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MATH 51 FINAL EXAM SOLUTIONS
Professors Clingher, Munson, and White March 15, 2004
1 Consider the matrices
A =
and^ R^ =
The matrix R is the row reduced echelon form of A. (You do not need to check this.)
1(a). Find a basis for the column space of A. Solution: R has pivots in columns 1, 2 and 4, so the corresponding columns of A form the basis:
1(b). Find a basis for the column space of R. Solution: R is its own rref, so again the pivot columns of R (that is, the first, second, and fourth columns) form the basis: e 1 , e 2 , and e 3.
1(c). Find a basis for the nullspace of A. The nullspace of A consists of all solutions of Ax = 0, or, equivalently, of Rx = 0. From Rx = 0, we see that the free variables are x 3 and x 5 and that
x =
−x 3 − x 5 x 3 − x 5 x 3 − 2 x 5 x 5
= x 3
Thus
and
form a basis for N (A).
- Find all solutions of
x 1 + 2 x 2 + x 3 + x 4 = 7 x 1 + 2 x 2 + 2 x 3 − x 4 = 12 2 x 1 + 4 x 2 + 6 x 4 = 4.
Solution: see example 6.4 on page 44 of the Levandosky text. The rref is
from which we get the solution:
x =
+^ x 2
+^ x 4
3(a). Find all eigenvalues of the matrix A =
. Solution:
0 = det(λI − A) =
λ − 5 0 0 − 1 λ − 2 − 1 − 1 − 1 λ − 2
= (λ − 5)
λ − 2 − 1 − 1 λ − 2
= (λ − 5)((λ − 2)^2 − (−1)^2 ) = (λ − 5)(λ − 3)(λ − 1)
so the eigenvalues are 5, 3, and 1.
3(b). The matrix M =
(^) has λ = 2 as one of its eigenvalues. (You
need not check this.) Let V be the eigenspace corresponding to this eigenvalue. (In other words, V consists of all eigenvectors with eigenvalue 2 together with the origin.) Find a basis for V.
Solution: 2I − M =
. Note that V is the nullspace of 2I − M. We
find this nullspace by solving (2I − M )v = 0. Solving by Gaussian elimination, we get the single equation x − 2 y − 2 z = 0
so y and z are free and
v =
x y z
2 y + 2z y z
(^) = y
(^) + z
Thus
(^) and
(^) form a basis for V.
- The velocity of a certain spaceship at time t is given by v(t) = (3t^2 , et−^1 , 6 t). At time t = 1, its position is (0, 0 , 7).
(a) Find the speed at time t. Solution:
‖v(t)‖ =
(3t^2 )^2 + (et−^1 )^2 + (6t)^2 =
9 t^4 + e^2 t−^2 + 36t^2.
(c). lim(x,y)→(0,0)
x^3 − y^3 x^2 + y^2
. Solution: In polar coordinates, the expression is
r^3 cos^3 θ − r^3 sin^3 θ r^2
= r(sin^3 θ − cos^3 θ)
which goes to 0 as r → 0. Hence the limit exists and is 0.
- Find the matrix derivative (i.e., the Jacobian matrix) DF (x, y, z) where
F (x, y, z) =
[
x + y^2 + z^3 ey^ + y sin z
]
[
1 2 y 3 z^2 0 ey^ + sin z y cos z
]
Solution:
- Find A−^1 , where A =
Solution: Row reduce [A : I]. An easy way to row reduce it is: subtract row 4 from each of the first 3 rows, then subtract row 3 (of the resulting matrix) from each the first two rows, then subtract row 2 from row 1:
1 2 3 4
(where in the last step we divided row 4 by 4, row 3 by 3, row 2 by 2.)
Hence A−^1 =
- Find the point or points (x, y) at which the function
f (x, y) =
x^4 4
− xy +
y^2 2
is a minimum. (You may assume that the minimum exists. Note the different exponents.)
Solution: At the minimum, 0 = fx = x^3 − y and 0 = fy = −x + y. So (from the second equation) y = x. Combining with the first gives 0 = x^3 − x = x(x^2 − 1) = x(x − 1)(x + 1), so x = −1, 0, or 1. Since y = x, the minimum must be at (− 1 , −1), (0, 0), or (1, 1). The corresponding values of f are − 1 /4, 0, and − 1 /4. Thus the minimum occurs at (− 1 , −1) and at (1, 1).
9(a). Consider the linear transformation T : R^2 → R^2 given by first reflecting across the line y = x, and then rotating counterclockwise around the origin by an angle of π/2. Find the matrix A for this linear transformation (with respect to the standard basis {e 1 , e 2 } of R^2 ).
