MATLAB®

Systems of Linear Equations

Computational Considerations

One of the most important problems in technical computing is the solution of systems of simultaneous linear equations.

In matrix notation, the general problem takes the following form: Given two matrices A and B, does there exist a unique matrix X so that AX = B or XA = B?

It is instructive to consider a 1-by-1 example. For example, does the equation

have a unique solution?

The answer, of course, is yes. The equation has the unique solution x = 3. The solution is easily obtained by division:

The solution is not ordinarily obtained by computing the inverse of 7, that is 7^-1 = 0.142857..., and then multiplying 7^-1 by 21. This would be more work and, if 7^-1 is represented to a finite number of digits, less accurate. Similar considerations apply to sets of linear equations with more than one unknown; the MATLAB^® software solves such equations without computing the inverse of the matrix.

Although it is not standard mathematical notation, MATLAB uses the division terminology familiar in the scalar case to describe the solution of a general system of simultaneous equations. The two division symbols, slash, /, and backslash, \, are used for the two situations where the unknown matrix appears on the left or right of the coefficient matrix:

You can think of "dividing" both sides of the equation AX = B or XA = B by A. The coefficient matrix A is always in the "denominator."

The dimension compatibility conditions for X = A\B require the two matrices A and B to have the same number of rows. The solution X then has the same number of columns as B and its row dimension is equal to the column dimension of A. For X = B/A, the roles of rows and columns are interchanged.

In practice, linear equations of the form AX = B occur more frequently than those of the form XA = B. Consequently, the backslash is used far more frequently than the slash. The remainder of this section concentrates on the backslash operator; the corresponding properties of the slash operator can be inferred from the identity:

(B/A)' = (A'\B')

The coefficient matrix A need not be square. If A is m-by-n, there are three cases:

The backslash operator employs different algorithms to handle different kinds of coefficient matrices. The various cases, which are diagnosed automatically by examining the coefficient matrix, include:

• Permutations of triangular matrices

• Symmetric, positive definite matrices

• Square, nonsingular matrices

• Rectangular, overdetermined systems

• Rectangular, underdetermined systems

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General Solution

The general solution to a system of linear equations AX = b describes all possible solutions. You can find the general solution by

1. Solving the corresponding homogeneous system AX = 0. Do this using the null command, by typing null(A). This returns a basis for the solution space to AX = 0. Any solution is a linear combination of basis vectors.

2. Finding a particular solution to the non-homogeneous system AX = b.

You can then write any solution to AX = b as the sum of the particular solution to AX = b, from step 2, plus a linear combination of the basis vectors from step 1.

The rest of this section describes how to use MATLAB to find a particular solution to AX = b, as in step 2.

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Square Systems

The most common situation involves a square coefficient matrix A and a single right-hand side column vector b.

Nonsingular Coefficient Matrix

If the matrix A is nonsingular, the solution, x = A\b, is then the same size as b. For example,

A = pascal(3);
u = [3; 1; 4];
x = A\u

x =
      10
     -12
       5

It can be confirmed that A*x is exactly equal to u.

If A and B are square and the same size, then X = A\B is also that size:

B = magic(3);
X = A\B

X =
      19    -3    -1
     -17     4    13
       6     0    -6

It can be confirmed that A*X is exactly equal to B.

Both of these examples have exact, integer solutions. This is because the coefficient matrix was chosen to be pascal(3), which has a determinant equal to one. A later section considers the effects of roundoff error inherent in more realistic computations.

Singular Coefficient Matrix

A square matrix A is singular if it does not have linearly independent columns. If A is singular, the solution to AX = B either does not exist, or is not unique. The backslash operator, A\B, issues a warning if A is nearly singular and raises an error condition if it detects exact singularity.

If A is singular and AX = b has a solution, you can find a particular solution that is not unique, by typing

P = pinv(A)*b

P is a pseudoinverse of A. If AX = b does not have an exact solution, pinv(A) returns a least-squares solution.

For example,

A = [ 1     3     7
     -1     4     4
      1    10    18 ]

is singular, as you can verify by typing

det(A)

ans =
     0

Note   For information about using pinv to solve systems with rectangular coefficient matrices, see Pseudoinverses.

