Solution:
To find the null space of A, augment A with the '0' vector and use Gauss Jordan
Elimination:
1 0 4 0 
1 0 4 0 
1 0 4 0 
1 2 3 0  2R1 R3  1 2 3 0  R1 R2  0 2 1 0 



2 0 8 0 

0 0 0 0 

0 0 0 0 

Writing out these equations, we get:
X1 - 4X3 = 0
2X2 - X3 = 0
So the null space is:
4x 3   4 
x
N(A)   3 2  1/ 2 x 3

 1 

 x 
 
3
where X3 is arbitrary.
To find the range space of A, augment A with an arbitrary 'b' vector and use Gauss
Jordan Elimination:
1 0 4 b1 
1 0 4
1 2 3 b2  2R1 R3  1 2 3


2 0 8 b3 

0 0 0
Writing out the last equation, we get:
-2b1 + b3 = 0
b1 
1 0 4
b2  R1  R2  0 2 1

2b1  b3 

0 0 0
b1 
b1  b2 
2b1  b3 

NAME:
So the range space is:
 b1
R(A)   b2

2b
1




where b1 and b2 are arbitrary.