Solution: The vector e 1 gets reflected to e 2 , which is then rotated to −e 1. So the first column of A is −e 1. The vector e 2 gets reflected to e 1 , which is then rotated to e 2. So the second column is e 2. Thus
A =
[
]
9(b). Find the matrix B for T with respect to the basis B consisting of v 1 = (1, 1) and v 2 = (− 1 , 1). (Note: it is possible to do part (b) directly, without doing any calculations and without using the answer to part (a).)
Solution: The vector v 1 gets reflected to itself, then rotated to v 2 = 0v 1 + 1v 2.
Thus the first column of B is
[
]
. The vector v 2 gets reflected to −v 2 , which is
then rotated to v 1 = 1v 1 + 0v 2. So the second column of B is
[
]
. Thus
B =
[
]
Another approach: The change of basis matrix is
C =
[
]
so
B = C−^1 AC =
[
] [
] [
]
[
]
- The temperature at point (x, y) on the floor of a room is given by f (x, y) = xy^2.
Plugging this into the constraint equation gives:
k^2 + k^2 /2 + 4k^2 = 22 (11/2)k^2 = 22 k^2 = 4
so k = 2 or k = −2. Hence (by (*)) (x, y, z) is either (2, 2 , 4) or (− 2 , − 2 , −4).
f (2, 2 , 4) = 2(2) + 2 + 4(4) = 22, f (− 2 , − 2 , −4) = − 22
so the maximum and minimum values are 22 and −22, respectively.
12(a) Suppose that x and y are vectors in Rn^ with ‖x‖ = 2 and ‖y‖ = 1. Suppose also that the angle between x and y is θ = arccos(1/4). Prove that the vectors x − 3 y and x + y are orthogonal.
Solution: x · y = ‖x‖ ‖y‖ cos θ = (2)(1)(1/4) = 1/ 2 ,
so
(x − 3 y) · (x + y) = ‖x‖^2 − 2 x · y − 3 ‖y‖^2 = 2^2 − 2(1/2) − 3(1^2 ) = 0.
Thus the vectors are perpendicular.
12(b). Prove there is no matrix A (with real entries) such that A^2 =
(Hint: use determinants.)
Solution: If the equation were true, then
det(A^2 ) =
But det(A^2 ) = det(AA) = (det A)(det A) = (det A)^2. So det A would have to be a (real) number whose square is −4. But that’s impossible.
13(a). Find a normal vector to S at p = (1, 2 , 3), where S is the surface
x + y + z + xyz = 12.
Solution: The surface is a level set of f (x, y, z) = x + y + z + xyz, so
∇f = (1 + yz, 1 + xz, 1 + xy)
is normal at (x, y, z). Plugging in (x, y, z) = (1, 2 , 3) gives a normal vector at p:
N = ∇f (p) = (7, 4 , 3).
13(b). Find an equation for the tangent plane to S at p (where S and p are in part (a).)
Solution: The equation is N · (v − p) = 0 (where v = (x, y, z)), or
(7, 4 , 3) · ((x, y, z) − (1, 2 , 3)) = 0
or (7, 4 , 3) · ((x − 1 , y − 2 , z − 3) = 0
or 7(x − 1) + 4(y − 2) + 3(z − 3) = 0.
14(a). Consider a function G : R^2 → R^2 such that
G(5, 2) =
[
]
and DG(x, y) =
[
(xy − 1)^1 /^3 (xy − 1)^1 /^3 sin(y − 2) x cos(y − 2)
]
Estimate G(5. 02 , 2 .001). (Hint: use linear approximation. Here DG denotes the Jacobian matrix, i.e., the matrix derivative.))
G(5. 02 , 2 .001) ∼= G(5, 2) + DG(5, 2)
[
]
Solution:
[
]
[
] [
]
[
]
[
]
[
]
14(b). Consider a function H : R^2 → R^2 such that:
H(0, 0) = (0, 0) and DH(0, 0) =
[
]
Suppose (x, y) is a point near (0, 0) such that H(x, y) = (. 07 , .06). Estimate x and y. (Hint: use linear approximation. Here DH denotes the Jacobian matrix, i.e., the matrix derivative.)
Solution:
[
]
= H(x, y) ∼= H(0, 0)+DH(0, 0)
[
x y
]
[
] [
x y
]
[
x + y 2 x
]
so
x + y ∼=. 07 2 x ∼=. 06
solving (just as if we had equality instead of ∼=) gives x ∼=. 03 and y ∼=..
- Let P be the plane given by
x 1 + x 2 + x 3 = 0.
