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Simultaneous Linear Equations

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The size of matrix   4   9 5 A.

B.

3  4 4  3 C.

D.

3  3

4

4

6 2 6 7 3 7 8 4   8   is

25% 25% 25% 25% 1.

2.

3.

4.

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10

The c

32

entity of the matrix [

C

]      4 .

1 9 5 A.

B.

C.

D.

61 2 6 .

3 2 3 6.3

does not exist 7 3 7 .

2 8 4 8 .

9    

25% 25% 25% 25% 1.

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2.

3.

4.

10

Given [

A

]    3 5  6 9 2 3   [

B

then if [C]=[A]+[B], c 12 = ]     2 8 6 9 .

2

33%

3 6  

33%

A. 0 B. 6 C. 12

33% 1.

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2.

3.

10

Given [

A

]    3 5  6 9 2 3   [ then if [C]=[A]-[B], c 23 =

B

]     2 8 9 6 .

2

33%

3 6  

33%

A.

B.

C.

-3 9 3

33% 2.

3.

Given      4 1 6  6 2  5   3 8 9      ,      4 9 4  3 7 5     then if [C]=[A][B], then c

31

= .

25% 25% 25% 25%

A.

B.

C.

D.

-57 -45 57 does not exist

1.

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2.

3.

4.

10

A square matrix [A] is lower triangular if

25% 25% 25% 25%

A.

B.

C.

D.

a a ij

 0 ,

i

a ij a ij ij j

 0 ,

j

   0 ,

i

0 ,

j

 

j i i

1.

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2.

3.

4.

10

A square matrix [A] is upper triangular if

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1.

2.

3.

4.

a a ij

 0 ,

i

a ij a ij ij j

 0 ,

j

   0 ,

i

0 ,

j

 

j i i

1.

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2.

3.

4.

An example of upper triangular matrix is A.

B.

C.

D.

  2   0 0   2   0 0   2   6 0 5   6 3   5 6 3     5   3 3   3 0 0 3 2 2 3 0 2 none of the above

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3.

4.

An example of lower triangular matrix is A.

B.

C.

D.

    2 3 4   2   3 4 0 0 5 9 2 5 0 0 6     0 0 6         2 3 9 5 6 0 0   0 0   none of the above

25% 25% 25% 25% 2.

3.

4.

An identity matrix [

I

] needs to satisfy the following A.

I ij

 0 ,

i

j

B.

I ij

 1 ,

i

j

C.

matrix is square D.

all of the above

25% 25% 25% 25% 1.

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2.

3.

4.

10

Given      1 0 0 0 1 0 1 .

0 0 01     then [A] is a matrix.

A.

B.

C.

D.

diagonal identity lower triangular upper triangular

25% 25% 25% 25% 1.

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2.

3.

4.

A square matrix [

A

]

n

n

is diagonally dominant if 1.

2.

3.

4.

a ii a ii

 

n

i j

  1

j n

i j

  1

j a ij

,

i a ij

 1 , 2 ,.....,

n

,

i

 1 , 2 ,.....,

n and a ii a ii

 

j n

  1

n

i j

  1

j a ij

,

i a ij

 ,

for any i

1 , 2 ,.....,

n and

 1 , 2 ,....,

n a ii a ii

 

j n

  1

j n

  1

a ij

,

a ij for any i

,

i

  1 , 2 ,...., 1 , 2 ,.....,

n n

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25% 25% 25% 25% 1.

2.

3.

4.

The following system of equations

x + y=2

6

x +

6

y=12

has solution(s).

1.

2.

3.

4.

no one more than one but a finite number of infinite

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1.

2.

3.

4.

10

PHYSICAL PROBLEMS http://nm.mathforcollege.com

Truss Problem

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Pressure vessel problem

a b a c

u

1 

c

1

r

c

2

r u

2 

c

3

r

c

4

r

b  4 .

2857       4 .

2857  6 0  .

5 10 10 7 7  9 .

2307  10 5  5 .

4619  10 5  0 .

15384 0 0  4 .

2857  10 7 6 .

5 4 .

2857  10 7  5 .

4619 0 .

0  10 15384 5 3 .

6057  10 5           

c c

c c

1 2 3 4               7 .

887 0 .

0  10 3 007 0      

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Polynomial Regression

We are to fit the data to the polynomial regression model (

T

1

1 )

,

(

T

2

2 ),

...,

(

T n-

1

,α n-

1 )

,

(

T n ,α n

)

α

a

0 

a

1

T

a

2

T

2               

i n

  1

i n

  1

n T i T i

2      

i n

  1

T i

 

i n

  1

T i

2  

i n

  1

T i

3        

i n

  1

T i

2  

i n

  1

T i

3  

i n

  1

T i

4                  

a

0

a

1

a

2             

i i n

  1

n

  1

i n

  1

T i T i

  2 

i i i

       

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END http://nm.mathforcollege.com

Simultaneous Linear Equations Gaussian Elimination

(Naïve and the Not That So Innocent Also) http://nm.mathforcollege.com

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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1.

2.

3.

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

33% 33% 33% http://nm.mathforcollege.com

1.

2.

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

33% 33% 33% http://nm.mathforcollege.com

1 2 3

Division by zero during forward elimination steps in

Naïve Gaussian elimination

for [A][X]=[C] implies the coefficient matrix [A]

33% 33% 33%

1.

