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MATH 51 MIDTERM 2 SOLUTIONS (AUTUMN 2001)
- Let
A =
Then
rref(A) =
(You do not need to verify this.)
(a) (3 points) Find a basis for C(A).
Solution. The first, third, and fourth columns of rref(A) have pivots, so
is a basis for C(A).
(b) (4 points) Express each column of A which is not in your basis for C(A) as a linear combination of your basis columns.
Solution. Denote the columns of A by v 1 through v 5. Then v 1 , v 3 and v 4 are the basis vectors from part (a). By inspection v 2 = 2v 1. Looking at the columns w 1 through w 5 of rref(A), it is clear that w 5 = − 5 w 1 − 4 w 3 + 2w 4. Therefore, since the columns of A and rref(A) share the same relations, it follows that v 5 = − 5 v 1 − 4 v 3 + 2v 4.
(c) (3 points) What is the maximum number of linearly independent vectors which can be found in N (A)?
Solution. 2. The rank of A is 3, so the nullity of A is 2 by the Rank-Nullity Theorem.
- (10 points) Answer each question True or False. No explanation is necessary.
(a) If V is a linear subspace of R^5 and V 6 = R^5 then any set of 5 vectors in V is linearly dependent.
Solution. True. (Any set of 5 linearly independent vectors in R^5 must span R^5 .)
(b) If A is a 4 × 7 and if the dimension of N (A) is 3 then for any b in R^4 the system Ax = b has infinitely many solutions.
Solution. True. (By the Rank-Nullity Theorem, rank(A) = 4, which implies that C(A) = R^4. Therefore, Ax = b has at least one solution for every b ∈ R^4. Since the null space of A is nontrivial, the set of solutions for each b is infinite.)
(c) If T : R^5 → R^3 is an onto linear transformation and if {v 1 , v 2 , v 3 } is a linearly independent set of vectors in R^5 , then {T(v 1 ), T(v 2 ), T(v 3 )} spans R^3.
Solution. False. (For example, let T be given by the matrix
A =
and let
v 1 =
v 2 =
v 3 =
Then T is onto since rank(A) = 3, and it is clear that {v 1 , v 2 , v 3 } is a linearly independent set. But
T(v 1 ) =
(^) T(v 2 ) =
(^) T(v 3 ) =
clearly do not span R^3 .)
(d) If A is an 11 × 5 matrix and the linear transformation T : R^5 → R^11 defined by T(x) = Ax is one to one, then rank(A) = 5.
Solution. True. (Since T is one to one, the nullity of A must be zero, so by the Rank-Nullity Theorem, rank(A) = 5.)
- (10 points) Let S denote the set of points in the face shown below.
Each figure below is the image of S under the some transformation. Find the corre- sponding matrix.
(a)
Solution. Since T(e 1 ) = −e 1 and T(e 2 ) = e 2 , the matrix is
[
]
(b)
(Clockwise rotation by angle 45◦.)
Solution. Since cos(− 45 ◦) = √^12 and sin(− 45 ◦) = − √^12 , the matrix is [ (^1) √ 2 √^12 − √^12 √^12
]
(c)
Solution. Since T(e 1 ) = 2e 1 and T(e 2 ) = e 2 , the matrix is
[
]
(d)
The transformation is the identity, whose matrix is
[
]
(e)
Solution. Since T(e 1 ) = −e 1 and T(e 2 ) = −e 2 , the matrix is
[
]
- (a) (5 points) Find the determinant of
A =
Solution. det(A) = − 12
(b) (5 points) Find the inverse of
B =
Solution. B−^1 =
- (10 points) In each part below, state which figure represents the parametric curve defined by the given formula.
Figure 1 Figure 2 Figure 3
−2 −1 0 1 2
−
−
0
1
2
−2 −1 0 1 2
−
−
0
1
2
−2 −1 0 1 2
−
−
0
1
2
Figure 4 Figure 5 Figure 6
−2 −1 0 1 2
−
−
0
1
2
−2 −1 0 1 2
−
−
0
1
2
−2 −1 0 1 2
−
−
0
1
2
(a) x(t) = (1 + cos t, sin t)
Solution. Figure 5.
(b) x(t) = (2 cos t, sin t)
Solution. Figure 3.
(c) x(t) = (2 cos t, 1 − cos t)
Solution. Figure 6.
(d) x(t) = (cos(2t), sin t)
Solution. Figure 1.
(e) x(t) = (cos t, sin(2t))
Solution. Figure 4.
- In each part, state whether or not the indicated limit exists. If it exists, evaluate the limit (you do not need to prove your answer). If it does not exist, explain why.
(a) (3 points) lim (x,y)→(1,2)
x^3 3 x + y
Solution. 15 (The function is rational and the denominator is nonzero at the point (1, 2), so just plug in.)
(b) (3 points) lim (x,y)→(0,0)
|x + y| |x| + |y|
Solution. The limit does not exist. Along the line x = 0 the limit is 1, and along the line y = −x the limit is zero.
(c) (4 points) lim (x,y)→(1,1)
ex+y^ ·
x − y ex−y^ − e−(x−y)^
Solution. The term ex+y^ goes to e^2. Making the substitution u = x − y, the rest becomes lim u→ 0
u eu^ − e−u^
= lim u→ 0
eu^ + e−u^
(using L’Hopital’s Rule). Thus the limit equals 12 e^2.
- (a) (4 points) Let f (x, y) = x sin y + exy. Compute
∂f ∂x
∂f ∂y
∂^2 f ∂y∂x
and
∂^2 f ∂x∂y
Solution.
∂f ∂x
= sin y + yexy,
∂f ∂y
= x cos y + xexy, and
∂^2 f ∂y∂x
∂^2 f ∂x∂y
cos y + exy(1 + xy).
(b) (6 points) Find an equation of the form z = ax+by+c which describes the tangent plane to the graph of the function g(x, y) =
32 − 3 x^2 − 5 y^2 at the point (1, 2 , 3) on the graph of g.
Solution. Since ∂g ∂x
(x, y) =
− 3 x √ 32 − 3 x^2 − 5 y^2
and
∂g ∂y
(x, y) =
− 5 y √ 32 − 3 x^2 − 5 y^2
we have ∂g∂x (1, 2) = −1 and ∂g∂y (1, 2) = −^103 and therefore the equation of the tangent plane is z = 3 − 1(x − 1) − 103 (y − 2), or equivalently z = −x − 103 y + 323.
