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EXAM II SOLUTIONS
Math 51, Spring 2002.
You have 2 hours.
No notes, no books, no calculators. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck!
Name
ID number
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- (/30 points)
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Bonus (/15 points)
Total (/150 points)
“On my honor, I have neither given nor received any aid on this examination. I have furthermore abided by all other aspects of the honor code with respect to this examination.”
Signature:
Circle your TA’s name:
Tarn Adams (2 and 6)
Mariel Saez (3 and 7)
Yevgeniy Kovchegov (4 and 8)
Heaseung Kwon (A02)
Alex Meadows (A03)
Circle your section meeting time:
11:00am 1:15pm 7pm
- (a) Without computing the inverse directly, and without row reducing, explain how you know that the matrix A given below is invertible.
A =
Solution: We can compute the determinant of this matrix using minors:
det A = 3 det
− 0 det
Since the determinant of the matrix is not zero, we know that it must be invertible.
(b) Find A−^1.
Solution: We row reduce (A|I):
(1st) − 2(3rd) (2nd) − 4(3rd) (3rd)
(1st)/ 3 (2nd)/ 2 (3rd)
The end result is (I|A−^1 ); so,
A−^1 =
(b) The row space R(M 2 ) contains the vectors
and the null space N (M 2 ) contains the vector
Solution: This is not possible, since vectors in the null space must always be perpendicular to vectors in the row space; whereas in this case, the first of the given row space vectors is clearly not perpendicular to the given null space vector, since their dot product is positive.
(c) The matrix M 3 is 4x2, the matrix M 4 is 2x4, and the 4x4 product M 3 M 4 has a 1-dimensional null space.
(Hint: What can you say about the dimension of N (M 4 )? And what can you say about the relationship between N (M 4 ) and N (M 3 M 4 )? )
Solution: Any vector −→x with M 4 −→x =
0 must also have M 3 M 4 −→x =
0 ; so, we conclude that N (M 4 ) ⊂ N (M 3 M 4 )
We can also draw conclusions about N (M 4 ) by using the Rank-Nullity Theorem, as follows: First note that we have
dim N (M 4 ) + dim C(M 4 ) = 4
Since C(M 4 ) ⊂ R^2 , we know that dim C(M 4 ) ≤ 2. Combining this with the equation above, we conclude that dim N (M 4 ) ≥ 2.
Combining this with our first observation, we see that dim N (M 3 M 4 ) ≥ 2. So, it is not possible to have dim N (M 3 M 4 ) = 1.
(b) Derive the matrices for the clockwise and counterclockwise rotations described above.
Solutions: Rotating counterclockwise, simple trig gives us
−→e 1 7 →
[
cos θ sin θ
]
−→e 2 7 →
[
− sin θ cos θ
]
The images of the standard basis vectors are the columns of our matrix, so we have
Rθ =
cos θ − sin θ sin θ cos θ
The result is similar for the clockwise rotation; this time, trig gives us
−→e 1 7 →
[
cos θ − sin θ
]
−→e 2 7 →
[
sin θ cos θ
]
Again these are the columns of our matrix, so we have
R−θ =
cos θ sin θ − sin θ cos θ
(c) Use the information from the first two parts to determine the matrix for the trans- formation Fθ.
Solution: As indicated in the statement of the question, the transformation Fθ is a composition of the three transformations from parts (a) and (b). So we conclude that Fθ = RθM R−θ =
cos θ − sin θ sin θ cos θ
cos θ sin θ − sin θ cos θ
cos θ − sin θ sin θ cos θ
cos θ sin θ sin θ − cos θ
cos^2 θ − sin^2 θ 2 sin θ cos θ 2 sin θ cos θ sin^2 θ − cos^2 θ
cos(2θ) sin(2θ) sin(2θ) − cos(2θ)
(b) Thm: The matrix
A =
a b c d
is invertible if and only if ad − bc 6 = 0. Solution: (⇐): If the determinant ad − bc is nonzero, then we easily check that
1 ad − bc
d −b −c a
is the inverse of A. So, A is invertible.
(⇒): We will prove the contrapositive; in particular, we will assume that ad − bc = 0, and then show that in that case, A is not invertible. We will accomplish this by demonstrating that the columns of A are dependent.
First, we consider the case where a and b are both zero. Clearly the rref(A) has only one pivot, so it is not invertible.
Now we proceed assuming that at least one of a and b is nonzero.
Given that ad − bc = 0, we observe that of course ad = bc. Then
b
[
a c
]
[
ab bc
]
[
ab ad
]
= a
[
b d
]
So, we see that at least one of the column vectors is a multiple of the other one. So, the column vectors of A are dependent, so A must not be invertible, as desired.
- Let
[−→v 1 ]S =
[
]
[−→v 2 ]S =
[
]
and consider the linear transformation T satisfying
[T (−→v 1 )]S =
[
]
[T (−→v 2 )]S =
[
]
(a) There exists a basis B = {−→w 1 , −→w 2 } such that each of the vectors −→w 1 , −→w 2 has positive integer components with respect to the standard basis S, and the matrix M = [T ]B has all integer entries, two of which are zero.
Find [−→w 1 ]S and [−→w 1 ]S , and write down M = [T ]B.
(Hint: You should be able to determine reasonable guesses for −→w 1 , −→w 2 by inspection, without doing any computations.)
Solution: First, we notice that the given information about the transformation T can be rewritten as
T (−→v 1 ) = 5−→v 2 T (−→v 2 ) = 3−→v (^1)
Since the vectors −→v 1 and −→v 2 appear in these equations multiple times, it seems reasonable to try using B = {−→v 1 , −→v 2 } as our basis – so we choose −→w 1 = −→v 1 , and −→w 2 =^ −→v
To represent T in terms of this basis, we make the columns of the matrix the images of our chosen basis vectors in terms of that same basis.
T (−→v 1 ) = 5 −→v (^2) = 0 −→v 1 + 5−→v (^2)
So [T (−→v 1 )]B =
[
]
T (−→v 2 ) = 3 −→v (^1) = 3 −→v 1 + 0−→v (^2)
and [T (−→v 2 )]B =
[
]
M = [T ]B =
So our chosen basis B = {−→w 1 , −→w 2 } = {−→v 1 , −→v 2 } satisfies the desired conditions.
Bonus Question: Prove or find a counterexample to the following claim:
Let A be an invertible matrix such that for each row of A, the elements in that row add up to 1.
Then A−^1 must have this same property – in other words, for each row of A−^1 , the elements in that row must add up to 1.
Hint: Consider the vector
Solution: The sum of the elements in a given row can be thought of as being the dot product of that row vector with the vector given above.
So, using the dot product interpretation of the definition of the matrix-vector product, the given condition on A can be restated as
A
Of course, when we then invert this equation, we obtain
A−^1
Again using the dot product interpretation of the definition of matrix-vector product, this means that A−^1 also satisfies the same condition, as desired.
