8 - Chapman MATLAB Programming for Engineers

Systems of linear equations can also be solved symbolically. Consider the following example that was previously solved numerically:

>>syms x

>>A = sym([3 2 -1; -1 3 2; 1 -1 -1])

A =

[1, -1, -1]

>>b = sym([10; 5; -1])

b = [ 10] [ 5] [ -1]

>>x = A\b

x = [ -2] [ 5] [ -6]

The results are the same as those obtained numerically. For this problem, there is little advantage to finding the result symbolically. However, solving a system of equations with respect to a parameter, the symbolic approach provides an advantage. Consider the following set of equations

2x1 − 3x2 = 3

5x1 + cx2 = 19

To solve for x1 and x2 as functions of the parameter c:

>>syms c

>>A = sym([2 -3; 5 c]);

>>b = sym([3; 19]);

>>x = A\b

x =

[ 3*(19+c)/(2*c+15)]

[23/(2*c+15)]

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