Machinery's Handbook, 31st Edition
Matrix Operations
127
The determinant of A is A
2 1 0 4 6 × – ( × ) 3 4 0 1 6 × – ( × ) – 5 4 4 1 1 × – ( × ) + = 2 0– 24 ( ) 3 0– 6 ( ) – 5 16– 1 ( ) + = = –48+18+75 = 45
The cofactors are
1+1 1 6 4 0 2+1 3 5 4 0 3+1 3 5 1 6
1+2 4 6 1 0 2+2 2 5 1 0 3+2 2 5 4 6
1+3 4 1 1 4 2+3 2 3 1 4
–1 ( )
–1 ( )
–1 ( )
=
=
=
=
–24
=
=
6
15
a 11
a 12
a 13
–1 ( )
–1 ( )
–1 ( )
=
=
=
=
20
=
=
–5
–5
a 21
a 22
a 23
3+3 2 3 4 1
–1 ( )
–1 ( )
–1 ( )
=
=
=
=
13
=
=
8
–10
a 31
a 32
a 33
–24 6 15 20 –5 –5 13 8 –10
–24 20 13 6 –5 8 15 –5 –10
The matrix of cofactors is
and the adjoint matrix is
Then the inverse of matrix A is
–24 20 13 6 –5 8 15 –5 –10
adj A ( ) A --------
1 45 ---
–1
=
=
A
Multiplying A - 1 by A results in the identity matrix:
–24
20
13
2
3
5
1
0
0
45
0
0
1 45
1 45 =
=
6
–5
8
4
1
6
0
1
0
0
45
0
15
–5
–10
1
4
0
0
0
1
0
0
45
Solving a System of Equations.— Matrices can be used to solve systems of simultaneous equations with a large number of unknowns. Geometrically, the solution of a system is the point in the plane or in space where the lines intersect. Variables may represent the unknowns in industrial and other engineering applications: a series of resistances in a circuit, for example, or, perhaps, factors in a manufacturing process, such as labor costs, materials, and equipment. Generally, this method is less cumbersome than using substitu- tion methods (see Solving a System of Linear Equations in ALGEBRA ). The coefficients of the equations are placed in matrix form. The matrix is then transformed by row and column operations into the identity matrix to yield a solution. The process, as described in the example, is called matrix reduction to row echelon form . It is done using any or all of three valid vector operations: 1) multiplying each entry in a row by a constant; 2) adding or subtracting two rows; 3) changing the order of rows. Example 9: Solve the system of linear equations in three dimensions using matrix operations. 4 x 1 – 8 x 2 12 x 3 + + = 16
3 x 1 x 2 – 2 x 3 + = 5 x 1 7 x 2 6 x 3 + + = 10
Solution: First, the equation coefficients and constants are placed into what is called an augmented matrix . The object is to transform the matrix of the original coefficients into the following form, thereby obtaining a solution ( x 1 , x 2 , x 3 ) to the system of equations.
Copyright 2020, Industrial Press, Inc.
ebooks.industrialpress.com
Made with FlippingBook - Share PDF online