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Changing 3 to 2 in the corner reduces the determinant F2nC2 by 1 times the cofactor
of that corner entry. This cofactor is the determinant of Sn 1 (one size smaller) which
is F2n . Therefore changing 3 to 2 changes the determinant to F2nC2 F2n which is
F2nC1 . Solutions to Exercises 55 23 (a) If we choose an entry from B we must choose an entry from the zero block; re sult zero. This leaves entries
from A
times entries from Dleading .det A/.det D/
to
10
00
01
00
(b) and (c) Take A D
,B D
,C D
,DD
. See #25.
00
10
00
01 24 (a) All L’s have det D 1I det Uk D det Ak D 2; 6; 6 for k D 1; 2; 3 26 27
28
29 30 1
.
3
0
AB
D 1 and det
D jAj times jD CA 1 B j
CD
CA 1
I
which is jAD ACA 1 B j. If AC D CA this is jAD CAA 1 B j D det.AD CB/.
If A is a row and B is a column then det M D det AB D dot product of A and B . If
A is a column and B is a row then AB has rank 1 and det M D det AB D 0 (unless
m D n D 1). This block matrix is invertible when AB is invertible which certainly
requires m n.
(a) det A D a11 C11 C C a1n C1n . Derivative with respect to a11 D cofactor C11 .
Row 1 2 row 2 C row 3 D 0 so this matrix is singular.
There are ﬁve nonzero products, all 1’s with a plus or minus sign. Here are the (row,
column) numbers and the signs: C .1; 1/.2; 2/.3; 3/.4; 4/ C .1; 2/.2; 1/.3; 4/.4; 3/
.1; 2/.2; 1/.3; 3/.4; 4/ .1; 1/.2; 2/.3; 4/.4; 3/ .1; 1/.2; 3/.3; 2/.4; 4/. Total 1.
The 5 products in solution 29 change to 16 C 1 4 4 4 since A has 2’s and 1’s: 25 Problem 23 gives det (b) Pivots 2; 3 ;
2 I .2/.2/.2/.2/ C . 1/. 1/. 1/. 1/ . 1/. 1/.2/.2/
.2/. 1/. 1/.2/: .2/.2/. 1/. 1/ 1 because the cofactor of P14 D 1 in row one has sign . 1/1C4 . The big
formula for det P has only one term .1 1 1 1/ with minus sign because three
exchanges
0I
D
take 4; 1; 2; 3 into 1; 2; 3; 4; det.P 2 / D .det P /.det P / D C1 so det
I0
01
is not right.
det
10 31 det P D 32 The problem is to show that F2nC2 D 3F2n F2nC2 D F2nC1 C F2n D F2n C F2n F2n 2 . Keep using Fibonacci’s rule:
1 C F2n D 2F2n C .F2n F2n 2 / D 3F2n F2n 2: 33 The difference from 20 to 19 multiplies its 3 by 3 cofactor D 1: then det drops by 1.
34 (a) The last three rows must be dependent (b) In each of the 120 terms: Choices
from the last 3 rows must use 3 columns; at least one of those choices will be zero.
35 Subtracting 1 from the n; n entry subtracts its cofactor Cnn from the determinant. That
cofactor is Cnn D 1 (smaller Pascal matrix). Subtracting 1 from 1 leaves 0. Problem Set 5.3, page 279
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
5ˇ
ˇ D 3; ˇ 1 5 ˇ D 6; ˇ 2 1 ˇ D 3 so x1 D 6=3 D 2 and x2 D
ˇ
ˇ2 4ˇ
ˇ1 2ˇ
4
3=3 D 1 (b) jAj D 4; jB1 j D 3; jB2 j D 2; jB3 j D 1: Therefore x1 D 3=4 and
x2 D 1=2 and x3 D 1=4.
ˇ
ˇ2 1 (a) ˇ
ˇ1 Solutions to Exercises 56
ˇ ˇˇ ˇ ˇ
ˇ ˇ bˇ
2 (a) y D ˇ a 1 ˇ = ˇ a d ˇ D c=.ad
c
c0 b c/ (b) y D det B2 = det A D .fg i d /=D . 3 (a) x1 D 3=0 and x2 D 2 =0: no solution
(b) x1 D x2 D 0=0: undetermined.
4 (a) x1 D det Œ b a2 a3 = det A, if det...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

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