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Appendix A
Simultaneous Equations
and Matrix Inversion
In circuit analysis, we often encounter a set of simultaneous equations
having the form
a11x1 a12 x2 p a1n xn b1
a21x1 a22 x2 p a2n xn b2
(A.1)
o
o
o
an1x1 an2 x2 p ann xn bn
where there are n unknown x1, x2, p , xn to be determined. Equation
(A.1) can be written in matrix form as
a11 a12 p a1n
x1
b2
a21 a22 p a2n
x2
b2
≥
¥ ≥ ¥ ≥ ¥
(A.2)
p
o
o
o
o
o
an1 an2 p ann
xn
bn
This matrix equation can be put in a compact form as
AX B
(A.3)
where
a11 a12
a21 a22
A ≥
o
o
an1 an2
p
p
p
p
a1n
a2n
¥,
o
ann
x1
x2
X ≥ ¥,
o
xn
b1
b2
B ≥ ¥
o
bn
(A.4)
A is a square (n n) matrix while X and B are column matrices.
There are several methods for solving Eq. (A.1) or (A.3). These
include substitution, Gaussian elimination, Cramer’s rule, matrix inversion, and numerical analysis.
A.1
Cramer’s Rule
In many cases, Cramer’s rule can be used to solve the simultaneous equations we encounter in circuit analysis. Cramer’s rule states that the solution
to Eq. (A.1) or (A.3) is
¢1
¢
¢2
x2 ¢
o
¢n
xn ¢
x1 A
(A.5)
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Appendix A
Simultaneous Equations and Matrix Inversion
where the ¢ ’s are the determinants given by
a11 a12 p a1n
a21 a22 p a2n
¢4
4,
o
o p o
an1 an2 p ann
¢2 4
¢1 4
b1
b2
o
bn
a12 p a1n
a22 p a2n
4
o p o
an2 p ann
o
p a1n
a11 a12
p a2n
a21 a22
p o 4, p , ¢ n 4 o
o
p ann
an1 an2
a11 b1
a21 b2
o
o
an1 bn
o
p b1
p b2
p o 4
p bn
(A.6)
Notice that ¢ is the determinant of matrix A and ¢ k is the determinant of the matrix formed by replacing the kth column of A by B. It
is evident from Eq. (A.5) that Cramer’s rule applies only when ¢ 0.
When ¢ 0, the set of equations has no unique solution, because the
equations are linearly dependent.
The value of the determinant ¢ , for example, can be obtained by
expanding along the first row:
a11 a12 a13 p a1n
a21 a22 a23 p a2n
¢ 5 a31 a32 a33 p a3n 5
o
o
o p o
an1 an2 an3 p ann
(A.7)
a11M11 a12 M12 a13 M13 p (1)1na1n M1n
where the minor Mi j is an (n 1) (n 1) determinant of
the matrix formed by striking out the ith row and jth column. The
value of ¢ may also be obtained by expanding along the first
column:
¢ a11 M11 a21M21 a31M31 p (1)n1an1 Mn1
(A.8)
We now specifically develop the formulas for calculating the determinants of 2 2 and 3 3 matrices, because of their frequent occurrence in this text. For a 2 2 matrix,
¢2
a11 a12
2 a11a22 a12 a21
a21 a22
(A.9)
For a 3 3 matrix,
a11 a12 a13
a22 a23
a12 a13
¢ 3 a21 a22 a23 3 a11(1)2 2
2 a21(1)3 2
2
a32 a33
a32 a33
a31 a32 a33
a31(1)4 2
a12 a13
2
a22 a23
a11(a22 a33 a32 a23) a21(a12 a33 a32 a13)
a31(a12 a23 a22 a13)
(A.10)
A-1
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Appendix A
A-2
Simultaneous Equations and Matrix Inversion
An alternative method of obtaining the determinant of a 3 3 matrix
is by repeating the first two rows and multiplying the terms diagonally
as follows.
a11 a12 a13
a21 a22 a23
¢
5 a31 a32 a33 5
a11 a12 a13 a21 a22 a23 a11a22 a33 a21a32 a13 a31a12 a23 a13 a22 a31 a23 a32 a11
a33 a12 a21
(A.11)
In summary:
The solution of linear simultaneous equations by Cramer’s rule boils
down to finding
xk ¢k
,
¢
k 1, 2, . . . , n
(A.12)
where ¢ is the determinant of matrix A and ¢ k is the determinant of
the matrix formed by replacing the k th column of A by B.
One may use other methods, such
as matrix inversion and elimination.
