finalfreak
Oct20-04, 07:53 PM
Hi everyone...First post here so be kind plz
In one of my assignments, I need to find all the solutions of the following system:
2 3 -5 1 x1 0
1 3 -2 5 x2 = 1
1 4 1 0 x3 0
A X = Y
I can easilly find the range space R{A}(Ind. Columns) and Null space N{A} (Ax =A~x= 0) using the echelon form of A. Since we have 3 leading entrys in A~ (Echelon), R{A} = R3, Dim(R{A}) = 3 thus Dim(N{a}) = 1 (single vector) which tells us that at least one solution is present. My problem is that I have no idea how to find that solution...I was thinking of taking the inverse of A and solving for A but since A is non-square...A is uninvertible.
I know that all the solution will have the following structure:
x = (the solution I cant find) + alpha(variable in R) * vector spanning N{A}
Please help me, Im really desperate!! I have an exam tomorow morning on this and I want to have a good grade ;)
thanks
Mike
In one of my assignments, I need to find all the solutions of the following system:
2 3 -5 1 x1 0
1 3 -2 5 x2 = 1
1 4 1 0 x3 0
A X = Y
I can easilly find the range space R{A}(Ind. Columns) and Null space N{A} (Ax =A~x= 0) using the echelon form of A. Since we have 3 leading entrys in A~ (Echelon), R{A} = R3, Dim(R{A}) = 3 thus Dim(N{a}) = 1 (single vector) which tells us that at least one solution is present. My problem is that I have no idea how to find that solution...I was thinking of taking the inverse of A and solving for A but since A is non-square...A is uninvertible.
I know that all the solution will have the following structure:
x = (the solution I cant find) + alpha(variable in R) * vector spanning N{A}
Please help me, Im really desperate!! I have an exam tomorow morning on this and I want to have a good grade ;)
thanks
Mike