Machinery's Handbook, 31st Edition
USING MATRICES TO SOLVE A SYSTEM OF EQUATIONS 129 Example 10: Use matrix operations to find the amperages of the currents ( I 1 , I 2 , I 3 ) in the following electrical network. By Kirchoff’s first law, concerning the sum of currents: that is, I 1 I 2 + I 3 = I 1 I 2 I 3 + – = 0 I 1 + − 40V I 3 I 2 2 Ω 5 Ω 10 Ω A
By Kirchoff’s second law, concerning the sum of voltages, and Ohm’s law, that voltage is the product of current and resistance:
− + 30V
B
2 I 1 5 I 3 + – 40 = 0 10 I 2 5 I 3 + – 30 = 0
By combining the three equations, a linear system of independent equations is formed. A system is linearly independent if no one equation is a constant multiple of any other. That is, the equations represent distinct lines, but they are not parallel (hence, they have a solution, that is, a point of intersection). Solve the system for the currents I 1 , I 2 , and I 3 :
I 1 I 2 I 3 + – = 0 2 I 1 5 I 3 + = 40 10 I 2 5 I 3 + = 30
Solution: If A is the matrix of coefficients of the currents, B is the matrix of the currents themselves (the variables), and C is the matrix of constants from the right side of the equations (the voltages), then the problem can be written in the following form: AB = C , or equivalently, B = A - 1 C , where A - 1 is the inverse of matrix A . Thus,
–1
I 1 I 2 I 3
I 1 I 2 I 3
1 1 –1 2 0 5 0 10 5
0 40 30
1 1 –1 2 0 5 0 10 5
0 40 30
=
=
=
and
=
A
B
C
Using the method of Example 8, the inverse of matrix A is
5 8 -- 1 8 --
3 16 ---
1 16 –---
–1
1 1 –1 2 0 5 0 10 5
50 15 –5 10 –5 7 –20 10 2
1 80 –---
7 80 --- 1 40 ---
1 16 –---
–1
=
=
=
A
1 8 --
1 4 –--
and finally, matrix B can be found by matrix multiplication:
5 8 -- 1 8 --
3 16 ---
1 16 –---
0 40 30
5.625 0.125 5.75
7 80 --- 1 40 ---
1 16 –---
–1
B A =
=
=
C
1 8 --
1 4 –--
Thus, I 1 = 5.625 amps, I 2 = 0.125 amp, and I 3 = 5.75 amps.
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