Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Matrices & Vectors Topic Test – Alpha Division
1. (C) A singular matrix has determinant equal to zero. Going through each of the answer
choices, we see that the only singular matrix is the one in choice C.
2. (C) Setting some of the corresponding entries equal to each other, we get 2 x  3  4 y  3
and 5 x  1  2 y  4 . Solving these two equations simultaneously yields x = 1 and y = –1, so
x2  y 2  2 .
3. (D) Expanding the matrix with respect to the first row yields
8
0
5
2 8
6 2
6 2 8  8
 0(something)  5
 8(16  48)  5(36  4)  312 .
6 8
2 6
2 6 8
x2
4. (B) If
3
x
 4 , then x 2  3 x  4 , or ( x  4)( x  1)  0 . The larger root is 4.
1
 3 9 6 3
5. (C) The element in the second row and first column of 

 is
 4 2 8 8
(4)(6)  (2)(8)  8 . We take this value and from it subtract the value in the second row and
 2 5 
first column of 
 , which is –3. So the answer is 8 – (–3) = 11.
 3 1 
6. (C) For a product of two matrices to be defined, the number of columns of the first matrix
must equal the number of rows of the second matrix. Note that ABC is not defined unless
n = m and a = b. If AT BC is defined, then a = b, which isn’t necessarily true; same thing
with BCAT . Checking choice C, we see that it is indeed defined; the result is a b  a matrix.
 8 8
 8 10 
7. (A) Say the matrices are A and B, then A  B  
and A  B  

 . Add the
11 2
 1 10
 0 18 
0 9 
8 1
two equations together to get 2 A  
, or A  
; consequently, B  


.
12 8
6 4 
5 6 
Thus, D | AB || A || B | (54)(53)  2862 , making | D | 2862 .
8. (D) Statements I, III, and IV are standard theorems. Statement II is easy to disprove by
 2 1
1 8 
B

finding a counterexample, say, A  
and

 2 9  . Note that tr(A)tr(B) = –24,
 1 1 


but tr(AB) = 0.
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Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Matrices & Vectors Topic Test – Alpha Division
9. (A) The characteristic polynomial is
x  4 1
 ( x  4)( x  2)  3  ( x  5)( x  1) . The
3 x  2
smaller root of this polynomial is –5.
 2 0   2  4
 2
10. (B) Since 
    2   , the vector in choice B is an eigenvector (with a



3 4  3   6 
3
corresponding eigenvalue of 2). It’s easy to check that the vectors in the other answer
choices do not yield a scalar multiple of the vector when multiplied with the given matrix.
11. (C) The system will not have a unique solution if the determinant of the coefficient matrix is
equal to 0. Thus, we have
4 3
1
2 1 3  8(9  1)  c(12  2)  3(4  6)  70  10c  0 .
8 c 3
The solution to this equation is, of course, –7.
MT  M
M  MT
12. (A) Let M  AB . Notice that
is a symmetric matrix and
is a skew2
2
symmetric matrix. The sum of the above matrices is
M T  M M  M T 2M


M.
2
2
2
So choice A is true. Now, to disprove the other choices. In choice B, the given summation is
the entry in the ith row and jth column of AT B , not AB. It’s easy to find a counter-example
for choice D; just multiply a 3  2 matrix with a 2  3 matrix (in that order) to obtain a 3  3
1 0 
.25 0 
matrix, for instance. For choice C, just take A  
and B  

 , for example. It’s
3 4 
 0 1
clear that | A || B | 1 , but A and B aren’t inverses of each other.
6 4 1
1
117
13. (C) By the standard matrix formula, the area is equal to
.
1 8 1
2
2
5 1 1
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Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Matrices & Vectors Topic Test – Alpha Division
14. (B) By the inverse “trick” for 2  2 matrices,
7 2 
4 1


1

1  1 2  1 2 
.

7  8  4 7   4 7 
Left-multiplying both sides of the given equation by this inverse yields
 1 2  6 9   6 7 
B


.
 4 7  0 8  24 20 
So the sum of the entries of B is – 6 – 7 + 24 + 20 = 31.
15. (B) Since a  c = –7, b  c = –36, and the dot product is distributive, we have
(3a  2b)  c  3(a  c)  2(b  c)  3(7)  2(36)  51 .
16. (B) By the standard formula, if  is the angle between the vectors, then
cos  
making sec  
[3, 4]  [15,8]
(3) 2  42 152  82

13
,
85
85
.
13
  cos 75  sin 75 
cos  sin  
17. (C) Let M  
and R  
 , the standard rotation matrix.
7
7 
 cos 5 
 sin  cos 
 sin 5
7
Notice that if  
, then M  ( R )T  ( R )T . So we have
5
M 2005   ( R )T 
2005
T
  cos(2807 )  sin(2807 )  1 0
   ( R ) 2005   ( R2005 )T  

.
 sin(2807 )  cos(2807 )  0 1 
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Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Matrices & Vectors Topic Test – Alpha Division
18. (C) Let’s find several values for An as well as corresponding values for  ( An ) and try to
find a pattern. Note that
1 1 
 2 1
 3 2
5 3 
, A2  
, A3  
, and A4  
A



