This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ge 3, then
edge 2 backwards. This corresponds to
Row 1 + Row 3 
Row being the 0 Row.
2
3 −1
x1
1
Setting x3 = 0, we have to solve
=
. Doing this by hand,
x2
0 −1
3
3/8
c
x1 = 3/8, x2 = 1/8, so = 1/8 + c gives the potentials.
x
0
c
Finally, the currents are given by 100
−1
10
3/8
−1
10
3/8
1/4
0 1 1/8 = − −2
0 2 1/8 = 3/4 .
−CA = 0 2 0 −1
x
002
0 −1 1
0
0 −2 2
0
1/4 10) A = −1
1
0
−1
0
1
0 −1
1
0 −1
0
0
0 −1 0
0
0
1
1 → 1 −1
0
−1
0
1
0 −1
1
0 −1
0
0
0 −1 0
0
0
1
1 → 1
0
0
0
0 −1
0
−1
1
−1
1
−1
0
0 −1 0
0
0
1
1 1 −1
00
1 −1
00
1 −1
00
1 −1
00
0 0 0 0
1 −1 0 1 −1 0 1 −1 0 1 −1 0 0
0 0 → 0
0
0 0 → 0
0 −1 1
→ 0 −1 −1 0 → 0 0 −1 0 0 0
01
0 −1 1
0 −1 1
0
00
0
0 −1 1
0
0 −1 1
0
0
00
0
0
00
an echelon form. From the second to last matrix (before the row swap to move zero rows to
the bottom) we have that edges 1, 2 and 4 are kept. The reason is from edges 1 and 2 we
can get edge 3. and from 3 and 4 we can get edge 5.
The other spanning trees correspond to the following triples of edges: {1, 2, 5}, {1, 3, 4},
{1, 3, 5}, {1, 4, , 5}, {2, 3, 4}, {2, 3, 5} and {2, 4, 5}. 1
0
1
0
1
0
1
0
12) (a) It’s almost a deﬁnition that A = so certainly AT A = .
1
0
1
0
1
0
1
0
T
As A A and A have the same rank and we know from exercise 10 that A hass rank 3, then
AT A has rank 3.
(b) For any matrix B , B T B is always positive semideﬁnite as T B T B = (B )T B =
x
x
x
x
2
B  ≥ 0.
x 1
0
1 0
Choosing B = A and noting A = , we see AT A is not positive deﬁnite.
1 0
1
0 ,...
View
Full
Document
This note was uploaded on 01/30/2014 for the course MATH 2940 taught by Professor Hui during the Fall '05 term at Cornell.
 Fall '05
 HUI
 Linear Algebra, Algebra

Click to edit the document details