-2 | 1 | 5 | 1 | 2 | 3 |
0 | 4 | 9 | -1 | -6 | -4 |
1 | -3 | 7 | 0 | 1 | -8 |
5 | 1 | -1 | 1 | 5 | 3 |
Addition | Scalar multiplication | |
---|---|---|
u + (v + w) = (u + v) + w | a×(b×u) = (ab)×u | |
Identitiy: u + 0 = 0 + u = u | (a + b)×u = a× u + b×u | |
Inverse: u + (-u) = (-u) + u = 0 | a×(u + v) = a×u + a×v | |
u + v = v + u | 1×u = u |
1 | -1 | 1 |
3 | -1 | 2 |
1 | 3 |
3 | -1 |
4 | 2 |
1 | 1/2 | 1/3 | 1/4 | 1/5 | 1/6 |
1/2 | 1/3 | 1/4 | 1/5 | 1/6 | 1/7 |
1/3 | 1/4 | 1/5 | 1/6 | 1/7 | 1/8 |
1/4 | 1/5 | 1/6 | 1/7 | 1/8 | 1/9 |
1/5 | 1/6 | 1/7 | 1/8 | 1/9 | 1/10 |
1/6 | 1/7 | 1/8 | 1/9 | 1/10 | 1/11 |
-1 |
2 |
3 |
4 |
-1 |
3 |
1 | -2 | 3 |
2 | -5 | 4 |
-1 | 3 | 4 |
L = M1-1 M2-1 = I + t1 e1T + t2 e2T =
1 | 0 | 0 |
2 | 1 | 0 |
-1 | -1 | 1 |
and the upper triangular matrix U = M2M1A is
U =
1 | -2 | 3 |
0 | -1 | -2 |
0 | 0 | 5 |
4 | -1 | 1 |
1 | 4 | 6 |
-2 | 3 | -7 |
2 | -2 | 1 |
-5 | 5 | -9 |
-4 | 3 | 1 |
2 | -2 | -1 |
-1 | -1 | 3 |
-4 | -6 | 2 |
A =
12 | 6 | 4 | 3 |
6 | 4 | 3 | 2 |
4 | 3 | 3 | 2 |
3 | 2 | 2 | 3 |
f ''(jh) = -h-2 A(j+1,:)*y + O(h), j = 0:(n-1),
where A is the matrix given in Problem 4.3.11.
B =
12 | 6 | 4 | 3 |
6 | 4 | 3 | 2 |
4 | 3 | 3 | 2 |
3 | 2 | 2 | 3 |
A =
3 | 1 | -4 | -3 |
-1 | -4 | 5 | 1 |
-1 | 0 | 1 | 1 |
1 | -2 | 1 | -1 |
1 | 3 | -4 | -1 |