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Chapter 04.
System of Equations
After reading this chapter, you should be able to:
- setup simultaneous linear equations in matrix form and vice-versa,
- understand the concept of the inverse of a matrix,
- know the difference between a consistent and inconsistent system of linear equations,
- andlearn that a system of linear equations can have a unique solution, no solution or infinite solutions. Matrix algebra is used for solving systems of equations. Can you illustrate this concept? Matrix algebra is used to solve a system of simultaneous linear equations. In fact, for many mathematical procedures such as the solution to a set of nonlinear equations, interpolation, integration, and differential equations, the solutions reduce to a set of simultaneous linear equations. Let us illustrate with an example for interpolation. Example 1 The upward velocity of a rocket is given at three different times on the following table. Table 5.1. Velocity vs. time data for a rocket Time, t Velocity, v (s) (m/s) 5 106. 8 177. 12 279. The velocity data is approximated by a polynomial as
v t at^2 bt c , 5 t12.
Set up the equations in matrix form to find the coefficients a , b , c of the velocity profile. Solution
The polynomial is going through three data points t 1 , v 1 , t 2 , v 2 , and t 3 , v 3 where from
table 5.1. t 1 5 , v 1 106. 8 t 2 8 , v 2 177. 2
04.05.2 Chapter 04.
t 3 12 , v 3 279. 2
Requiring that v t at^2 bt c passes through the three data points gives
v t 1 v 1 at 12 bt 1 c
v t 2 v 2 at 22 bt 2 c
v t 3 v 3 at 32 bt 3 c
Substituting the data t 1 , v 1 , t 2 , v 2 , and t 3 , v 3 gives
a 5 2 b 5 c 106. 8
a 8 2 b 8 c 177. 2
a 12 2 b 12 c 279. 2
or 25 a 5 b c 106. 8 64 a 8 b c 177. 2 144 a 12 b c 279. 2 This set of equations can be rewritten in the matrix form as
a b c
a b c
a b c
The above equation can be written as a linear combination as follows
a b c
and further using matrix multiplication gives
c
b
a
The above is an illustration of why matrix algebra is needed. The complete solution to the set of equations is given later in this chapter. A general set of m linear equations and n unknowns, a 11 (^) x 1 a 12 x 2 a 1 n xn c 1 a 21 (^) x 1 a 22 x 2 a 2 n xn c 2 …………………………………… ……………………………………. am (^) 1 x 1 am 2 x 2 ........ amnxn c m can be rewritten in the matrix form as
04.05.4 Chapter 04.
12 34 yx 46 is a consistent system of equations as it has a unique solution, that is, yx^ 11 . b) The system of equations 12 24 yx ^63 is also a consistent system of equations but it has infinite solutions as given as follows.Expanding the above set of equations,
2 3
x y
x y
you can see that they are the same equation. Hence, any combination of x , y that satisfies
2 x 4 y 6
is a solution. For example x , y 1 , 1 is a solution. Other solutions include
x , y ( 0. 5 , 1. 25 ), x , y ( 0 , 1. 5 ), and so on.
c) The system of equations 12 24 yx 46 is inconsistent as no solution exists. How can one distinguish between a consistent and inconsistent system of equations?
A system of equations A X C is consistent if the rank of A is equal to the rank of the
augmented matrix A C
A system of equations A X C is inconsistent if the rank of A is less than the rank of
the augmented matrix A C .
But, what do you mean by rank of a matrix? The rank of a matrix is defined as the order of the largest square submatrix whose determinant is not zero. Example 3 What is the rank of
A?
Solution The largest square submatrix possible is of order 3 and that is [ A ] itself. Since det( A ) 23 0 ,the rank of [ A ] 3.
System of Equations 04.05.
Example 4 What is the rank of
A?
Solution The largest square submatrix of [ A ] is of order 3 and that is [ A ] itself. Since det( A ) 0 , the rank of [ A ] is less than 3. The next largest square submatrix would be a 2 2 matrix. One of the square submatrices of [ A ] is
B 23 01
and det( B ) 2 0. Hence the rank of [ A ] is 2. There is no need to look at other 2 2 submatrices to establish that the rank of [ A ] is 2.
