Download Show that when the unit vector j is multiplied by the following

Solution
20.1104 Intro to Eng. Analysis, Studio version
Fall 1998 – Test #2 Oct. 9, 1998
Instructions: 60-min time limit. You can use Maple to do the problems and check your
answers, but clearly show your setup and solution for each problem on the test pages – read each
problem to see what we want specifically.
1 (10 points) i and j are the unit vectors along x and y axes. Write the vector sum (i + j) as a column
matrix with numerical entries. Then do matrix multiplication by hand or hand calculator (show steps)
to find the new vector by appropriate multiplication between (i + j) and the following 3 x 3 matrix:
1 0 0 
0 0  1


0 1 0 
1   0   1 
i  j  0  1  1
0 0 0
1 0 0  1  (1)(1)  (0)(1)  (0)(0)  1
0 0  1 1  (0)(1)  (0)(1)  (1)(0)  0

  
  
0 1 0  0  (0)(1)  (1)(1)  (0)(0)  1
2 (10 points) Use Gauss-Jordan elimination to find the inverse of matrix B shown below. Show your
steps by writing Maple commands such as mulrow(B,1,-1), addrow(B,1,2,1), etc.
 2 3 
B

 3  5
Augment B with the identity matrix, I, and row-reduce.
 2 3
 3 5

1
0
0  addrow( B,2,1,1)  1
    

1 
3
(B 2,2,1,2)  1
addrow

 
0
0
1
 5  3
 3  2
2
1
5
0
Thus:
1  addrow( B1,1,2,3)  1
     
1 
0
2
1
1
 3  2
1
  5  3
1
B    3  2


3 (20 points) Does the following system of equations have a solution? Show your work and explain
your answer.
x1  3x2  2 x3  2 x5  0
2 x1  6 x2  5 x3  2 x4  4 x5  3x6  1
5 x3  10 x4  15 x6  5
2 x1  6 x2  8 x4  4 x5  18 x6  6
1
Solution
With 6 unknowns and only four equations, there are not enough constraints to uniquely determine all
of the variables, x1 – x6. Thus there are either infinite solutions or no solutions. To look for
inconsistencies, set-up the augmented matrix and row-reduce.






1
3
2
2
6
5 2
0
0
5
2
6
0
0
2
0
4
3
10
0
15
8
4
18
0


 1 MAPLE rref 
   

5


6

1
3
0
4
2
0
0
0
1
2
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0 
0 
1 / 3

0 
The last row has 0 = 0, which is always true and offers no additional useful information. There are no
inconsistencies (i.e. 0 = 1) in any other row. Therefore there are an infinite number of solutions to this
system of equations.
4. (10 points) Consider the following augmented matrix for a system of equations. Write out the
equations represented by this augmented matrix and then solve the system.
1 0 0 0 
0 1 2 0


0 0 0 1
From:
1st row: x1 = 0.
2nd row: x2 + x3 = 0.
3rd row: 0 = 1.
The system is inconsistent. From the third row 0 = 1, which can never be true,
thus there are no solutions.
5. (10 points) Matrix A equals matrix B below. Set up a matrix expression that allows you to solve
for a, b, c, and d. Then solve that system using a matrix method of your choice.
bc 
 a b
8 1
A 
,B  


3d  c 2a  4d 
7 6
Since the matrices are equal, each element of the matrices must be equal. Therefore:
a–b=8
b+c=1
3d + c = 7
2a – 4d = 6.
So we have a system of four equations with four unknowns. Setting this up as a matrix yields,
1  1
0 1

0 0

2 0
0   a  8 
1 0   b  1

1 3   c  7 
   
0  4  d  6 
0
2
Solution
Setting up the augmented matrix and row-reducing yields:






1
1
0
0
0
1
1
0
0
0
1
3
2
0
0
4
8


1  rref 


7


6

1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
5
 3
4

1
Thus:
a=5
b = -3
c=4
d=1
6. (20 points) Determine the force developed in each cable used to support the 40-lb crate shown
below. The unit vector along cable AB is – 0.318i – 0.424j + 0.848k. The unit vector along cable AC
is – 0.318i + 0.424j + 0.848k. The unit vector along cable AD is i.
Solve by: drawing a free body diagram (5 points), setting up the appropriate equations in matrix form
[a][x]=[b] (12 points), and solving the system of equations by a matrix method of your choice (3
points).
Free Body Diagram
Vectors:
FAB = -0.318 FAB i - 0.424 FAB j + 0.848 FAB k
FAC = -0.318 FAC i + 0.424 FAC j + 0.848 FAC k
FAD = FAD i
W = -40 lb k
3
Solution
Writing the equilibrium equations yields:

Fx = 0: -0.318 FAB - 0.318 FAC + FAD = 0
Fy = 0: -0.424 FAB + 0.424 FAC = 0
Fy = 0: 0.848 FAB + 0.848 FAC = 40 lb
Setting this up in matrix form:
  0.318  0.318 1  F AB   0 
  
 0.424 0.424 0 

  F AC    0 
 0.848
0.848 0  F AD  40
Augmenting and row-reducing yields,
 0.318  0.318 1 0 
1 0 0 23.6
 0.424 0.424 0 0  rref



 0 1 0 23.6
 0.848
0 0 1 15.0 
0.848 0 40
Thus:
FAB = 23.6 lb
FAC = 23.6 lb
FAD = 15 lb
7. (20 points) Draw appropriate free body diagrams (10 points) for the following problem from
Chapter 3 of our textbook. Then answer the following True-False question, with a short explanation
of your answer (10 points): “T or F: We can solve for the 3 unknown tensions and 1 unknown angle
 using Gauss-Jordan elimination because the system is a set of linear equations in 4 unknowns.”
Free Body Diagrams
FALSE: This problem will not yield a system of linear equations, due to the presence of an unknown
angle, a nonlinear term, cos , is introduced into the equations. Thus the system can not be solved by
Gauss-Jordan elimination, which only works on linear equations.
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