2012 Mu Alpha Theta National Convention
Alpha Matrices and Vectors Solutions
 3 1
1.  2 7

 2 1
 3 1
2.  2 7

 2 1
0   1 3 1   3  1 1  3 0  1   4 4 1 
4   0 2 0    2  0 7  2 4  0   2 9 4 . A
1   1 5 4  2  1 1  5 1  4   3 6 5 
0   1 3 1   3  0  0
9  2  0
3  0  0   3 11 3 




4  0 2 0    2  0  4 6  14  20 2  0  16   2 28 18 . D
1   1 5 4  2  0  1
625
2  0  4   3 13 2 
 2  a11  a12  a13  
 a11 a12 a13 




3. Let A  a21 a22 a23 . Then Ax   2  a21  a22  a23   . The sum of the entries of A is then


 2  a31  a32  a33  
 a31 a32 a33 
1
 4  4  16   8 . C
2
4.
1,8, 4  12  82   4   81  9 . Then the vector
2
2
2 16 8
1,8, 4  , , 
has the
9
9 9
9
same direction as 1,8, 4 with magnitude 2. B
5. Let x be the price of an apple, y be the price of an orange, and z be the price of a banana.
Then  x, y, z  is the solution to the system of equations 2 x  y  3z  19 , x  3 y  z  10 , and
19 1 3
10 3 1
4x  2z  14 . Using Cramer’s Rule, x 
14 0 2 90

 3. C
2 1 3
30
1 3 1
4 0 2
6. Expanding the determinant by minors down the third column, the determinant is equal to
1 1 1
 1
43
2
3
3    3  12  2  12  3  2   28 . A
4
1
1
a b 
1 1  a b   a  c b  d 
 a b  1 1
7. Let B  
. AB  
. AB  








c d 
1 1   c d   a  c b  d 
 c d  1 1 
 a  b a  b 
c  d c  d  . These two matrices are equal when a  c  a  b , b  d  a  b , a  c  c  d ,


 a b
and b  d  c  d . This occurs when a  d and b  c . Then B is of the form B  

 b a 
and B  a2  b2 . Since a and b are real numbers, the smallest possible value of this determinant
is 0. B
Mu Matrices and Vectors Solutions
2010  National Convention
8. The parallelogram is defined by the vectors 3, 2,0 and 4,6,0 (when viewing the xy-plane
as a subspace of three-dimensional space). The area of the parallelogram is given by
i j k
3, 2, 0  4, 6, 0  3 2 0  10k  10 .B
4 6 0
9. Using properties of determinants, AT BT  AT  BT  A  B  2  3  6 . E
10. The matrix will not be invertible when its determinant is zero.
x 1 3
2 0 x  0  4 x  12  0  2 x 2  14  2 x 2  4 x  2  0 . x 2  2 x  1  0 . The only solution is
4 2 7
x  1. C
11. The vector 2, 3,6 is normal to the plane.
2, 3,6  22   3  62  7 . The vector
2
1
2 3 6
2, 3, 6  ,  ,
is a unit vector that is normal to the plane. D
7
7 7 7
12. det A is given by the ratio of the area of the transformed triangle to the area of the original
triangle. The transformed triangle has area 6 while the original triangle has area 1. det A  6 .
D or E accepted
13. A2  3 A  2 I n . A3  3 A2  2 A  3  3 A  2I n   2 A  7 A  6I n . C
 1
3


cos  60   sin  60    2
2

 rotates a vector counter-clockwise 60 .
14. The matrix 

sin
60

cos
60






3
1 




 2
2 
 1

3
3  3


 2 3   3 
  
2 
2  2 . B
 2




 3
1   7 
1  1  


 3  2   2 
 2
2 
15. 3u  v  3 3, 2,5  1,0, 2  9, 6,15  1,0, 2  8, 6,17 . C
16. u  v  3 1   2  0  5   2  3 10  7 . A
i
j
17. u  v  3 2
1
0
k
5  4i  5 j  0k  2k  0i  6 j  4,11, 2 . D
2
18. Notice that the i-th row is equal to the first row times i . Therefore all n rows are scalar
multiples of the first row, and hence the rank of A is one. A
19. BT is a 5  3 matrix and A1 is a 3  3 . Then BT A1B is a 5  5 matrix. D
1
1
 2 1
 2 1  7 6  1  3 1  7 6  1 3
1  3 1
20. 
. A
 


 
 


 . The
2  4 2 
 4 3
 4 3 19 12 2  4 2  19 12 5 0 
sum of the entries is then 3. B
2010  National Convention
Mu Matrices and Vectors Solutions
21. Let  be the angle the two vectors. Then cos   
u v
3  2  1 4
10



2
2
2
2
u v
10 20
3 1 2  4
2
. Since  is the smaller angle between the vectors,   45 . A
2
22. Since all of the rows of A are equal, the rank of A is one and hence A  0 . A
23. The trace of a matrix is the sum of the elements on the diagonal. Therefore the trace of A is
27. D
 x  y
24. The matrix that reflects across the line y  x is the matrix A such that A      , so
 y  x
0 1 
A
. D
1 0 
2
25. Notice that in general matrix multiplication is not commutative, so  A  B  
A2  AB  BA  B2  A2  2 AB  B 2 in general. Similarly,  AB   ABAB  A2 B 2 . In general
2
there is no simple formula to calculate
statements are false. E
3 2
26. The row reduction yields  2 1

 4 5
 A  B
1
 A1  B 1 .
 AB 
1
 B 1 A1  A1 B 1 . All four
5 3 2 5 3 2 5
8   0 7 14   0 1 2  . Therefore the third
2 0 7 14  0 0 0 
equation in the system can be expressed in terms of the first two equations. The system reduces
to a system of two equations in two unknowns, and hence there is one solution. B
27. Let  be the angle the two vectors. Since u  v  u v cos   , the dot product will be
negative when cos    0 . Therefore 90    270 . Since u is in the first quadrant, v can be
in any quadrant except for quadrant I. C
  2 2
1 0  
28. The value of  will have to satisfy  
 

  x  022 . Since x is non-zero, this
0 1  
 1 3
2
2
  2    3     2   2  5  4  0 . Then either   1 or
can only happen when
1
3
  4. B
0 1 2  0 1 2  0 0 4 
0 0 4  0 1 2  0 0 0 






2
3
29. A  0 0 1 0 0 1  0 0 0 . A  0 0 0  0 0 1   0 0 0 .


 



 

0 0 0  0 0 0  0 0 0 
0 0 0  0 0 0  0 0 0
Therefore An  033 for n  3 and hence A2012  033 . E
30. A vector quantity has both a magnitude and direction. Density is a scalar quantity but it has
no direction. B