Finding the general solution of this nonhomogeneous linear system

[PainterGuy]

Homework Statement

I'm trying to find the general solution of nonhomogeneous linear system.

Relevant Equations

I've presented my work below.

Hi,

I was trying to do the following problem.

My attempt.

Finding the reduced row echelon form for the system above.

I do not see any way to proceed any further. The following is the solution presented in solution manual. How do I proceed to get the following answer?

 

[abdv17]

You just need to use the following two facts :

1) If x' is a non-trivial solution to the homogeneous system Ax=0, then ax' is also a solution for this system where a is any scalar.

2) Let H be the general solution of the system Ax=o. Then, the general solution N for the system Ax=b is given by : N = x' + H, where x' is any specific solution for Ax=b.

 

[PainterGuy]

Thank you!

I'm already aware of the two facts you mentioned above. I do not understand how they get the following value for x_p.

 

[PainterGuy]

Homework Statement:: I'm trying to find the general solution of nonhomogeneous linear system.
Relevant Equations:: I've presented my work below.

Hi,

I was trying to do the following problem.

View attachment 279668

My attempt.

Finding the reduced row echelon form for the system above.

View attachment 279669I do not see any way to proceed any further. The following is the solution presented in solution manual. How do I proceed to get the following answer?

View attachment 279670

I see that they are setting x3=0 to get the following result but I don't understand the reason for setting x3=0. Could you please help me?

 

[abdv17]

Since the rank of the coefficient matrix (by definition the rank of a matrix is the number of non-zero rows in the reduced row echelon form of the matrix) equals the rank of the augmented matrix (both are 3) and is one less than the number of variables in the system (4), the solution of the system has 4 - 3 = 1 degree of freedom.

And because we have a degree of freedom in the solution (that is to say we are free to assign any value we wish to anyone of x1, x2, or x3, and then the other two will be determined through the equations x1 - x3 = -1, and x2 + 2x3 = 0), we can set x3 = 0 to obtain the given solution.

 

[PainterGuy]

Since the rank of the coefficient matrix (by definition the rank of a matrix is the number of non-zero rows in the reduced row echelon form of the matrix) equals the rank of the augmented matrix (both are 3) and is one less than the number of variables in the system (4), the solution of the system has 4 - 3 = 1 degree of freedom.

And because we have a degree of freedom in the solution (that is to say we are free to assign any value we wish to anyone of x1, x2, or x3, and then the other two will be determined through the equations x1 - x3 = -1, and x2 + 2x3 = 0), we can set x3 = 0 to obtain the given solution.

Thanks a lot! I really appreciate it.I think they specifically chose x3=0 because the third column which belong to x3 is a dependent column; column_3 = -1(column_1) + 2(column_2). Do I make sense?

Are there three parameters? How is a parameter differentiated from a variable in this case?

Another Question:

This time they are setting x3=-1 in their solution which is shown at the bottom. Any particular reason for doing this? By the way, what is that symbol which looks like "α" in their solution?

Their solution:

 

[abdv17]

Edited by mentor to fix up quoted text.

Thanks a lot! I really appreciate it.

I think they specifically chose x3=0 because the third column which belong to x3 is a dependent column; column_3 = -1(column_1) + 2(column_2). Do I make sense?

Since the solutions we get by 1) choosing x1 = -1 , or, 2) choosing x2 = 0, or, 3) choosing x3 = 0 are identical, I don't find it particularly meaningful to say that they chose x3 = 0. In any case column 1 is dependent on columns 2 & 3 ; likewise column 2 is dependent on columns 1 & 3.

Are there three parameters? How is a parameter differentiated from a variable in this case?

View attachment 279682

No, since there is only one free variable (i.e. only one degree of freedom) in the solution, just one parameter suffices to describe the general solution.

Regarding variable vs parameter : There are four variables in the system (the components of the vector x : x1, x2, x3, & x4) since it is their values which determine if the system is satisfied or not. The solution of the system has one parameter since only one variable (alpha) is required to describe the complete solution set given 1) the general solution of the homogeneous system for the same matrix, and 2) some specific solution for the given system.

Another Question:

This time they are setting x3=-1 in their solution which is shown at the bottom. Any particular reason for doing this?

Nope.

By the way, what is that symbol which looks like "α" in their solution?

I think it's "alpha" too (it is just a variable that can take any real value).

View attachment 279683Their solution:

View attachment 279684

 

[Mark44]

@PainterGuy, for your other problem, the one that is a homogeneous system, a good way to get a basis is as follows.

From the reduced row-echelon matrix, we have these equations:
$x_1 - x_3 = 0$
$x_2 + 2x_3 = 0$
$x_4 = 0$

Written another way:
$x_1 = x_3$
$x_2 = -2x_3$
$x_3 = x_3$
$x_4 = 0$
This formulation makes it more obvious that an arbitrary vector that is a solution of the homogeneous system (i.e., is in the nullspace of the matrix) is
$x_3\begin{bmatrix}1 \\ -2 \\ 1 \\ 0 \end{bmatrix}$, where $x_3$ is some arbitrary value.

If you choose $x_3 = 1$, then the vector is
$\begin{bmatrix}1 \\ -2 \\ 1 \\ 0 \end{bmatrix}$

In fact, every solution of the homogeneous system is some scalar multiple of the vector above, which makes this vector a basis for the nullspace of the homogeneous system.

 

[PainterGuy]

Question:

EDIT:
I just noticed that in case of x_h was evaluated at x3=α=-1 but at the end, they still included α in the general solution. Please compare their general solution with mine, you'd understand what I'm trying to understand. Thank you!

 

[Mark44]

I just noticed that in case of x_h was evaluated at x3=α=-1 but at the end, they still included α in the general solution.

No, to get $x_h$, they set $\alpha$ to zero. This produced the vector <-1, 0, 0, 1>. A particular solution $x_p$ is obtained by setting $\alpha$ to whatever value is convenient, but they chose $\alpha = -1$. The general solutions consist of $x_h$ plus all possible scalar multiples of $x_p$.

 

[vela]

View attachment 279742

Note that you can rewrite your expression for $x_p$ as
$$ x_p = \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} + \alpha \begin{bmatrix} -1 \\ 2 \\ -1 \\ 0\end{bmatrix},$$ which is actually the general solution to the problem.

When you solve for $x_p$, you're looking for A particular solution, not THE particular solution. Any vector that satisfies the non-homogeneous system will do. Why did they choose $x_3 = 0$? Because it's a convenient value that makes the equations you found really easy to solve. (Zeros are your friends.) They could have just as easily chosen $x_3 = 1$ and come up with a different vector for $x_p$. Similarly, they could have chosen to set $x_1 = 0$ and solved for $x_2$ and $x_3$. But because of the equations you ended up with, setting $x_3=0$ minimized the work so that's what they did.

The homogeneous solution will be the part of the solution that has the arbitrary constants in it. In this case, it's the second term above. So in this case, you don't need to solve the homogeneous and inhomogeneous systems separately. You can do what you did and come up with a solution with an $\alpha$ floating around in it, and identify the constant part as a particular solution.

The basic idea is you want to recognize that you can always split up the solution to a linear system this way. In other places where these concepts apply, like in differential equations, it turns out it's often easy to write down the solution to the homogeneous system but quite difficult to find a particular solution. So you learn to solve for those things separately and then combine them to get the general solution.