Only Cramer’s method is covered
here, because of its simplicity and
also because of the availability of
powerful calculators.
Example A.1
You may not find much need to use Cramer’s method described in
this appendix, in view of the availability of calculators, computers, and
software packages such as MATLAB, which can be used easily to solve
a set of linear equations. But in case you need to solve the equations
by hand, the material covered in this appendix becomes useful. At any
rate, it is important to know the mathematical basis of those calculators and software packages.
Solve the simultaneous equations
4x1 3x2 17,
3x1 5x2 21
Solution:
The given set of equations is cast in matrix form as
c
4 3 x1
17
d c d c
d
3
5 x2
21
The determinants are evaluated as
¢2
4 3
2 4 5 (3)(3) 11
3
5
¢1 2
17 3
2 17 5 (3)(21) 22
21
5
¢2 2
4
17
2 4 (21) 17 (3) 33
3 21
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Appendix A
Simultaneous Equations and Matrix Inversion
A-3
Hence,
x1 ¢1
22
2,
¢
11
x2 ¢2
33
3
¢
11
Practice Problem A.1
Find the solution to the following simultaneous equations:
3x1 x2 4,
6x1 18x2 16
Answer: x1 1.833, x2 1.5.
Determine x1, x2, and x3 for this set of simultaneous equations:
25x1 5x2 20x3 50
5x1 10x2 4x3 0
5x1 4x2 9x3 0
Solution:
In matrix form, the given set of equations becomes
25 5 20 x1
50
£ 5 10 4 § £ x2 § £ 0 §
5 4
9 x3
0
We apply Eq. (A.11) to find the determinants. This requires that we
repeat the first two rows of the matrix. Thus,
25 5 20
¢ 3 5 10 4 3 5 4
9
25 5 20
5 10 4
5 5 4
95
25 5 20
5 10 4
50 5 20
¢ 1 3 0 10 4 3 0 4
9
50 5 20
0 10 4
5 0 4
95
50 5 20
0 10 4
25(10)9 (5)(4)(20) (5)(5)(4)
(20)(10)(5) (4)(4)25 9(5)(5)
2250 400 100 1000 400 225 125
Similarly,
4500 0 0 0 800 0 3700
Example A.2
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Appendix A
A-4
Simultaneous Equations and Matrix Inversion
25 50 20
25 50 20
5 0 4
¢ 2 3 5 0 4 3 5 5 0
95
5 0
9
25
50
20
5 0 4
0 0 1000 0 0 2250 3250
25 5 50
25 5 50
5 10 0
¢ 3 3 5 10 0 3 5 5 4 0 5
5 4 0
25 5 50
5 10 0
0 1000 0 2500 0 0 3500
Hence, we now find
¢1
3700
29.6
¢
125
¢2
3250
x2 26
¢
125
¢2
3500
x3 28
¢
125
x1 Practice Problem A.2
Obtain the solution of this set of simultaneous equations:
3x1 x2 2x3 1
x1 6x2 3x3 0
2x1 3x2 6x3 6
Answer: x1 3 x3, x2 2.
A.2
Matrix Inversion
The linear system of equations in Eq. (A.3) can be solved by matrix
inversion. In the matrix equation AX B, we may invert A to get X,
i.e.,
X A1B
(A.13)
where A1 is the inverse of A. Matrix inversion is needed in other
applications apart from using it to solve a set of equations.
By definition, the inverse of matrix A satisfies
A1A AA1 I
(A.14)
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Appendix A
Simultaneous Equations and Matrix Inversion
where I is an identity matrix. A1 is given by
adj A
det A
A1 (A.15)
where adj A is the adjoint of A and det A 0A 0 is the determinant of
A. The adjoint of A is the transpose of the cofactors of A. Suppose we
are given an n n matrix A as
A ≥
a11 a12 p a1n
a21 a22 p a2n
o
¥
(A.16)
an1 an2 p ann
The cofactors of A are defined as
C cof (A) ≥
c11 c12 p c1n
c21 c22 p c2n
¥
o
(A.17)
cn1 cn2 p cnn
where the cofactor ci j is the product of (1)ij and the determinant
of the 1n 12 1n 12 submatrix is obtained by deleting the
ith row and jth column from A. For example, by deleting the first
row and the first column of A in Eq. (A.16), we obtain the cofactor
c11 as
c11 (1)2 4
a22 a23 p a2n
a32 a33 p a3n
4
(A.18)
o
an2 an3 p ann
Once the cofactors are found, the adjoint of A is obtained as
c11 c12 p c1n T
c21 c22 p c2n
adj (A) ≥
¥ CT
o
cn1 cn2 p cnn
(A.19)
where T denotes transpose.