Example 5 How do I now use the concept of rank to find if
3
2
1 x
x
x
is a consistent or inconsistent system of equations? Solution The coefficient matrix is
A
and the right hand side vector is
C
The augmented matrix is
B
Since there are no square submatrices of order 4 as [ B ] is a 3 4 matrix, the rank of [ B ] is at most 3. So let us look at the square submatrices of [ B ] of order 3; if any of these square submatrices have determinant not equal to zero, then the rank is 3. For example, a submatrix of the augmented matrix [ B ] is
System of Equations 04.05.
E
det( E ) 0
F
det( F ) 0
G
det( G ) 0 All the square submatrices of order 3 3 of the augmented matrix [ B ] have a zero determinant. So the rank of the augmented matrix [ B ] is less than 3. Is the rank of augmented matrix [ B ] equal to 2?. One of the 2 2 submatrices of the augmented matrix [ B ] is
H 6425 85
and det( H ) 120 0
So the rank of the augmented matrix [ B ]is 2. Now we need to find the rank of the coefficient matrix [ B ].
A
and det( A ) 0 So the rank of the coefficient matrix [ A ] is less than 3. A square submatrix of the coefficient matrix [ A ] is
J 85 11
det( J ) 3 0 So the rank of the coefficient matrix [ A ] is 2. Hence, rank of the coefficient matrix [ A ] equals the rank of the augmented matrix [ B ]. So the system of equations [ A ] [ X ] [ C ]is consistent.
04.05.8 Chapter 04.
Example 7 Use the concept of rank to find if
3
2
1 x
x
x
is consistent or inconsistent. Solution The augmented matrix is
B
Since there are no square submatrices of order 4 4 as the augmented matrix [ B ] is a 4 3 matrix, the rank of the augmented matrix [ B ] is at most 3. So let us look at square submatrices of the augmented matrix ( B ) of order 3 and see if any of the 3 3 submatrices have a determinant not equal to zero. For example, a square submatrix of order 3 3 of [ B ]
D
det( D ) = 0 So it means, we need to explore other square submatrices of the augmented matrix [ B ]
E
det( E 0 12. 0 0. So the rank of the augmented matrix [ B ]is 3. The rank of the coefficient matrix [ A ] is 2 from the previous example. Since the rank of the coefficient matrix [ A ] is less than the rank of the augmented matrix [ B ] , the system of equations is inconsistent. Hence, no solution exists for [ A ] [ X ] [ C ].
If a solution exists, how do we know whether it is unique? In a system of equations [ A ] [ X ] [ C ] that is consistent, the rank of the coefficient matrix [ A ] is the same as the augmented matrix [ A C ]. If in addition, the rank of the coefficient matrix [ A ] is same as the number of unknowns, then the solution is unique; if the rank of the coefficient matrix [ A ] is less than the number of unknowns, then infinite solutions exist.
04.05.10 Chapter 04.
Solution While finding out whether the above equations were consistent, we found that the rank of the
coefficient matrix [ A ] equals the rank of augmented matrix A C equals 2
Since the rank of [ A ] 2 < number of unknowns = 3, infinite solutions exist.
If we have more equations than unknowns in [A] [X] = [C], does it mean the system is inconsistent?
No, it depends on the rank of the augmented matrix A C and the rank of [ A ].
a) For example
3
2
1 x
x
x
is consistent, since rank of augmented matrix = 3 Now since the rank of (rank of coefficient matrix = 3 A ) = 3 = number of unknowns, the solution is not only consistent but also unique. b) For example
3
2
1 x
x
x
is inconsistent, since rank of augmented matrix = 4rank of coefficient matrix = 3 c) For example
3
2
1 x
x
x
is consistent, since rank of augmented matrix = 2 rank of coefficient matrix = 2 But since the rank of [ A ] = 2 < the number of unknowns = 3, infinite solutions exist.
Consistent systems of equations can only have a unique solution or infinite solutions.Can a system of equations have more than one but not infinite number of solutions?
No, you can only have either a unique solution or infinite solutions. Let us suppose [ A ] [ X ] [ C ]has two solutions [ Y ]and [ Z ]so that
System of Equations 04.05. [ A ] [ Y ][ C ] [ A ] [ Z ][ C ] If r is a constant, then from the two equations
r A Y r C
1 r A Z 1 r C
Adding the above two equations gives
r A Y 1 r A Z r C 1 r C
A r Y 1 r Z C
Hence
r Y 1 r Z
is a solution to
A X C
Since r is any scalar, there are infinite solutions for [ A ] [ X ] [ C ]of the form
r Y 1 r Z
Can you divide two matrices?
