20.9 First-order differential equations 585
The first stage is to reduce Ato upper triangular form by Gauss elimination:
The second stage is to reduce the left half of the augmented matrix to diagonal form
by further elimination steps:
The left half is reduced to the unit matrix by dividing row 2 by 5 and row 3 by.
Then
and the right half of the augmented matrix is the inverse matrixA
− 1:
and
0 Exercises 32, 33
20.9 First-order differential equations
We saw in Chapters 11 to 14 that some important differential equations can be
solved by analytical methods to give solutions that are expressed in terms of known
functions. Many differential equations cannot however be solved in this way, either
AA AA I
−−==
11A
−=
−
−−
−
11
6
111
4108
175
100
010
001
161616235343167656
−
−−
−
=
( )
−IA|
165row row 12
100
050
00
256516161−×
→
−
6610325320315751
−−
−
row row
row row
13
23
120
05
52203−×
−×
→ 00
00
1
653272521032532031575
−−
−−
−
row row 32
12 3
05 8
00
100
210
1
75651575+×
→
−
row row
row row
22 1
33 1
12 3
05 8
07
+×
−×
→
−− 110
100
210
− 301