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202 ALGEBRA
There exist various methods for the construction of LU-decompositions. In particular
if the following conditions hold:
a 11 a 12
a 11 ≠ 0, ≠ 0, ... , det A ≠ 0,
a 21 a 22
then the elements of the desired matrices L and U can be calculated by the formulas
j–1
⎧
⎪
⎨ a ij – for i ≥ j,
l ij = l is u sj
s=1
0 for i < j,
⎪
⎩
⎪ 1
⎧ i–1
a ij – for i < j
⎪
⎨ l is u sj
u ij = l ii s=1
⎪ 1 for i = j,
⎪
0 for i > j.
⎩
5.5.2-3. Solutions of a square system with different right-hand sides.
◦
1 . One often has to solve a system of linear equations with a given basic matrix A
(1)
and different right-hand sides. For instance, consider the systems AX (1) = B , ... ,
AX (m) = B (m) .These m systems can be regarded as a single matrix equation AX = B,
where X and B are matrices of size n × m whose columns coincide with X (j) and B (j)
(j = 1, 2, ... , m).
Example 5. Suppose that we have to solve the equation AX = B with the given basic matrix A and
different right-hand sides:
7
( 1 2 –3 ) ( ) ( 10 )
A = 3 –2 1 , B (1) = 1 , B (2) = 6 .
–2 1 3 5 –5
Using the Gauss-Jordan procedure, we obtain
1 2 –3 710 12 –3 7 10 10 –1/2 2 4 10 0 34
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎝ 3 –21 16 ⎠ → ⎝ 0 –810 –20 –24 ⎠ → ⎝ 01 –5/4 5/23 ⎠ → ⎝ 01 0 53 ⎠ .
–21 3 5 –5 05 –3 19 15 00 13/4 1/32 0 00 1 20
Therefore,
3
4
( ) ( )
(1) (2)
X = 5 , X = 3 .
2 0
2 .If B = I,where I is the identity matrix of size n × n, then the solution of the matrix
◦
–1
equation AX = I coincides with the matrix X = A .
Example 6. Find the inverse of the matrix
2 1 0
( )
A = –307 .
–541
Let us transform the augmented matrix of the system, using the Gauss-Jordan method. We get
⎛ ⎞ ⎛ 1 1/2 0 1/2 00 ⎞
21 0 100
–30 7 010 0 3/2 7 3/2 10
⎝ ⎠ → ⎝ ⎠ →
–54 –1 001 0 13/2 –1 5/2 01
10 –7/3 0 –1/3 0 10 0 14/47 –1/94 –7/94
⎛ ⎞ ⎛ ⎞
01 14/3 1 2/3 0 01 0 19/47 1/47 7/47 ⎠ .
→ ⎝ ⎠ → ⎝
00 –94/3 –4 –13/3 1 00 1 6/47 13/94 –3/94