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ARBITRARY UNKNOWNS
The echelon form of the augmented matrix confirms the existence of arbitrary unknowns, i.e. a consistent system of equations in which one or more variables can be chosen arbitrarily There are several ways to establish if indeed a certain variable can be included in the subset of arbitrary unknowns. A few simple examples illustrate the point.
Example 1 For the system of equations below, establish the existence of 1 arbitrary unknown and determine if x, y, and z can each be arbitrary.
x y z
x y z
x y z
x y z
b^ A|b g =^
F
H
G
G
G
G
I
K
J
J
J
J
F
H
G
G
G
G
I
K
J
J
J
J
F
H
G
G
G
G
I
K
J
J
J
J
The 4 by 4 system has been reduced to a 2 by 3 system. That is, there are only 2 linearly independent equations in the 3 variables (unknowns) and hence one of the variables is arbitrary. These equations are
x y z
y z
Now, suppose we try to make z the arbitrary unknown. This requires that solutions for x and y be expressed in terms of z. This can be attempted in one of two ways. One approach is to simply transfer all terms involving z over to the right side of the equations and consider z as a parameter. This yields
x y z
y z
All that remains is substituting y = -1 - z into the first equation and then solving for x. The final solution can be expressed as
x = 1, y = -1- z , z = arbitrary
Clearly, an infinite number of solutions exist since there is a different solution for each arbitrarily assigned value for z. A slightly different approach involves reformulating the reduced equations from the echelon form as
Ax^ = b ^ A ^ = F b ^ x
HG^
I
KJ^
F
HG^
I
KJ^
F
HG
I
KJ
where , = , =
z z
x y
i.e. as a new system in matrix form with modified coefficient matrix A , constant vector b^ , and vector of unknowns x . The identical solution as given above is easily obtained from the modified augmented matrix
d A|b ^ i =^ − ~
F
HG^
I
KJ^ −^ −
F
HG^
I
KJ
z z z
x = 1, y = -1- z , z = arbitrary
The second approach is somewhat more instructive because the modified coefficient matrix A determines whether the variables placed on the right hand side ( z in this example) are arbitrary. When A , which will always be a square matrix, is nonsingular, Ax ^ = b has a unique solution and the variables moved to the right hand side are indeed arbitrary.
Consider what happens when x is selected to be the arbitrary variable. The reduced 2 by 3 system with (^) x as the arbitrary unknown becomes
Ax^ ^ = b ^ A ^ = F b ^ x
HG^
I
KJ^
F
HG^
I
KJ^
F
HG
I
KJ
where , = , =
x y z
and attempting to solve for a solution by Gauss-Jordan gives
d A|b^ ^ i =^ − ~
F
HG^
I
KJ^ −
F
HG^
I
KJ
x x
x x x x x x
x x x
x x x x
x x x
x
1 2 3 4 5 6
2 4 6
3 4 5 6
4 5 6
5
Clearly x 5 is not arbitrary. The modified matrix A obtained from the columns of the echelon form with the x 5 column omitted is shown below. Quite obviously its singular, as expected, confirming that (^) x 5 is not arbitrary.
A^ =
F
H
G
G
G
G
G
I
K
J
J
J
J
J
However, there may be other non-arbitrary variables in addition to x 5 which are not as obvious. For example, suppose we proceed to solve the reduced 5 by 6 system using Gauss-Jordan with x 2 selected as the arbitrary unknown. Observe what happens.
d^ A|b i =
F
H
G
G
G
G
G
I
K
J
J
J
J
J
F
H
G
G
G
G
G
I
K
J
J
J
J
J
F
H
G
G
G
G
G
I
K
J
J
J
J
J
2 2
2
2
2
2 2
x x
x
x
x
x x
2
2 2 2
F
H
G
G
G
G
G
I
K
J
J
J
J
J
x
x x x
The bottom row of zeros in the first 5 columns signifies that a solution for x 6 is impossible when x 2 is chosen to be arbitrary and the Gauss-Jordan method terminates without a solution. Furthermore, for consistency the last row implies that x 2 must be zero, further evidence it can’t be arbitrary. The 4th^ row represents the equation
x (^) 5 = 1 − x 2
which is consistent with the last row of the echelon form which states that x 5 =1.