2.

3.

is invertible is not invertible cannot be determined to be invertible or not

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1.

2.

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

33% 33% 33% http://nm.mathforcollege.com

1.

2.

3.

Using 3 significant digit with

chopping

at all stages, the result for the following calculation is

x

1  6 .

095  3 .

456  1 .

99 8

25% 25% 25% 25%

A. -0.0988

B. -0.0978

C. -0.0969

D. -0.0962

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A.

B.

C.

D.

Using 3 significant digits with

rounding-off

at all stages, the result for the following calculation is

x

1  6 .

095  3 .

456  1 .

99 8

25% 25% 25% 25%

A. -0.0988

B. -0.0978

C. -0.0969

D. -0.0962

A.

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B.

C.

D.

Simultaneous Linear Equations LU Decomposition

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You thought you have parking problems. Frank Ocean is scared to park when __________ is around.

25% 25% 25% 25%

A. A$AP Rocky B. Adele C. Chris Brown D. Hillary Clinton

A.

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B.

C.

D.

Truss Problem

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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

11 

a

22  ...

a ii

 ...

a

nn i n

  1

a ii

Using forward elimination to transform [A] nxn upper triangular matrix, [U] nxn .

   

n

n

n

n

det    det   to an

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If you have

n

equations and

n

unknowns, the computation time for forward substitution is approximately proportional to

33% 33% 33%

A. 4n B. 4n

2

C. 4n

3

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A.

B.

C.

If you have a

n

x

n

matrix, the computation time for decomposing the matrix to LU is approximately proportional to

33% 33% 33%

A. 8n/3 B. 8n

2 /3

C. 8n

3 /3

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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

33% http://nm.mathforcollege.com

1.

33% 2.

33% 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. Now you try to use the Gaussian Elimination method to accomplish the same task. It will now take approximately ____ seconds.

25% 25% 25%

A.

4

25%

B.

64 C.

D.

6800 27200

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1 2 3 4

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% 25% 25% 25%

A.

4 B.

6 C.

D.

8 12

1 http://nm.mathforcollege.com

2 3 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)=at 2 +bt+c

to approximate the velocity profile.

t v

(s) m/s 0 0 14 227.04

15 362.78

20 517.35

30 602.97

35 901.67

25% 25% 25% 25%

A.

B.

C.

D.

    176 225 400 14 15 20 1   1 1      

a b c

         227 .

04 362 .

78 517 .

35       225   400 900     0 225 400     400 900 1225 15 20 30 0 15 20 20 30 35 1   1 1    

a

 

b c

         362 517 602 .

.

78 .

35 97     1   1 1      

a b c

         0 362  517 .

.

78 35     1   1 1      

b a

 

c

       517 602 901 .

.

.

35 97 67    

1.

2.

3.

4.

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Three kids-Jim, Corey and David receive an inheritance of $2,253,453. The money is put in three trusts but is not divided equally to begin with. Corey’s trust is three times that of David’s because Corey made and A in Dr.Kaw’s class. Each trust is put in and interest generating investment. The total interest of all the three trusts combined at the end of the first year is $190,740.57 . The equations to find the trust money of Jim (J), Corey (C) and David (D) in matrix form is

25% 25% 25% 25%

A.

    1 0 0 .

06 1 3 0 .

08 0 .

1  1 011        

J C D

         2 , 253 190 , 0 , 453 740 .

57     B.

C.

    0 1 0 .

06 1 1 0 .

08 0 .

 1 3 011        

J C D

         2 , 253 190 0 , 453 , 740 .

57     D.

  1   0 6 1 1 8  1 3 11        

J C D

         2 , 253 190 0 , 453 , 740 .

57       1   0 6 1 3 8 1  1 11        

J C D

         2 , 253 190 0 , 453 , 740 .

57    

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1.

2.

3.

4.

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.

2.

3.

4.

The condition number of a

n×n

diagonal matrix [A] is A.

B.

max

a ii

,

i

 1 ,...,

n

min  max

a ii a ii

,

i

,

i

 1 ,...,

n

 1 ,...,

n

2  C.

D.

2 min 1

a ii

,

i

 1 ,...,

n

 max

a ii

,

i

 1 ,...,

n

 min

a ii

,

i

 1 ,...,

n

25% 25% 25% 25% http://nm.mathforcollege.com

1.

2.

3.

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 B. machine epsilon C. product of the condition number of coefficient matrix and machine epsilon D. norm of the coefficient matrix

25% 25% 25% 25% http://nm.mathforcollege.com

1.

2.

3.

4.

If

cond

  and 

mach

 0 .

119  10  6 , then in [A][X]=[C], at least these many significant digits are correct in your solution,

25% 25% 25% 25%

A. 0 B. 1 C. 2 D. 3

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1.

2.

3.

4.

THE END

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Consider there are only two computer companies in a country. The companies are named

Dude

and

Imac

. Each year, company

Dude

keeps 1/5 th of its customers, while the rest switch to

Imac

. Each year,

Imac

customers, while the rest switch to keeps 1/3

Dude.

If in 2003,

Dude

had 1/6 th rd of its of the market and

Imac

had 5/6 th of the marker, what will be share of

Dude

computers when the market becomes stable?

25% 25% 25% 25%

1.

2.

3.

4.

37/90 5/11 6/11 53/90

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1.

2.

3.

4.