Exact Solutions.   For b =[5;2;12], the equation AX = b has an exact solution, given by

pinv(A)*b

ans =
    0.3850
   -0.1103
    0.7066

You can verify that pinv(A)*b is an exact solution by typing

A*pinv(A)*b

ans =
    5.0000
    2.0000
   12.0000

Least Squares Solutions.   On the other hand, if b = [3;6;0], then AX = b does not have an exact solution. In this case, pinv(A)*b returns a least squares solution. If you type

A*pinv(A)*b

ans =
   -1.0000
    4.0000
    2.0000

you do not get back the original vector b.

You can determine whether AX = b has an exact solution by finding the row reduced echelon form of the augmented matrix [A b]. To do so for this example, enter

rref([A b])
ans =
    1.0000         0    2.2857         0
         0    1.0000    1.5714         0
         0         0         0    1.0000

Since the bottom row contains all zeros except for the last entry, the equation does not have a solution. In this case, pinv(A) returns a least-squares solution.

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Overdetermined Systems

Overdetermined systems of simultaneous linear equations are often encountered in various kinds of curve fitting to experimental data. Here is a hypothetical example. A quantity y is measured at several different values of time, t, to produce the following observations:

Enter the data into MATLAB with the statements

t = [0 .3 .8 1.1 1.6 2.3]';
y = [.82 .72 .63 .60 .55 .50]';

Try modeling the data with a decaying exponential function:

The preceding equation says that the vector y should be approximated by a linear combination of two other vectors, one the constant vector containing all ones and the other the vector with components e^–t. The unknown coefficients, c₁ and c₂, can be computed by doing a least squares fit, which minimizes the sum of the squares of the deviations of the data from the model. There are six equations in two unknowns, represented by the 6-by-2 matrix:

E = [ones(size(t)) exp(-t)]

E =
      1.0000    1.0000
      1.0000    0.7408
      1.0000    0.4493
      1.0000    0.3329
      1.0000    0.2019
      1.0000    0.1003

Use the backslash operator to get the least squares solution:

c = E\y

c =
      0.4760
      0.3413

In other words, the least squares fit to the data is

The following statements evaluate the model at regularly spaced increments in t, and then plot the result, together with the original data:

T = (0:0.1:2.5)';
Y = [ones(size(T)) exp(-T)]*c;
plot(T,Y,'-',t,y,'o')

You can see that E*c is not exactly equal to y, but that the difference might well be less than measurement errors in the original data.

A rectangular matrix A is rank deficient if it does not have linearly independent columns. If A is rank deficient, the least squares solution to AX = B is not unique. The backslash operator, A\B, issues a warning if A is rank deficient and produces a least squares solution that has at most rank(A) nonzeros.

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Underdetermined Systems

Underdetermined linear systems involve more unknowns than equations. The solution to such underdetermined systems is not unique. The matrix left division operation in MATLAB finds a basic solution, which has at most m nonzero components.

Here is a small, random example:

R = [6 8 7 3; 3 5 4 1]
R =
     6     8     7     3
     3     5     4     1

rand('state', 0);
b = fix(10*rand(2,1))
b =
       9
       2

The linear system Rx = b involves two equations in four unknowns. Since the coefficient matrix contains small integers, it is appropriate to use the format command to display the solution in rational format. The particular solution is obtained with

format rat
p = R\b
p =
       0
      -3/7
       0
      29/7

One of the nonzero components is p(2) because R(:,2) is the column of R with largest norm. The other nonzero component is p(4) because R(:,4) dominates after R(:,2) is eliminated.

The complete solution to the underdetermined system can be characterized by adding an arbitrary vector from the null space, which can be found using the null function with an option requesting a "rational" basis:

Z = null(R,'r')
Z =
      -1/2           -7/6     
      -1/2            1/2     
       1             0      
       0             1 

It can be confirmed that R*Z is zero and that any vector x where

x = p + Z*q

for an arbitrary vector q satisfies R*x = b.

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Matrices in the MATLAB® Environment

Inverses and Determinants