Transcript www.eng.usf.edu

Simultaneous Linear Equations
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A farmer had 196 cows. When he
rounded them up, he had 200 cows.
(Reader’s Digest)
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The name of the person in the picture is
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A. A$AP Rocky
B. Kid Cudi
C. MC Hammer
D. T.I.
E. Vanilla Ice
A.
20%
B.
20%
20%
C.
D.
20%
E.
4
9
The size of matrix 
5
A.
3 4
B.
43
C.
3 3
D.
4 4
6
7
2
6
3
7
8
4 is
8 
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2.
3.
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4.
10
The c32 entity of the matrix
4.1 61

[C ]   9
2
 5 6.3
A.
B.
C.
D.
7
3
7.2
8 

4 
8.9
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2.
3.
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2
3
6.3
does not exist
1.
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4.
10
Given
3 6
[A]  
5  9
2
3
6
2
[B]  
 8 9.2
then if [C]=[A]-[B], c23=
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3
6
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33%
A. -3
B. 3
C. 9
1.
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2.
3.
4  6 3 
4 3 




Given A  1 2  8, B   9 7 
6  5  9
4  5
then if [C]=[A][B], then c31=
A.
B.
C.
D.
-57
-45
57
does not exist
.
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2.
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3.
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4.
10
A square matrix [A] is upper triangular if
1.
2.
3.
4.
aij  0, i  j
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2.
3.
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aij  0, j  i
aij  0, i  j
aij  0, j  i
1.
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4.
An identity matrix [I] needs to satisfy the
following
A.
Iij  0, i  j
B.
Iij  1, i  j
C.
matrix is square
D.
all of the above
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2.
3.
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4.
10
The following system of equations
x + y=2
6x + 6y=12
has
solution(s).
1.
2.
3.
4.
no
one
more than one but a finite
number of
infinite
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2.
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10
PHYSICAL PROBLEMS
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Truss Problem
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Pressure vessel problem
c2
u1  c1 r 
r
a
c
b
c4
u 2  c3 r 
r
a
b
4.2857 107

7
4
.
2857

10


 6.5

0

 9.2307 105
0
 5.4619 105
 4.2857 107
 0.15384
6.5
0
4.2857 107
  c1   7.887 103 
  

5.4619 105  c 2  
0


0.15384  c3   0.007 
  

 3.6057 105  c 4  
0

0
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Polynomial Regression
We are to fit the data to the polynomial regression model
(T1,α1 ),(T2 ,α2 ), ...,(Tn-1 ,αn-1 ),(Tn ,αn )
α  a0  a1T  a2T 2

