
B
10
10
3
:
4
3
01 2
2
3
:
1
3
001:
1
4
2
6
6
6
6
6
6
6
6
4
3
7
7
7
7
7
7
7
7
5
; R
3
! 2
3
32
R
3
B
100:
1
2
010:
1
2
001:
1
4
2
6
6
6
6
6
6
6
6
4
3
7
7
7
7
7
7
7
7
5
; R
1
! R
1
2
10
3
R
3
and R
2
! R
2
1
2
3
R
3
‘ The required solution is x 5
1
2
; y 5
1
2
; z 5
1
4
.
3.12 Inverse Matrix Method
Consider the follow ing system of linear equations:
a
1
x 1 b
1
y 5 c
1
a
2
x 1 b
2
y 5 c
2
:
This system can also be written in the form
a
1
b
1
a
2
b
2
x
y
5
c
1
c
2
.
So, we have AX 5 B, where A 5
a
1
b
1
a
2
b
2
; X 5
x
y
; and B 5
c
1
c
2
.
Now
AX 5 B
.A
21
ðAXÞ5 A
21
B
.ðA
21
AÞX 5 A
21
B
.IX 5 A
21
B
.X 5 A
21
B
For a system of three linear equations:
a
1
x 1 b
1
y 1 c
1
z 5 d
1
a
2
x 1 b
2
y 1 c
2
z 5 d
2
a
3
x 1 b
3
y 1 c
3
z 5 d
3
8
<
:
46 Mathematical Formulas for Industrial and Mechanical Engineering