# Changing varibles in a Matrix to Values

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Sharyn Fallick on 2 Mar 2015
below is the problem I am trying to solve in MatLab. Everything works for computation of these matrix, but once I try to replace the variables with numbers I am receiving errors, or the answers are still in fraction form and pi isn't replaces by the number. Instead it still says cos(2*pi) for example. I would appreciate any help with this. Thanks for your time
syms r E G b h t A Ay Iz Iy alpha theta
LambdaQBt=[cos(alpha),sin(alpha),0;-sin(alpha),cos(alpha),0;0,0,1]
LambdaQB=transpose(LambdaQBt)
HQB=[1,0,0;0,1,0;-r*(1-cos(alpha)),r*sin(alpha),1]
HQBt=transpose(HQB)
TQB=(LambdaQBt*HQB)
fuq=[1/(E*A),0,0;0,1/(G*Ay),0;0,0,1/(E*Iz)]
TQBt=(HQBt*LambdaQB)
Product=(TQBt*fuq*TQB)
fBA=int(Product,alpha,0,theta)
r=3 E=200*10^9 G=77*10^9 b=.1 h=.0866 t=.01 Ay=.0049 Iz=(t*(b^3))/12 Iy=0 A=3*b*t
F=[10000;-5000;10000]
vB=fBA*F
Steven Lord on 16 Aug 2018
To extend Stephen's answer, if the symbolic expression DOES contain any symbolic variables without values you can use the vpa function to approximate the numeric parts.
syms x
oneThird = sym(1)/3;
y = oneThird*x + x^2
vpa(y)

Tiasa Ghosh on 16 Aug 2018
I guess the answer vB is still a symbolic expression is due to the fact that you have assigned numerical values to r,E,G,b and so on, after the fBA has already been evaluated as a symbolic expression. So in the last line when you compute fBa*F , the answer is obviously a symbolic expression.

Christopher Creutzig on 3 Dec 2018
Assigning values to MATLAB variables does not affect symbolic expressions that have variables of the same name inside. Use the subs command instead. Or start with the concrete values instead of symbolic indeterminates.