Solving Linear Parametric RHS
System of Equations with Applications

This site is a part of the JavaScript E-labs learning objects for decision making. Other JavaScript in this series are categorized under different areas of applications in the MENU section on this page.

Professor Hossein Arsham   

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In many applied decision making problems such as sensitivity analysis of linear optimization, one needs to solve a linear parametric right-hand-side (RHS) system of equations. Often varieties of parametric RHS systems of equations can be reduced to the following standard form:

AX = b + Dr,

where A is an n by n non-singular matrix, the column matrix b the numerical value of the RHS, and diagonal matrix D contains the coefficients of the parameters r = [r1, r2, …rn]T, respectively.

The following JavaScript solve parametric systems of up to 7 equations. Starting at the upper left table, enter matrix A and then the RHS column, followed by an n by n diagonal matrix D, then click on Calculate button. The JavaScript returns the constant term in the solution as the elements of the (n+1) column, followed by coefficient of r as a square matrix.

Application 1: Suppose we wish to solve the following RHD parametric systems of 2 equations with 2 unknowns:

2 X1 + X2 = 40 + r1
     X1 + 2 X2 = 50 + r2

By entering the matrix [A, b, D]:

2    1    40    1    0
1    2    50    0    1

and after clicking on the Calculate button, we get the following parametric solution:

  X1 = 10 + 2/3r1 - 1/3 r2
X2 = 20 - 1/3r1 + 2/3r2

Application 2: Under what conditions on values of r1 and r2 the solution to the following parametric system of equations remains non-negative?

2 X1 + X2 = 40 + .4r1
X1 + 2 X2 = 50 + .5r2

Following the above data entering instruction, the entering table is [A, b, D]:

2    1    40    0.4    0
1    2    50    0    0.5

The parametric solution is:

X1 = 10 + 4/15r1 - 1/6r2
 X2 = 20 - 2/15r1 + 1/3r2

Therefore, in order to have non-negative solution, the feasible region for r1 and r2 is the set:

S = {r1, r2 | 10 + 4/15r1 -1/6r2 ³ 0, 20 -2/15r1 + 1/3r2 ³ 0}.

Application 3: Note that one may use this JavaScript to solve the ordinary, i.e., non-parametric system of equations by setting the matrix D to zero matrix, and further more not entry matrix D in the data matrix.

A straightforward JavaScript for solving systems of equations up to 10 equations and be found at a Solving System of Linear Equations .

Application 4: To find the inverse of a square matrix A enter the matrix A and then an identity matrix of the same size, then click on Calculate button. The following numerical example illustrates the process:

Suppose we wish to find the inverse (A^-1 ) of the following matrix (if it exists) A:

		2	1
A	=
		1	-1

Entring the matrix A and then the 2 by 2 identity matrix, the output is: This gives X1 = 1/3 , X2 = -2/3. Therefore, A^-1 is:

		1/3	1/3
A-1	=
		1/3	-2/3

For an alternative approach, use Solving System of Linear Equations which computes the inverse of up-to 10 by 10 matrix.

Warning: In all applications and cases, after clicking on the Calculate button, the output must contain an identity matrix appearing on the left-hand-side of the table. Otherwise, the entering matrix might have been a singular matrix.

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