Reputation: 15
I'm trying to solve a system of 10 linear equations out of which the middle 8 equations look alike. They look like this :
t_i-1 - 2.3086*(t_i) + t_i+1 == -7.7160
where i = 2:9
so I decided to construct the coefficient matrix and the constant matrix(array) for the system of equations through looping.This is what I did.
T = sym('t' , [1 10]); %% Creates a vector T = [ t1 t2 .... t10]
A_10 = zeros(10,10);
b_10 = zeros(10,1);
for i = 2:9 %% This loop generates the equations and arranges them in the matrices A_10 and B_10.
T(i-1) - 2.3086*T(i) + T(i+1) == -7.7160;
[A_10(i,i-1:i+1),b_10(i,1)] = equationsToMatrix(ans)
end
Everything except for the ninth row(last but one) is correct in the Matrix A_10. This is what A_10 looks like
A_10 =
Columns 1 through 9
0 0 0 0 0 0 0 0 0
1.0000 -2.3086 1.0000 0 0 0 0 0 0
0 1.0000 -2.3086 1.0000 0 0 0 0 0
0 0 1.0000 -2.3086 1.0000 0 0 0 0
0 0 0 1.0000 -2.3086 1.0000 0 0 0
0 0 0 0 1.0000 -2.3086 1.0000 0 0
0 0 0 0 0 1.0000 -2.3086 1.0000 0
0 0 0 0 0 0 1.0000 -2.3086 1.0000
0 0 0 0 0 0 0 1.0000 1.0000
0 0 0 0 0 0 0 0 0
Column 10
0
0
0
0
0
0
0
0
-2.3086
0
The last three elements of the row nine should be 1 , -2.3086 , 1 like the previous rows but it shows 1, 1, -2.3086. What am I doing wrong here?
This is what the iteration looks like in the loop
ans = t8 - (11543*t9)/5000 + t10 == -1929/250
The equation is correct too. I can't figure out what the problem is.
Upvotes: 1
Views: 61
Reputation: 8401
Without the second input vars
, equationsToMatrix
uses symvar
to determine the variable list.
Using symvar
directly with the last equation gives
>> i = 9;symvar(T(i-1) - 2.3086*T(i) + T(i+1) == -7.7160)
ans =
[ t10, t8, t9]
So for whatever reason, symvar
produced the incorrect ordering for only the last equation (possibly because 1 < 9). To remedy the situation, pass your intended ordering using the second input
eqn = T(i-1) - 2.3086*T(i) + T(i+1) == -7.7160;
[A_10(i,i-1:i+1),b_10(i,1)] = equationsToMatrix(eqn,T(i-1:i+1));
You'll also noticed I assigned the equation to an explicit variable eqn
. This is better practice than relying on ans
.
Also, since you're producing a numeric array anyway, you can produce A
without the Symbolic Toolbox in a number of ways. For example:
n = 10;
A = full(spdiags(ones(n,1)*[1,-2.3086,1],[-1,0,1],n,n));
A([1,end],:) = 0;
Upvotes: 3