Due at noon, Friday, December 5, 1997
Extended to 1:00pm, Monday, December 8, 1997
You may do the calculations necessary for these problems either by hand or with Maple. Please submit a handwritten solution, or email an ASCII/Postscript/html document to the TA. (You may also submit through email which attached your Maple worksheet.)
Solution:
C(n) = 1.387891671 x ln( x ) - 0.955521423 x
+ 0.56617708.
Maple worksheet(Maple text)
Solution:
(a) Yes.
(b) No.
L( x1, y1, z1) +
L( x2, y2, z2) =
(1, 2, -1) + (1, 2, -1) = (2, 4, -2)
not equivalent to
L( x1 + x2, y1 + y2,
z1 + z2) = (1, 2, -1).
(c) No.
L(1, 1, 1) + L(1, 1, 1) = (12 + 1, 1 - 1) +
(12 + 1, 1 - 1) = ( 2, 0)
not equivalent to L(1 + 1, 1 + 1, 1 + 1)= L(2, 2, 2) =
(22 + 2, 2 - 2) = (6, 0).
| [ | 5 | 0 | -4 | ] |
| [ | 0 | -3 | 0 | ] |
| [ | -4 | 0 | -1 | ] |
(a) - ( lambda - 7) ( lambda + 3 )2.
(b) 7, -3.
(c) {(-2, 0, 1)} when lambda = 7 ;
{(1, 0, 2), (0, 1, 0)} when lambda =-3.
| [ | x1 | ] | [ | x2 | ] | ||||
| < | [ | y1 | ] | , | [ | y2 | ] | >a -> | x1x2 + 2 y1y2 + 3 z1z2 |
| [ | z1 | ] | [ | z2 | ] |
(b)
< , >b : R3 × R3 ->
R
| [ | x1 | ] | [ | x2 | ] | ||||
| < | [ | y1 | ] | , | [ | y2 | ] | >b -> | x1z2 + y1y2 + z1x2 |
| [ | z1 | ] | [ | z2 | ] |
(a) Yes. || (x, y, z) || = ( x2 +
2y2 + 3z2 )1/2.
(b) No. When vT = [ 1, 21/2, -1 ] != 0,
< v, v >b = 0.
| [ | 1 | 2 | 1 | 0 | ] | ||
| A = | [ | 2 | 3 | 4 | 7 | ] | . Let LA(v) = A·v. Find: |
| [ | 3 | 5 | 5 | 7 | ] | ||
| [ | 3 | 4 | 7 | 14 | ] |
(a) The null space of A, the vector space spanned by {(-5, 2, 1, 0),
(-14, 7, 0, 1)}.
(b) The column space of A, Span([1, 2, 3, 3]T, [2, 3, 5,
4]T, [1, 4, 5, 7]T, [0, 7, 7,
14]T).
| [ | 3 | 3 | ] |
| [ | 4 | 2 | ] |
(a) The general solution:
| [ | 1 | ] | [ | 3/2 | ] | |||
| C1 | [ | -2 | ] | e-5t + C2 | [ | 1 | ] | e3t |