Lesson 7

Last modified:January 1, 1970 GMT
An Introduction to Matlab(tm): Lesson 7
SOLVING SIMULTANEOUS LINEAR EQUATIONS
One of the most common applications of matrix algebra occurs
in the solution of linear simultaneous equations. Consider a set of
n equations for which the unknowns are x1, x2, ....... xn.
a11x1 + a12x2 + a13x3 + ............... a1nxn = b1
a21x1 + a22x2 + a23x3 + ............... a2nxn = b2
a31x1 + a32x2 + a33x3 + ............... a3nxn = b3
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an1x1 + an2x2 + an3x3 + ............... annxn = bn
The matrix format for these equations is
[a][x] = [b]
where
a11 a12 a13 .... a1n

x1

b1

a21 a22 a23 .....a2n

x2

b2

[a] = a31 a32 a33 .... a3n
. . .
.
.
. . .
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. . .
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. . .
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.
an1 an2 an3 .... ann

[x] = x3
.
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.
xn

[b] = b3

bn

Note that only when the equations are linear in the unknown xi's is
the matrix formulation possible.
The classical matrix solution to this equation is based on the
definition of the inverse of a matrix. The inverse of a matrix is
that matrix which when multiplied by the original matrix, results
in the identity matrix. Then
a-1 a = I

where a-1 denotes the 'inverse of matrix a' and I denotes the
identity matrix of the same order as matrix 'a'. If 'a' is a known
coefficient matrix and if 'b' is a column vector of known terms,
the problem is one of finding the n vales of x1, x2, .... xn that
satisfy these simultaneous equations. Pre-multiplying both sides of
the equation by the inverse of 'a' gives
[a-1][a][x] = [a-1][b]
or
[x] = [a-1][b]
Thus if we find the inverse of the coefficient matrix, the product
of the inverse times the matrix of non-homogeneous terms, b, yields
the unknown matrix, x. To observe the simple property of the
inverse, define the 4x4 matrix, C
C = [1 -4 3 2; 3 1 -2 1; 2 1 1 -1; 2 -1 3 1]
calculate the inverse using the 'matlab' command, inv().
C_inverse = inv(C)
and note that
C_inverse * C = [1 0 0 0;0 1 0 0;0 0 1 0;0 0 0 1]
where the right hand side is the 4x4 identity matrix.
We leave to a study of linear algebra, the method and steps
required to calculate the inverse of a matrix. Suffice it to say
here that for systems of equations numbering in the hundreds or
thousands, which often occur in complex engineering problems, the
calculation of the matrix inverse is a time consuming operation
even for modern computers.
As an example of solving a system of equations using the
matrix inverse, consider the following system of three equations.
x1 - 4x2 + 3x3 = -7
3x1 + x2 - 2x3 = 14
2x1 + x2 + x3 = 5
Using 'matlab', define the two matrices for [a] and [b].

a = [ 1 -4 3; 3 1 -2; 2 1 1];
b = [ -7; 14; 5];
Find the solution vector, x, by
x = inv(a)*b
giving
x=[3
1
-2 ]
LEFT AND RIGHT MATRIX 'DIVISION' IN MATLAB
In contrast to matrix inversion, the preferred methods of
solution for large-scale systems of simultaneous equations are
methods of back substitution whereby the equations can be
rearranged so that first xn is solved, then xn-1 and so on until
finally x1 is obtained in order. Thus in back substitution we never
solve the equations simultaneously, rather we solve them
sequentially. For example, the above set of equations can be
rearranged through appropriate combination into the below
equations. (You should carry out this algebra to ensure you
understand the steps matlab takes in solving with back
substitution. In the general case, the steps or algorithm is much
more complex than that required for 3 equations.)
x1 - 4x2 + 3x3 = -7
13x2 - 11x3 = 35
(34/13)x3 = (68/13)
In this format, we see from the third equation the solution for x3
= 2 and knowing this, the second equation yields x2 = 1, and
knowing both x3 and x2, the first equation gives x1 = 1. This is an
application of back substitution, whereby each unknown is obtained
by simple sequential solution to a single equation. Methods of back
substitution are employed by 'matlab' when we invoke the 'right
division' and 'left division' commands.

