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EXAM II SOLUTIONS
Math 51, Spring 2003.
You have 2 hours.
No notes, no books, no calculators. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck!
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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:
Byoung-du Kim (2 and 6)
Ted Hwa (3 and 7)
Jacob Shapiro (4 and 8)
Ryan Vinroot (A02)
Michel Grueneberg (A03)
Circle your section meeting time:
11:00am 1:15pm 7pm
- (a) Find bases for the kernel and the image of the linear transformation given by
T
x y z
3 x − y + z y + 2z 3 y + 6z
Solution: Of course the kernel of the linear transformation is just the null space of the corresponding matrix; and similarly, the image is the column space. So we will solve this problem in terms of matrices.
The matrix in question has columns which are the images of the standard basis vectors:
T (e 1 ) =
(^) T (e 2 ) =
(^) T (e 3 ) =
So the matrix is
A =
This we can row reduce to its reduced row echelon form
rref(A) =
The solutions to the homogeneous system are
x y z
−z − 2 z z
(^) = z
and so a basis for the null space (and thus the kernel of the linear transformation) is
And since there are pivots in the first and second columns of rref(A), a basis for the column space (and thus the image of the linear transformation) is
- (a) Suppose that A is a 4x3 matrix, and that C(A) has dimension 2. What is the di- mension of N(A)?
Solution: The matrix A is a function from R^3 to R^4. So, the Rank-Nullity theorem tells us that dim C(A) + dim N(A) = 3
Since the dimension of C(A) is given to be 2, we know that N(A) must have dimen- sion 1.
(b) Suppose that N(M 2 ) = {
0 }. Show that N(M 2 M 1 ) = N(M 1 ).
Solution: We need to show that
M 2 M 1 −→x =
0 ⇐⇒ M 1 −→x =
(⇐) If M 1 −→x =
0 , then M 2 M 1 −→x = M 2 (M 1 −→x ) = M 2
(⇒) Now suppose that M 2 M 1 −→x =
- Of course this means that M 2 (M 1 −→x ) =
Since we are given that the null space of M 2 is just the zero vector, this means that (M 1 −→x ) =
0 , as desired.
(c) Suppose that M 2 is a 3x2 matrix, M 1 is a 2x4 matrix, the rank of M 2 is 2, and the rank of M 2 M 1 is 1. What is the dimension of C(M 1 )?
Suggestions: Use part (b) and the Rank-Nullity Theorem to determine the dimen- sions of N(M 2 ) and N(M 2 M 1 ), and then deduce the dimension of N(M 1 ).
Solution: We have that M 1 : R^4 → R^2 , and M 2 : R^2 → R^3. So, we conclude that M 2 M 1 : R^4 → R^3. We then can use the Rank-Nullity Theorem as suggested to conclude
rankM 2 = 2 =⇒ dim C(M 2 ) = 2 =⇒ dim N(M 2 ) = 0 =⇒ N(M 2 ) = {
rankM 2 M 1 = 1 =⇒ dim C(M 2 M 1 ) = 1 =⇒ dim N(M 2 M 1 ) = 3
Since N(M 2 ) = {
0 }, part (b) then tells us that N(M 1 ) = N(M 2 M 1 ), and so we conclude that dim N(M 1 ) = 3.
Applying the Rank-Nullity Theorem to M 1 , we conclude that dim C(M 1 ) = 1.
If we were to row reduce instead, the first thing we would do is add appropriate multiples of the top three rows to the bottom two rows to eliminate the first three components of the bottom two rows; since adding multiples of a row to another row does not change the determinant, we get simply
= − det
= − det
Using the linearity of determinant in each row, we get
= −24 det
det
Again adding multiples of rows to other rows, this becomes
det
det
= −24 det
(b) Is the above matrix A invertible? Why or why not?
Solution: The determinant of the matrix is non-zero, so the matrix IS invertible.
(c) Suppose B is a 2x2 matrix with determinant 3, and that S is a set in R^2 such that the area of B(S) is 10.
What is the area of S?
Solution:
area(S) · | det B| = area(B(S))
area(S) · | 3 | = 10
area(S) =
(b) Is the above enough information to determine the second column of B−^1? If so, find that second column; if not, explain why you cannot find it.
Solution: We were able to find the first column of B−^1 (in other words, B−^1 (−→e 1 )) because −→e 1 is a linear combination of the two vectors for which we know the image of B−^1 ; namely,
However, we immediately note that −→e 2 is NOT a linear combination of those vec- tors, because they are both zero in their second coordinate.
So, we cannot use this method to determine the image of −→e 2.
In fact, some experimenting shows that B−^1 e 2 is definitely not uniquely determined by the given information. For example:
i. If
B =
then B(−→e 3 ) = −→e 2 , and thus the second column of B−^1 is B−^1 −→e 2 = −→e 3.
ii. However, if
B =
then B(−→e 3 ) = −−→e 2 , and thus the second column of B−^1 is B−^1 −→e 2 = −−→e 3.
So, we clearly cannot come to any firm conclusions about the second column of B−^1.
- For this problem, we define the following linear transformations in the plane:
Rθ = rotation counter-clockwise around the origin by an angle θ Fθ = flip (reflection) over the line Lθ obtained by rotating the x-axis counterclockwise by the angle θ F = F 0 = flip over the x-axis itself
v
0 (^0) v
F(v)
L 0
v
F (v) 0
x
y
x
y
x
y
0
R (v)
(a) Write down the matrices that correspond to Rθ and F.
Solution: The columns of Rθ are the images of the standard basis vectors. If we rotate the vector
[
]
counterclockwise by the angle θ, we get
[
cos θ sin θ
]
; and if we
rotate the vector
[
]
counterclockwise by the angle θ, we get
[
− sin θ cos θ
]
. So the matrix is Aθ =
cos θ − sin θ sin θ cos θ
For the flip F , we have that the image of −→e 1 is −→e 1 , and the image of −→e 2 is −−→e 2. So the matrix is B =
Bonus Question: Use linear algebra to help you find the area of the shaded region below, bounded by an ellipse centered at the origin, and two lines (make sure to be explicit in how you are using linear algebra results!).
( )3 3^2
Solution: An ellipse is just a stretched circle; in particular, the ellipse in question is the image of the unit circle by the linear transformation whose matrix is
A =
And the points indicated on the given ellipse are the images by the same transformation of the points (^) [√ 3 / 2 1 / 2
]
[
√^1 /^2
]
which of course correspond to the angles π/6 and π/3 on the unit circle.
So, we conclude that the indicated area is in fact just the image under this same linear trans- formation of the sector S of the unit circle between the angles π/6 and π/3; and of course S is just one twelfth of the unit circle.
So, the area of the shaded region A(S) is then
area(A(S)) = | det A| · area(S)
= | 21 |
π
7 π 4
