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
We have
A 5
a
1
b
1
c
1
a
2
b
2
c
2
a
3
b
3
c
3
2
4
3
5
; X 5
x
y
z
2
4
3
5
; and B 5
d
1
d
2
d
3
2
4
3
5
So, the relation X 5 A
21
B can be used to determine the values of x, y, and z that
satisfy the system.
Example 1
Solve the system
4x 1 3y 5 13
3x 1 y 524
Solution
The given system can be written in the form
AX 5 B
where
A 5
45
31

; X 5
x
y

; and B 5
13
24

The solution is given by X 5 A
21
B.
Let’s first find A
21
. Here
1 25
234

and A
jj
5 4 3 1 2 3 3 5 5211
A
21
5
1
A
jj
1
211
1 25
234

Now
X 5 A
21
B 5
1
211
1 25
234

13
24

52
1
11
33
255

5
23
5

.
x
y

5
23
5

Hence the required solution is x 523 and y 5 5.
47Linear Algebra
Example 2
Find the solution, if any, of the following system:
x 1 3y 2 z 523
3x 2 y 1 2z 5 1
2x 2 y 1 2z 521
8
<
:
Solution
The given system can be written as
13 21
3 212
2 212
2
4
3
5
x
y
z
2
4
3
5
5
23
1
21
2
4
3
5
So, we have AX 5 B, where
A 5
13 21
3 212
2 212
2
4
3
5
; X 5
x
y
z
2
4
3
5
; and B 5
23
1
21
2
4
3
5
Let’s find A
21
first. The cofactors of the elements of A are
A
11
5
212
212
5 0; A
12
52
32
22
522; A
13
5
3 21
2 21
521
A
21
52
3 21
212
525; A
22
5
1 21
22
5 4; A
23
52
213
2 21
527
A
31
5
3 21
212
5 5; A
32
52
1 21
32
525; A
33
5
13
3 21
5210
The matrix of cofactors is
0 22 21
254 7
5 25 210
2
4
3
5
Again
A
jj
5 a
11
A
11
1 a
12
A
12
1 a
13
A
13
5 1 3 0 1 3 3 ð22Þ1 ð21Þ3 ð21Þ525 0
A
21
5
1
A
jj
1
2 5
0 255
224 25
217 210
2
4
3
5
48 Mathematical Formulas for Industrial and Mechanical Engineering

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