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MATH 51 MIDTERM 2 SOLUTIONS
November 13, 2008
1(a). Find the matrix for for the linear map T given by T
([
x y
])
7 x − 2 y x + 3y 5 y
Solution: By inspection, matrix is
. Alternatively, one can find the matrix by calculating
T (e 1 ) and T (e 2 ) to get columns 1 and 2, respectively.
1(b). Find the matrix for reflection in R^2 about the line y = −x.
Solution: First column is T (e 1 ) = −e 2 =
[
]
. Second column is T (e 2 ) = −e 1 =
[
]
. Thus
the matrix is
[
]
1(c). Find the matrix for T : R^3 → R^3 , where T is rotation by 180◦^ about the y-axis, followed by rotation by 180◦^ about the z-axis.
Solution:
e 1 → −e 1 → e 1 e 2 → e 2 → −e 2 e 3 → −e 3 → −e 3 ,
so the columns of the matrix are e 1 , −e 2 , and −e 3. Thus the matrix is
- Find the determinant of the matrix C =
Solution:
∣ = 1(−5)^ −^ 4(−3) = 7, where we added row
1 to row 2 and subtracted 2 times row 1 from row 3, and then expanded using the first column. Instead of expanding using the first column we could also add 3 times row 2 to row 3:
∣ ∣ ∣ ∣ ∣ ∣ 1 2 3 0 1 4 0 − 3 − 5
- Find the inverse of the matrix A =
Solution:
so A−^1 is
4(a). Find all eigenvalues of the matrix A =
0 = det(λI − A) =
λ − 1 − 4 − 3 − 1 λ − 1 − 2 0 0 λ − 7
= (λ − 7)
λ − 1 − 4 − 1 λ − 1
Solution:
= (λ − 7)((λ − 1)^2 − (−1)(−4)) = (λ − 7)((λ − 1)^2 − 4)
Thus λ = 7 or (λ − 1)^2 = 4. In the latter case, λ − 1 = ±2, so λ = 1 ± 2, so λ = 3 or λ = −1. Thus the eigenvalues are 7, 3, and −1.
4(b). The matrix B =
[
]
has characteristic polynomial p(λ) = (λ − 15)(λ + 5). Find a
basis of R^2 consisting of eigenvectors of B.
Solution: For λ = 15, we find a nonzero vector in the nullspace of (15 I − B)v = 0. Note that
v 1 =
[
]
is such a vector. (Any nonzero scalar multiple of v 1 would also do.)
We can get the second basis eigenvector in either of two ways:
(1) By symmetry of B, we know any eigenvector with eigenvalue −5 must be orthogonal to v 1.
Thus we can just rotate v 1 by 90◦^ to get our second eigenvector: v 2 =
[
]
(2) For λ = −5, we find a nonzero vector in the nullspace of − 5 I − B =
[
]
. Note that
v 2 =
[
]
is such a vector.
4(c). Suppose that C is a symmetric 2x2 matrix with determinant 7. Suppose that the vector
v =
[
]
is an eigenvector of C with eigenvalue 5. Find an eigenvector w that is not a scalar
multiple of v, and find its eigenvalue. Explain.
Solution: w = 1v 1 + 2v 2 =
6(b). Find [v]B (the expression for v in the B-coordinate system) for the vector v =
Solution: [v]B =
[
c 1 c 2
]
where v = c 1 v 1 + c 2 v 2 , i.e., where
(^) = c 1
(^) + c 2
By Gaussian elimination (or just by inspection), c 1 = 2 and c 3 = −3. Thus [v]B =
[
]
7(a). Find the equation of the tangent plane to the surface z = x^2 y + y^3 at the point (x, y, z) = (2, 1 , 5).
Solution: Let f (x, y, z) = x^2 y + y^3. Note that ∂f∂x = 2xy and ∂f∂y = x^2 + 3y^2. Thus ∂f∂x (2, 1) = 4
and ∂f∂y (2, 1) = 7. The equation of the tangent plane is
z = f (2, 1) +
∂f ∂x
(2, 1)(x − 2) +
∂f ∂y
(2, 1)(y − 1)
or z = 5 + 4(x − 2) + 7(y − 1).
Alternate method: The surface is a level set of the function g(x, y, z) = x^2 y + y^3 − z. Note that ∇g = (2xy, x^2 + 3y^2 , −1). Thus ∇g(2, 1 , 5) = (4, 7 , −1) is a normal to the surface at the point (2, 1 , 5).
The equation for the tangent plane is thus
(4, 7 , −1) · ((x, y, z) − (2, 1 , 5)) = 0
or 4(x − 2) + 7(y − 1) − (z − 5) = 0.
7(b). Find the matrix DG(x, y) for the map G given by G(x, y) =
x^2 y y^3 x sin y
Solution: The first column is ∂G∂x or
2 xy 0 sin y
. The second column is ∂G ∂y or
x^2 3 y^2 x cos y
. Thus
DG(x, y) =
2 xy x^2 0 3 y^2 sin y x cos y
- Suppose that F : R^2 → R^2 is a map such that F (0, 0) =
[
]
and such that DF (0, 0) =
[
]
8(a). Estimate F (. 002 , .003).
Solution: Using the linear approximation at (0, 0) gives
F (. 002 , .003) ' F (0, 0) + DF (0, 0)
[
]
[
]
[
] [
]
[
]
[
]
[
]
8(b). Find a point (x, y) near (0, 0) so that
(*) F (x, y) '
[
]
Solution: If we use linear approximation to estimate F (x, y), then (*) becomes
[ 3 1
]
[
] [
x y
]
[
]
so (^) [ 1 1 2 1
] [
x y
]
[
]
Solving by Gaussian elimination gives x ' .003 and y ' .001.
9(a). Suppose v is an eigenvector of A with eigenvalue λ. Prove that v is also an eigenvector of I + A^2 , and find its eigenvalue.
Solution: (I + A^2 )v = Iv + A^2 v = v + A(Av) = v + A(λv) = v + λ(Av) = v + λ^2 v = (1 + λ^2 )v. Thus v is an eigenvector of I + A^2 with eigenvalue 1 + λ^2.
9(b). Suppose that A and B are similar matrices, i.e., that B = C−^1 AC for some invertible matrix C. Suppose that λ is a real number. Prove that λI − A and λI − B are also similar.
C−^1 (λI − A)C = C−^1 (λI)C − C−^1 AC = λC−^1 IC − B = λC−^1 C − B = λI − B
Thus λI − A and λI − B are similar.