In problems of this type, the prudent thing to do is verify that the modified coefficient matrix A is nonsingular before proceeding to find a solution. In the previous example, when x 2 was assumed to be arbitrary, A became
A^ =
F
H
G
G
G
G
G
I
K
J
J
J
J
J
From MATLAB, its easy to verify that (^) A is singular and therefore (^) x 2 should not be chosen as arbitrary.
A =
1 1 1 1 - 0 0 1 0 - 0 1 -3 -1 11 0 0 1 -1 - 0 0 0 1 0
EDU» det(A)
ans =
0
The same approach applies when more than one variable is arbitrary. To illustrate, consider the system of equations
In the 3 by 5 system of equations that correspond to either echelon form, there must be 2 arbitrary unknowns. To check if say x 4 and x 5 can be arbitrary, we look at the modified coefficient matrix A that results when the columns for x 4 and x 5 are removed from the first echelon form.
A^ ^ = A
F
H
G
G
I
K
J
J
Since A is a nonsingular matrix, there is a unique solution to Ax (^) = b
where x ^ = and b
F
H
G
G
I
K
J
J =
F
H
G
G
I
K
J
J
x x x
x x x x x x
1 2 3
4 5 4 5 4 5
The solution is x ^ A ^ b 1 = =
F
H
G
G
I
K
J
J
F
H
G
G
I
K
J
J
F
H
G
G
I
K
J
J
−
d i
4 5 4 5 4 5
4 5
4 5
x x x x x x
x x
x x
i.e. (^) x 1 = 1 + 2 x 4 + (^) x 5 , (^) x 2 = 2, (^) x 3 = - x 4 + 2 x 5 , (^) x 4 = arbitrary, (^) x 5 = arbitrary
Since we know from the previous solution that x 2 is not arbitrary, it should come as no surprise that any 3 by 3 submatrix formed from the first five columns of either echelon matrix (minus the zero rows) is destined to be singular if it excludes the second column, the one corresponding to x 2. The resulting 3 by 3 matrices obtained from the first echelon form are given below and the reader should verify that they are all singular.
− − − − −
F
H
G
G
I
K
J
J
F
H
G
G
I
K
J
J
− − −
F
H
G
G
I
K
J
J
F
H
G
G
I
K
J
J
1 1 3 1 1 2 1 1 2
1 1 3 0 1 2 0 1 2
1 2 2 3
2 4 2 5
x x x x
x x x x
and columns removed and columns removed
and colums removed and columns removed
1 -1 - 0 1 - 0 1 -
1 1 - 0 1 1 0 1 1
The row reduced echelon form is even more explicit as to why x 2 cannot be arbitrary. It is clear from this echelon form that x 2 = 2 and hence not arbitrary.
Furthermore, removing the x 2 column from the row reduced echelon form leaves the following matrix
1 0 -2 -1 1 0 0 0 0 2 0 1 1 -2 0
F
H
G
G
I
K
J
J
Regardless of which additional column is removed for the 2nd^ arbitrary variable, the resulting 3 by 3 modified coefficient matrix A will be singular, confirming that x 2 cannot be one of the two arbitrary unknowns.
In summary, either echelon form of the original augmented matrix will reveal the existence of arbitrary unknowns. The original m by n system of m equations in n unknowns will be reduced to an m 1 by n system where m 1 ≤ m indicating that m - m 1 equations from the original system were redundant. If m 1 is less than n, the existence of n - m 1 arbitrary unknowns is assured. A particular subset of n - m 1 unknowns is arbitrary provided the m 1 by m 1 submatrix of the echelon matrix obtained by removing the columns corresponding to the n - m 1 unknowns is nonsingular.
The row reduced echelon form in MATLAB will identify the arbitrary variables directly, i.e. any row with a single 1 in the first n columns implies the variable associated with that column is not arbitrary. For example, in the following 4 by 6 row-reduced echelon form, x 1 and x 5 cannot be arbitrary unknowns.
rref ( A|b ) =
F
H
G
G
G
G
G
I
K
J
J
J
J
J
