Section |
Questions |
1.2 |
6,
7, 8, 9 |
1.3 |
2,
5, 7, 9, 11, 15 |
1.4 |
1,
2, 5, 7, 9, 10, 11 |
1.5 |
1,
3, 4, 7, 9, 11, 12, 13, 22, 23, 24 |
|
|
2.2 |
1,
2, 3, 4, 5, 6, 7, 8, 16 |
|
|
3.1 |
1,
2, 3, 4, 5, 6 |
3.2 |
1,
2, 3, 4, 5, 7, 8 |
3.3 |
3,
4, 6, 7, 8, 11, 12, 13, 15, 16, 17, 18 |
3.4 |
1,
2, 3, 4, 5, 11, 14, 21, 22, 23, 24 |
|
|
4.1 |
1,
2, 4, 6, 16, 17 |
4.4 |
1,
2, 3, 4, 5, 7, 8, 9, 13, 14, 15, 16, 17, 19, 23 |
4.5 |
2,
3, 4, 5, 6, 8, 9, 10 |
|
|
Extra
Problems.
Show that for any non-zero vector x, xTAx
> 0, and thus, A is positive definite.
Hint. Let f(x1,
x2, x3) = xTAx. Then write it as a sum of squares. Use the LU Decomposition, if you wish, to help you get the expression
to look like a sum of squares.
Maple Assignments.