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Do it by hand? You can use the normal way of solving simultaneous equations, but there's also a way to do it using matrices (though I can't remember what it is right now).

Yeah, use augmented matrices. You just reduce to reduced Row echelon from. A graphing calculator can also help solve this equation for you, but keeping track of matrix variables (prevents mistakes) and eases doing row operations.

Look up 'linear programming' on Google, and 'solving systems of linear equations'.

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Why is this in programming? :/

Likely because it is considered linear programming, and they did not see the context, but I do not know for sure.

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You can also use Cramer's rule and do it with determinants. I've never learnt about augmented matrices, so don't know about those.

An augmented matrix, is one that is just separated, typically to represent equality on one side. Cramer's rule and determinants of matrices are actually more advanced topics than augmented matrices

By the way, I am also curious where the equality symbols are here (only one as mentioned). Am I to assume the others are equal to zero, or that the last number in each equation is the Right Hand Side?

Reduced Row Echelon example (giving answers). Say each row is X1,X2,X3, then given answer matrix:

Code:

[1 0 0 | A]
[0 1 0 | B]
[0 0 1 | C]

X1 = A
X2 = B
X3 = C

But in your example, with 5 variables, and 4 equations, you will not get full answers, you will get answers in terms of themselves, as yours in an underdetermined system. Try it out and see!

You can also use Cramer's rule and do it with determinants. I've never learnt about augmented matrices, so don't know about those.

Augmented matrices is just a shorthand way of using Gaussian elimination (which, by the way, is much less computationally intensive than Cramer's rule).

ok try to use that algo :
note: it is an algo without pivoting!!!!!!!!

Code:

Quote:

function A=MEG(A,b)
[n,n] = size(A);
for k=1:n-1
for i=k+1:n
m=A(i,k)/A(k,k) ;
for j = k :n
A(i,j)=A(i,j)-m *A(k,j);
b(i)=b(i)-m*b(k);
end
end
end

when ur A will be U which is an upper triangular matrice use that algo to find ur x vector:

Code:

Quote:

function[x]=subs_retro(A,b)
[n,n]=size(A);
x=zeros(n,1);
x(n)=b(n)/A(n,n);//b here is the new b
for i=n-1:-1:1
x(i)=(b(i)-A(i,i+1:n)*x(i+1:n))/A(i,i);//b here is the new b
end

Hi all.
Yes, you can use it for such SOE. If I understand correctly, 12x8 SOE have an infinite set of solutions -- linear space. Axiom give you the basis of this space and a partial solution. Any linear function of basis-vectors will be a solution of uniform equation A*x=0. Any solution of nonuniform SOE A*x=b looks like x = x_part + c*x_base.
Example:

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