If [ A ] [ B ] [ C ]is defined, it might seem intuitive that [ A ] C^ B , but matrix division is not
defined like that. However an inverse of a matrix can be defined for certain types of square matrices. The inverse of a square matrix [ A ] , if existing, is denoted by [ A ] ^1 such that [ A ] [ A ]^1 [ I ][ A ]^1 [ A ] Where [ I ]is the identity matrix. In other words, let [ A ] be a square matrix. If [ B ] is another square matrix of the same size such that [ B ][ A ] [ I ], then [ B ] is the inverse of [ A ]. [ A ] is then called to be invertible or nonsingular. If [ A ] ^1 does not exist, [ A ] is called noninvertible or singular. If [ A ] and [ B ] are two n n matrices such that [ B ][ A ] [ I ], then these statements are also true [ B ] is the inverse of [ A ] [ A ] is the inverse of [ B ] [ A ] and [ B ] are both invertible [ A ] [ B ]=[ I ]. [ A ] and [ B ] are both nonsingular all columns of [ A ] and [ B ]are linearly independent all rows of [ A ] and [ B ] are linearly independent. Example 10 Determine if [ B ] (^) 53 32 is the inverse of
System of Equations 04.05.
n n nn
n
n
a a a
a a a
a a a A 1 2
21 22 2
11 12 1 [ ]
' 1 ' 2 '
21 ' 22 ' ' 2
11 ' 12 ' 1 ' [ ]^1 n n nn
n
n
a a a
a a a
a a a A
[ I ]
Using the definition of matrix multiplication, the first column of the [ A ] ^1 matrix can then be found by solving
' 1
' 21 11 '
1 2
21 22 2
11 12 1
n n nn n
n
n
a
a
a
a a a
a a a
a a a
Similarly, one can find the other columns of the [ A ] ^1 matrix by changing the right hand side accordingly. Example 11 The upward velocity of the rocket is given by Table 5.2. Velocity vs time data for a rocket Time, t (s) Velocity, v (m/s) 5 106. 8 177. 12 279. In an earlier example, we wanted to approximate the velocity profile by
v t at^2 bt c , 5 t 12
We found that the coefficients a , b ,and c in v t are given by
04.05.14 Chapter 04.
c
b
a 144 12 1
First, find the inverse of
A
and then use the definition of inverse to find the coefficients a , b ,and c. Solution If
31 ' 32 ' 33
' 21 ' 22 23 ' 11 ' 12 ' 13 ' 1 a a a
a a a
a a a A
is the inverse of [ A ] , then
31 ' 32 ' 33 '
21 ' 22 ' ' 23
11 ' 12 ' 13 ' a a a
a a a
a a a
gives three sets of equations
31 '
21 '
11 ' a
a
a
' 32
' 22 12 ' a
a
a
33 '
23 '
13 ' a
a
a
Solving the above three sets of equations separately gives
31 '
21 '
11 ' a
a
a
32 '
22 '
12 ' a
a
a
04.05.16 Chapter 04.
T
1 2
21 22 2
11 12 1
n n nn
n
n
C C C
C C C
C C C
adjA
where Cij are the cofactors of a (^) ij. The matrix
n nn
n
n
C C
C C C
C C C
1
21 22 2
11 12 1
itself is called the matrix of cofactors from [ A ]. Cofactors are defined in Chapter 4. Example 12 Find the inverse of
A
Solution From Example 4.6 in Chapter 04.06, we found
det A 84
Next we need to find the adjoint of [ A ]. The cofactors of A are found as follows. The minor of entry a 11 is
144 12 1
M 11
^81
The cofactors of entry a 11 is
C 11 1 1 ^1 M 11
M 411
The minor of entry a 12 is
144 12 1
M 12
System of Equations 04.05.
^641
The cofactor of entry a 12 is
C 12 1 1 ^2 M 12
( M 8012 )
Similarly C 13 384 C 21 7 C 22 119 C 23 420 C 31 3 C 32 39 C 33 120 Hence the matrix of cofactors of [ A ] is
C
The adjoint of matrix [ A ] is [ C ]T,
adj A C T
Hence
A 1 det^1 A adj A
