COURSE
MATH 3191: Applied Linear Algebra. 
Fall 1995. University of Colorado Denver

INSTRUCTOR 
Dr. Andrew Knyazev
Office: CU (Dravo) 620G. Phone: 556-8102.
Office hours: Tue 3pm - 6pm

TEXTBOOK
Elementary Linear Algebra. 7th ed. Howard Anton and Chris Rorres. Wiley, 1994.


Exam #2     (Chapter 2)         Monday, October 23, 1995
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This is not a very simple test! Please, think, use your knowledge,
and be attentive. Good luck! Andrew.
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           1 0 1
1. Let A = 0 1 0   . Then det(A)   =
           1 0 2

[A] 1   [B]  2    [C] 3   [D] 4     [E]   None

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              2 0 0 2
2. Let A  =   1 1 0 1   Then det(A)   =
              1 2 2 1
              1 0 0 1

[A] 0   [B] 1   [C] 2   [D] 3     [E] 4

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            2 0 
3. Let A =  0 1 . Then det(adj(A))  =

[A] 1   [B]  2   [C] 3   [D] 4     [E]  None

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            1 1 
4. Let A =  0 1 and its eigenvector x = (x  x ) has

the component x  =  1.  Then another component x  = 

[A] 1   [B]  2   [C] 3   [D]  any number     

[E]  such  eigenvector does not exist

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            1 1 
5. Let A =  0 1 and its eigenvector x = (x  x ) has

the component x  =  1.  Then the corresponding eigenvalue is 

[A] 1   [B]  2   [C]  3   [D]  any number     

[E]  such  eigenvector does not exist
 
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6. Let A be a square matrix. Then  det(A+A  ) = 2 det(A) 

[A] for 1x1 matrix A only 
[B] for any symmetric matrix A 
[C] if A is a zero matrix only  
[D] always
[E] The correct answer is not given by [A],[B],[C] or [D]

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7. The expression +a  a  a  a  a  a  a    is a signed elementary

product from a 7x7 matrix A if 

[A] i=1, j=2    [B]  i=2, j=6    [C]  i=6, j=6   [D]  i=6, j=2  [E] None

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8. Which of the following statements MUST be true:

[A] If det(A) = det(B) = 0, then det(A+B)=0. 
[B] det(ABC)=det(BCA)=det(CAB) for square matrices A,B, 
and C of the same sizes. 
[C] det(A)=det(A  )
[D] det(A)=det(A  )
[E] None of the above is correct

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           1 0 1
9. Let A = 0 1 0   . Then its cofactor C    =
           1 0 1

[A]   1   [B]    2   [C]   3   [D]  4     [E]   None

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10. Let det(A)=1 and det(B)=2. Then det((A  B )  ) =

[A]   1   [B]    2   [C]   3   [D]  4     [E]   None