/

left division
right division

Understand that matrices cannot be divided. The operation is not

defined. The syntax of 'right division' and 'left division' simply
invoke back substitution methods for obtaining a solution vector
from a system of simultaneous equations. Whether 'right division'
(/) or 'left division' () is appropriate depend on how the matrix
equation is posed.
Left Division. If the matrix equation is
[a][x] = [b]
then we have a nxn matrix [a] pre-multiplying a nx1 matrix [x],
resulting in a nx1 matrix [b]. The solution for x is obtained by
using the left division operation.
[x] = [a][b]
If the matrix equation is cast in this format, the use of right
division to obtain the solution vector x will result in an error
message.
Returning to 'matlab', recall we have defined the matrices [a]
and [b] in the format of the matrix equation appropriate for left
division.
a = [ 1 -4 3; 3 1 -2; 2 1 1];
b = [ -7; 14; 5];
Now enter the command
x1 = ab
and obtain the result
x1 = [ 3
1
-2 ]
which i s the same result found earlier for x when solving by
matrix inversion.
Right Division. The matrix format in which right division is
employed is less common but no less successful in solving the
problem of simultaneous equations. Right Division is invoked when
the equations are written in the form

[x][A] = [B]
Note that in this form [x] and [B] are row vectors rather than
column vectors as above. This form represent a 1xn matrix (x) premultiplying a nxn matrix (A), resulting an a 1xn matrix (B). Again,
recalling the set of equations,
x1 - 4x2 + 3x3 = -7
3x1 + x2 - 2x3 = 14
2x1 + x2 + x3 = 5
These equations can be written in matrix format
[x][A] =[B]
if
x = [x1 x2 x3]

B = [ -7 14 5]

and
1 3 2
[A] = -4 1 1
3 -2 1
Note that [A] is equal to the transpose of [a].
A = a'
Having so defined A and B, the solution for x can be obtained by
right division.
x = B/A
results in
x=[3
1
-2 ]

PRACTICE PROBLEMS

1. Find the solution to
r + s + t + w = 4
2r - s

+ w = 2

3r + s - t - w = 2
r - 2s - 3t + w = -3
using the matrix inverse and left and right division.
2. Find the solution to
2x1 + x2 - 4x3 + 6x4 + 3x5 - 2x6 = 16
-x1 + 2x2 + 3x3 + 5x4 - 2x5

= -7

x1 - 2x2 - 5x3 + 3x4 + 2x5 + x6 = 1
4x1 + 3x2 - 2x3 + 2x4

+ x6 = -1

3x1 + x2 - x3 + 4x4 + 3x5 + 6x6 = -11
5x1 + 2x2 - 2x3 + 3x4 + x5 + x6 = 5
using the matrix inverse and left and right division.
3. Back substitution is facilitated by recognizing that any
nonsingular symmetric matrix can be decomposed or factored into the
product of two matrices, called here L and U.
A=LU
This is called LU factorization. U is an upper triangular matrix
(all 0's below the diagonal). L is called a permutation of a lower
triangular matrix in which 1's occur at every element which is a
permutation of the two numbers defining the size of the matrix.
These matrices can be obtained in 'matlab' using the command
[L,U] = lu(A)
Using the matrix A defined in Prob 2 above, find the upper and
lower triangular factor matrices for the matrix of coefficients, A.
If these matrices are called AU and AL, show that

AL * AU * x = B
where
x [ 2 -1 1 0 3 -4 ]' and B = [16 -7 1 -1 -11 5]'
Back to Matlab In-House Tutorials

Lesson 7

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