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MATH 51 MIDTERM 2 SOLUTIONS
March 9, 2006
1(a). Find the determinant of the matrix C =
Solution: for example,
∣ ∣ ∣ ∣ ∣ ∣ 1 2 3 − 1 − 1 1 2 1 1
1(b). Find the inverse of the matrix M =
. Ans:
so
M −^1 =
- Let v 1 =
[
]
and v 2 =
[
]
. Let T : R^2 → R^2 be the linear transformation
such that
(i) T (v 1 ) = 3v 1 + 2v 2
(ii) T (v 2 ) = −v 1 − v 2.
2(a). What is the matrix for T with respect to the basis B = {v 1 , v 2 }?
Solution: The first column of the matrix is [T (v 1 )]B, or
[
]
. The second column
is [T (v 2 )]B, or
[
]
. Thus the matrix is B =
[
]
2(b). What is the matrix for T with respect to the standard basis {e 1 , e 2 }?
Solution: The change of basis matrix is C =
[
]
(the matrix whose columns
are v 1 and v 2 .) The matrix for T in standard coordinates is A = CBC−^1. Now
C−^1 =
[
]
, so
A =
[
] [
] [
]
[
] [
]
[
]
3(a). Find all eigenvalues of the matrix A =
Solution:
0 = det(λI − A) =
λ − 1 − 4 − 3 − 1 λ − 1 − 2 0 0 λ − 7
= (λ − 7)
λ − 1 − 4 − 1 λ − 1
= (λ − 7) ((λ − 1)(λ − 1) − 4) = (λ − 7)(λ^2 − 2 λ − 3) = (λ − 7)(λ − 3)(λ + 1)
so the eigenvalues are 7, 3, and −1.
3(b). The matrix B =
[
]
has characteristic polynomial
p(λ) = (λ − 15)(λ + 5).
Find a basis for R^2 consisting of eigenvectors of B.
Solution:[ For λ = 15, we find a nonzero vector in the nullspace of 15I − B = 16 − 8 − 8 4
]
. Note v 1 =
[
]
is such a vector. (Any nonzero multiple 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 or- thogonal 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 v 2 =
[
]
is such a vector.
5(a). Find the following partial derivatives:
∂u ∂x
= 3 x^2 y^2 ∂u ∂y
= 2 x^3 y
∂v ∂x
= −y/x^2 ∂v ∂y
= 1 /x
∂^2 u ∂x∂y
= 6 x^2 y
5(b). Suppose f (u, v) is a function of u and v such that
∂f ∂u
= 12−v and
∂f ∂v
= 1−u.
Find ∂ ∂x
f (u(x, y), v(x, y))
at the point (x, y) = (1, 2).
Solution:
∂ ∂x
f (u(x, y), v(x, y)) =
∂f ∂u
∂u ∂x
∂f ∂v
∂v ∂x = (12 − v)
∂u ∂x
∂v ∂x = (12 − y/x)(3x^2 y^2 ) + (1 − x^2 y^2 )(−y/x^2 ) = 36x^2 y^2 − 3 xy^3 − y/x^2 + y^3.
Thus at the point (x, y) = (1, 2):
∂f ∂x
= 36(1)^2 (2)^2 − 3(1)(2)^3 − 2 /(1)^2 + 2^3 = 144 − 24 − 2 + 8 = 126
- Find the following:
6(a). The matrix for T : R^3 → R^3 , where T is rotation by 180◦^ about the y-axis followed by 180◦^ rotation about the z-axis.
Solution: Let Ry and Rz denote the 180◦^ rotations about the y and z axes, respectively. Then
T (e 1 ) = Rz (Ry e 1 ) = Rz (−e 1 ) = e 1 T (e 2 ) = Rz (Ry e 2 ) = Rz (e 2 ) = −e 2 T (e 3 ) = Rz (Ry e 3 ) = Rz (−e 3 ) = −e 3
so the matrix is (^)
6(b). The matrix for the linear map T given by
T
([
x y
])
3 x + 7y 13 y x − 4 y
Solution: (^)
6(c). The matrix for reflection in R^2 about the line y = −x.
Solution: e 1 7 → −e 2 and e 2 7 → −e 1 , so the matrix is
[
]
6(d). The product
[
] [
]
Solution:
[
]
- Find the equation of the tangent plane to the surface
xy + yz + zx = 11
at the point (1, 2 , 3).
Solution: The surface is a level set of the function f (x, y, z) = xy + yz + xz.
∇f (x, y, z) = (fx, fy , fz ) = (y + z, x + z, x + y)
so N = ∇f (1, 2 , 3) = (2 + 3, 1 + 3, 1 + 2) = (5, 4 , 3)
is a normal vector to the surface (and therefore also to the tangent plane) at (1, 2 , 3). Since the tangent plane passes through the point (1, 2 , 3), the equation is
(5, 4 , 3) · ((x, y, z) − (1, 2 , 3)) = 0
or (equivalently)
5(x − 1) + 4(y − 2) + 3(z − 3) = 0
or 5 x + 4y + 3z = 22.