In addition to using the cofactors to find the adjoint of A, they are
also used in finding the determinant of A which is given by
n
0A 0 a ai j ci j
(A.20)
j1
where i is any value from 1 to n. By substituting Eqs. (A.19) and (A.20)
into Eq. (A.15), we obtain the inverse of A as
CT
0A 0
(A.21)
a b
d
c d
(A.22)
A1 For a 2 2 matrix, if
A c
A-5
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Appendix A
A-6
Simultaneous Equations and Matrix Inversion
its inverse is
A1 1
d b
1
d b
c
d c
d
0A 0 c a
ad bc c a
(A.23)
For a 3 3 matrix, if
a11 a12 a13
A £ a21 a22 a23 §
a31 a32 a33
(A.24)
we first obtain the cofactors as
c11 c12 c13
C £ c21 c22 c23 §
c31 c32 c33
(A.25)
where
c11 2
a22 a23
2,
a32 a33
c21 2
c31 2
a21 a23
2,
a31 a33
c13 2
a11 a13
2,
a31 a33
c23 2
c12 2
a12 a13
2,
a32 a33
c22 2
a12 a13
2,
a22 a23
c32 2
a11 a13
2,
a21 a23
c33 2
a21 a22
2,
a31 a32
a11 a12
2,
a31 a32
a11 a12
2
a21 a22
(A.26)
The determinant of the 3 3 matrix can be found using Eq. (A.11).
Here, we want to use Eq. (A.20), i.e.,
0 A 0 a11c11 a12 c12 a13c13
(A.27)
The idea can be extended n 7 3, but we deal mainly with 2 2 and
3 3 matrices in this book.
Example A.3
Use matrix inversion to solve the simultaneous equations
2x1 10x2 2,
x1 3x2 7
Solution:
We first express the two equations in matrix form as
c
2 10 x1
2
d c d c d
1 3 x2
7
or
AX B ¡ X A1B
where
A c
2 10
d,
1 3
X c
x1
d,
x2
2
B c d
7
The determinant of A is 0A 0 2 3 10(1) 16, so the inverse
of A is
A1 1 3 10
c
d
16 1
2
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Appendix A
Simultaneous Equations and Matrix Inversion
A-7
Hence,
X A1B 1 3 10 2
1 64
4
c
d c d c
d c
d
16 1
2 7
16
16
1
i.e., x1 4 and x2 1.
Practice Problem A.3
Solve the following two equations by matrix inversion.
2y1 y2 4,
y1 3y2 9
Answer: y1 3, y2 2.
Determine x1, x2, and x3 for the following simultaneous equations using
matrix inversion.
x1 x2 x3 5
x1 2x2 9
4x1 x2 x3 2
Solution:
In matrix form, the equations become
1 1
1 x1
5
£ 1 2
0 § £ x2 § £ 9 §
4 1 1
x3
2
or
AX B ¡ X A1B
where
1 1
1
A £ 1 2
0§,
4 1 1
x1
X £ x2 § ,
x3
5
B £ 9§
2
We now find the cofactors
c11 2
2
0
1
0
1 2
2 2, c12 2
2 1, c13 2
2 9
1 1
4 1
4 1
c21 2
c31 2
1
1
1
1
2 2, c22 2
2 5,
1 1
4 1
1 1
2 2,
2 0
c32 2
c23 2
1 1
23
4 1
1 1
1 1
2 1, c33 2
23
1 0
1 2
Example A.4
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Appendix A
A-8
Simultaneous Equations and Matrix Inversion
The adjoint of matrix A is
2 1 9 T
2
2 2
adj A £ 2 5
3 § £ 1 5 1 §
2 1
3
9
3
3
We can find the determinant of A using any row or column of A. Since
one element of the second row is 0, we can take advantage of this to
find the determinant as
0A 0 1c21 2c22 (0)c23 1(2) 2(5) 12
Hence, the inverse of A is
A1 2
2 2
1
£ 1 5 1 §
12
9
3
3
2
2 2
5
1
1
XA B
£ 1 5 1 § £ 9 § £ 4 §
12
9
3
3 2
2
1
i.e., x1 1, x2 4, x3 2.
Practice Problem A.4
Solve the following equations using matrix inversion.
y1 y3 1
2y1 3y2 y3 1
y1 y2 y3 3
Answer: y1 6, y2 2, y3 5.