 n


 n
  Ti 
 i n1 
 T 2 
i
 
 i1 
 n

  Ti 
 i 1 
 n 2
  Ti 
 i1 
 n 3
  Ti 
 i1 
n
 n 2 

  Ti  
i


 i1   a   i1
0
n
n



a    T 
3
T
  i   1 

i
i

i 1
 i1    
a2   n
n


2
4

T
  Ti  

i i
 i1
 i 1  
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







END
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Simultaneous Linear Equations
Gaussian Elimination
(Naïve and the Not That So Innocent Also)
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You know Lady Gaga; Who is Shady Gaga
A. Lady Gaga’s sister!
B. A person who looks bad with
their sunglasses on.
C. A person who looks good with
sunglasses on but bad once
he/she takes the sunglasses off
D. That is what Alejandro calls
Lady Gaga
E. Slim Shady’s sister!
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A.
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B.
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C.
D.
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E.
10
The goal of forward elimination steps in
Naïve Gauss elimination method is to
reduce the coefficient matrix to a (an)
_________
matrix.
A.
B.
C.
D.
diagonal
identity
lower triangular
upper triangular
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2.
3.
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4.
One of the pitfalls of Naïve Gauss
Elimination method is
A.
B.
C.
large truncation error
large round-off error
not able to solve
equations with a
noninvertible coefficient
matrix
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1.
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2.
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3.
Increasing the precision of numbers from
single to double in the Naïve Gaussian
elimination method
A.
B.
C.
avoids division by zero
decreases round-off error
allows equations with a
noninvertible coefficient matrix to
be solved
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2
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3
Division by zero during forward elimination
steps in Naïve Gaussian elimination for
[A][X]=[C] implies the coefficient matrix [A]
1.
2.
3.
is invertible
is not invertible
cannot be determined to be invertible
or not
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2.
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3.
Division by zero during forward elimination
steps in Gaussian elimination with partial
pivoting of the set of equations [A][X]=[C]
implies the coefficient matrix [A]
1.
2.
3.
is invertible
is not invertible
cannot be determined to be invertible
or not
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33%
1.
33%
2.
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3.
Using 3 significant digit with chopping
at all stages, the result for the following
calculation is
6.095  3.456  1.99
x1 
8
A.
B.
C.
D.
25%
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25%
B.
C.
25%
-0.0988
-0.0978
-0.0969
-0.0962
A.
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D.
Using 3 significant digits with rounding-off
at all stages, the result for the following
calculation is
6.095  3.456  1.99
x1 
8
A.
B.
C.
D.
25%
25%
25%
B.
C.
25%
-0.0988
-0.0978
-0.0969
-0.0962
A.
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D.
Determinants
If a multiple of one row of [A]nxn is added or subtracted
to another row of [A]nxn to result in [B]nxn then
det(A)=det(B)
The determinant of an upper triangular matrix
[A]nxn is given by detA  a11  a22  ...  aii  ...  ann  a
n
ii
i 1
Using forward elimination to transform [A]nxn to an
upper triangular matrix, [U]nxn.
Ann  U  nn
det  A  det U 
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Simultaneous Linear Equations
LU Decomposition
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You thought you have parking problems.
Frank Ocean is scared to park when
__________ is around.
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B.
C.
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A. A$AP Rocky
B. Adele
C. Chris Brown
D. Hillary Clinton
A.
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D.
Truss Problem
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If you have n equations and n unknowns,
the computation time for forward
substitution is approximately proportional
to
33%
A. 4n
B. 4n2
C. 4n3
33%
33%
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A.
B.
C.
If you have a nxn matrix, the computation
time for decomposing the matrix to LU is
approximately proportional to
33%
A. 8n/3
B. 8n2/3
C. 8n3/3
33%
33%
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A.
B.
C.
LU decomposition method is computationally more
efficient than Naïve Gauss elimination for solving
A.
B.
C.
a single set of simultaneous linear
equations
multiple sets of simultaneous linear
equations with different coefficient
matrices and same right hand side
vectors.
multiple sets of simultaneous linear
equations with same coefficient
matrix and different right hand side
vectors
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2.
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3.
For a given 1700 x 1700 matrix [A], assume that it takes
about 16 seconds to find the inverse of [A] by the use of
the [L][U] decomposition method. The approximate
time in seconds that all the forward substitutions take
out of the 16 seconds is
25%
A.
B.
C.
D.
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2
3
25%
4
6
8
12
1
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4
The following data is given for the velocity of the rocket as a function of time.
To find the velocity at t=21s, you are asked to use a quadratic polynomial
v(t)=at2+bt+c to approximate the velocity profile.
t
(s)
0
14
15
20
30
35
v
m/s
0
227.04
362.78
517.35
602.97
901.67
25%
1.
2.
3.
4.
25%
25%
2.
3.
25%
176 14 1 a  227.04
 225 15 1 b   362.78

  

400 20 1  c  517.35
 225 15 1 a  362.78
400 20 1 b   517.35

  

900 30 1  c  602.97
0 1 a   0 
 0
 225 15 1 b   362.78

  

400 20 1  c  517.35
 400 20 1  a  517.35
 900 30 1 b   602.97

  

1225 35 1  c  901.67
1.
4.
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THE END
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4.09
Adequacy of Solutions
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The well or ill conditioning of a system of
equations [A][X]=[C] depends on the
A. coefficient matrix only
B. right hand side vector only
C. number of unknowns
D. coefficient matrix and the
right hand side vector
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1.
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2.
3.
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4.
The condition number of a n×n diagonal matrix
[A] is
A.
max aii , i  1,...,n
min aii , i  1,...,n
B. max a
C.
, i  1,...,n 
25%
2
ii


1


 min a , i  1,...,n 
ii


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25%
2.
3.
25%
2
D. maxaii , i  1,...,nmin aii , i  1,...,n
1.
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4.
The adequacy of a simultaneous linear system of
equations [A][X]=[C] depends on (choose the most
appropriate answer)
A. condition number of the
coefficient matrix
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25%
2.
3.
25%
B. machine epsilon
C. product of the condition
number of coefficient matrix
and machine epsilon
D. norm of the coefficient matrix
1.
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4.
If cond A and  mach  0.11910 , then
in [A][X]=[C], at least these many
significant digits are correct in your
solution,
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6
25%
A. 0
B. 1
C. 2
D. 3
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2.
3.
4.
THE END
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