.
1 0 
1 1
2 1
3 2 
Thus, we have  ( A)  3 ,  ( A2 )  5 ,  ( A3 )  8 , and  ( A4 )  13 . Now the pattern is a lot
clearer:  ( An )  Fn 3 , the (n + 3)rd Fibonacci Number, where F1  F2  1 and
Fn  Fn1  Fn2 for integers n > 2. Our desired sum is then equal to
9
 ( A
n 1
n
9
12
3
n 1
n 1
n 1
)   Fn 3   Fn   Fn  ( F14  1)  (1  1  2)  F14  5 ,
N
where we use the Fibonacci identity
F
n 1
n
 FN  2  1 to simplify the sum. Calculating the
Fibonacci sequence recursively, we find that F14  377 , and so the answer is F14  5  372 .
19. (B) Let the eigenvalues of A be a, b, and c; we know that these are precisely the roots of the
characteristic polynomial. Using the standard formulas relating the roots of a polynomial to
its coefficients, we have a  b  c  0 , and ab  bc  ac  12 . Thus, we have
a 2  b2  c2  (a  b  c)2  2(ab  bc  ac)  02  2(12)  24 .
20. (A) The total amount of money the man spent on stock x is the dot product of the vector that
represents the price of stock x in each of the days and the vector that represents the number of
shares of stock x the man bought in each of the days. The first vector is the xth column of A
while the second is the xth row of B. Since multiplication of two matrices takes rows from
the first matrix and computes its dot product with columns from the second matrix, the
desired entry is in the xth row and xth column of AT B . The transpose on A is necessary to
turn the column we desire into a row.
21. (E) Given a square matrix of order greater than 2 whose rows and columns form arithmetic
progressions, we can always use an appropriate set of row operations to get one row (or
column) to equal another row, and consequently, we can make any row or column in A
consist entirely of zeroes. Thus, the determinant of A is 0.
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Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Matrices & Vectors Topic Test – Alpha Division
1 2 3 
a b c 


22. (B) Direct calculation of the entries shows that A  1 4 9 . Let B   d e f  . The




1 8 27 
 g h i 
trace of AB is (a  2d  3g )  (b  4e  9h)  (c  8 f  27i) . This quantity will be maximized
when the larger values are substituted for the variables with the larger coefficients.
Proceeding in this manner, we get 27(9)  9(8)  8(7)  4(6)  3(5)  2(4)  3  2  1  424 .
23. (E) Expanding, we obtain (a + 2b)  (3a + 4b) = 3(a  a) + 4(a  b) + 6(b  a) + 8(b  b).
Since the cross product of a vector with itself is the zero vector and that switching the order
of the cross product negates the result, we have 4(a  b) – 6(a  b) = –2(a  b) = –2v.
24. (A) Using properties of the cross product, we have
 (a  b)  b  (a  c)  c  c  b   (a  c)   c   a   (b)   c  a  b   c  c  c  0 .
25. (B) If u and v are the two vectors, then we know that u  v  u v sin  and
u  v  u v cos , where  is the angle between the two vectors. Combining these two
formulas gives us
u v
uv

u v sin 
u v cos
 tan  . So the answer is
192  (6) 2  (2) 2
401
tan  

.
3
3
i
26. (A) We have a  b  0
j k
9 8 , but notice that we don’t need to compute the entire cross
4 0 2
product, as the dot product of any 3d vector with [0, 1, 0] produces the y-component of the
0 8
 32 .
vector. So we only need the y-component of a  b, which is 
4 2
27. (B) The expression in choice A doesn’t make sense because somewhere it involves
subtracting a vector from a scalar. This also happens in choices C and D except a scalar is
being subtracted from a vector. The only meaningful expression is the one in choice B.
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Mu Alpha Theta National Convention: Hawaii, 2005
Solutions – Matrices & Vectors Topic Test – Alpha Division
28. (B) We can plug in each of the points in each answer choice to find that choice B is the
correct answer. Alternatively, setting (1, 1, 1) as the origin and using the other two points,
we find that two vectors that span the plane are [0, 1, –1] and [–2, 1, 0]. Thus, a normal
vector to the plane is [0, 1, –1]  [–2, 1, 0] = [1, 2, 2], so the equation of the plane is
1( x  1)  2( y  1)  2( z  1)  0 , or after some rearranging, x + 2y + 2z = 5.
29. (C) Pick (3, –1, 2) as an origin and make vectors with the other three points. This produces
the vectors [–5, –6, 1], [2, 0, 4], and [1, 7, 0]. Using the triple scalar product, the volume of
the tetrahedron is none other than
5 6 1
1
130 65
.
2 0 4 

6
6
3
1 7 0
30. (D) The direction vectors for the given lines are [5, –2, –4] and [–8, 3, 7]. By the anglebetween-vectors formula,
cos  
[5,  2,  4]  [8,3, 7]
5  (2)  (4)
2
2
2
(8)  3  7
2
2
2

74
3 610
.
The above result is for the obtuse angle between the vectors, but since the cosines of
74
supplementary angles are negatives of each other, the answer is
.
3 610
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