Finding all solutions of Ax = y

[finalfreak]

Hi everyone...First post here so be kind please

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 can't find) + alpha(variable in R) * vector spanning N{A}

Please help me, I am really desperate! I have an exam tomorow morning on this and I want to have a good grade ;)

thanks

Mike

 

[Galileo]

Row-reduce the equation until A is in row-echelon form.
You can basically read the answer from that form.

 

[M98Ranger]

Hi Mike,

First of all, welcome to the forum! Finding all solutions of a system of equations can be a bit tricky, but I'll do my best to explain it to you.

To find all solutions of Ax = y, we first need to determine the rank of A. The rank of a matrix is the number of linearly independent rows or columns in the matrix. In this case, the rank of A is 3, since there are 3 linearly independent rows (as you mentioned, the dimension of R{A} is 3).

Next, we need to determine the number of variables in our system. In this case, there are 3 variables (x1, x2, and x3). This means that we will have 3 unknowns in our solution.

To find the solution, we will use the method of back substitution. We will start by setting one variable (x3) equal to a free parameter (let's call it t). This means that x3 = t. Now, we can use this value of x3 to solve for the other variables.

Using the first row of the matrix, we can write the equation 2x1 + 3x2 - 5(t) = 0. Solving for x1, we get x1 = (5t - 3x2)/2.

Using the second row, we can write the equation x1 + 3x2 - 2(t) = 1. Substituting the value of x1 from the first equation, we get (5t - 3x2)/2 + 3x2 - 2(t) = 1. Simplifying, we get x2 = (3 - t)/2.

Now, we have expressions for x1 and x2 in terms of t. Putting these values into our original system, we get the following solution:

x1 = (5t - 3((3-t)/2))/2
x2 = (3 - t)/2
x3 = t

This is the general solution to the system Ax = y. To find all solutions, we can choose different values of t and plug them into these equations. Each different value of t will give us a different solution to the system.

I hope this helps! Good luck on your exam tomorrow. Remember to check your work and make sure your solutions satisfy the original system of equations. Let me know if you have