Finding all solutions of Ax = y

  • Thread starter finalfreak
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In summary, the conversation is about finding the solutions of a system of equations for a given assignment. The speaker mentions using echelon form to find the range space and null space, but is struggling to find the specific solution. They mention the inverse of A, but it is not possible since A is non-square. They ask for help and mention an upcoming exam.
  • #1
finalfreak
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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
 
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  • #2
Row-reduce the equation until A is in row-echelon form.
You can basically read the answer from that form.
 
  • #3


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
 

1. What is the meaning of "finding all solutions" in the context of Ax = y?

When we say "finding all solutions" of Ax = y, it means finding all possible values of x that satisfy the equation, where A represents a matrix and y represents a vector. In other words, we are looking for the set of all possible solutions that make the equation true.

2. How do you solve for all solutions of Ax = y?

The process of finding all solutions of Ax = y involves using various techniques such as Gaussian elimination, inverse matrices, or matrix decomposition methods. These methods help in reducing the equation to a simpler form, making it easier to solve for all possible solutions.

3. Can there be more than one solution for Ax = y?

Yes, there can be more than one solution for Ax = y. In fact, there can be infinitely many solutions for this type of equation, depending on the dimensions of the matrix and vector. This is because matrices can have infinite combinations of values that satisfy the equation.

4. How do you know if a solution to Ax = y is unique?

A solution to Ax = y is unique if and only if the matrix A is invertible. This means that there is only one set of values for x that will make the equation true. If the matrix is not invertible, there will be infinitely many solutions or no solutions at all.

5. Why is finding all solutions of Ax = y important?

In many scientific and mathematical applications, finding all solutions of Ax = y is crucial for solving complex problems. It allows us to understand the behavior and relationships between different variables in a system. Additionally, it helps in finding the best solution or combination of solutions for a given equation, leading to more accurate and efficient results.

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