Determine all the solutions of the system

[Hall]

Homework Statement

$$
5x +2y -6z +2u =-1$$
$$
x- y +z -u =-2$$

Relevant Equations

Gauss-Jordan Elimination.

(I don't know how to make augmented matrices in latex, so what I would do is to use an equal to sign)
$$
\begin{bmatrix}
5 &2&-6&2 \\
1&-1&1&-1 \\
\end{bmatrix}
= \begin{bmatrix}
-1\\
-2\\
\end{bmatrix}$$
$R_1 \to R_1 +2R_2$
$$
\begin{bmatrix}
7 &0&-4&0\\
1&-1&1&-1\\
\end{bmatrix}
= \begin{bmatrix}
-5\\
-2
\end{bmatrix} $$

Now, I would like to stop the Elimination process and put the variable in:
$z = (7x+5)/4$
$x -y +(7x+5)/4 -u =-2$

How to go further on? And what the question meant by "all the solutions"? Did they mean the general solution?

 

[PeroK]

How to go further on? And what the question meant by "all the solutions"? Did they mean the general solution?

Yes, they want a mathematical description of all possible solutions. Can you see how many dimensions the solution space has?

 

[Hall]

Yes, they want a mathematical description of all possible solutions. Can you see how many dimensions the solution space has?

I always disliked the system of Linear equations and never paid a good attention to it. All I know from my high school studies is that when the number of unknowns exceed the number of Equations we usually have infinite number of solutions.

So, it seems to me the dimension will be infinite.

 

[PeroK]

So, it seems to me the dimension will be infinite.

The dimension isn't infinite. A line would be an infinite number of solutions, but that is one dimensional.

 

[Hall]

The dimension isn't infinite. A line would be an infinite number of solutions, but that is one dimensional.

To be honest I don't know how do we find the dimension, but it seems all the variables shall be expressed in terms of x, therefore one dimensional.

 

[PeroK]

To be honest I don't know how do we find the dimension, but it seems all the variables shall be expressed in terms of x, therefore one dimensional.

You have $z$ in terms of $x$. But, what about $y$ in terms of $x$?

Hint: two equations and four variables ...

 

[sysprog]

I always disliked the system of Linear equations and never paid a good attention to it. All I know from my high school studies is that when the number of unknowns exceed the number of Equations we usually have infinite number of solutions.

So, it seems to me the dimension will be infinite.

Can you solve for 2 of the variables and identify the 2 others as free?

 

[Hall]

You have $z$ in terms of $x$. But, what about $y$ in terms of $x$?

Hint: two equations and four variables ...

y can be written in terms of x and u. So, I think the dimension is 2.

 

[Hall]

Can you solve for 2 of the variables and identify the 2 others as free?

Yes, x and u are free and z and y can be written in terms of them.

 

[PeroK]

y can be written in terms of x and u. So, I think the dimension is 2.

Can you finish it off? Take $x$ and $u$ as free variables (or $x$ and $y$, which is what I've done!) and express the other variables in terms of them.

 

[Hall]

Can you finish it off? Take $x$ and $u$ as free variables (or $x$ and $y$, which is what I've done!) and express the other variables in terms of them.

$z = (7x +5)/5$
$y = (11x -4u +13)/4$

 

[PeroK]

$z = (7x +5)/5$
$y = (11x -4u +13)/4$

Looks good. A 2D plane in 4D space. Not easy to visualise!

 

[Hall]

Looks good. A 2D plane in 4D space. Not easy to visualise!

Not hard for you I think. Can you give help me visualizing it? (Invoking Relativity?)

 

[PeroK]

Not hard for you I think. Can you give help me visualizing it? (Invoking Relativity?)

I can't visualise it. I rely on using the number of free parameters, or the number of linearly independent vectors from a point in the plane.

 

[Mark44]

(I don't know how to make augmented matrices in latex
$$
\begin{bmatrix}
5 &2&-6&2 \\
1&-1&1&-1 \\
\end{bmatrix}
= \begin{bmatrix}
-1\\
-2\\
\end{bmatrix}$$

Like this:
$\begin{bmatrix} 5 &2&-6&2 & | & -1 \\ 1&-1&1&-1&| & -2\\ \end{bmatrix}$

$R_1 \to R_1 +2R_2$

The above row operation is not useful for Gauss-Jordan elimination. One possibility is to add a multiple of row 1 to row 2 so as to eliminate the leading entry of row 2. (Or you could add a multiple of row 2 to row 1 so as to eliminate the leading entry of row 1.)

$$
\begin{bmatrix}
7 &0&-4&0\\
1&-1&1&-1\\
\end{bmatrix}
= \begin{bmatrix}
-5\\
-2
\end{bmatrix} $$

Now, I would like to stop the Elimination process and put the variable in:
$z = (7x+5)/4$
$x -y +(7x+5)/4 -u =-2$

How to go further on? And what the question meant by "all the solutions"? Did they mean the general solution?

Yes.

Not hard for you I think. Can you give help me visualizing it? (Invoking Relativity?)

This has nothing to do with relativity or spacetime. The solution is just a plane embedded in R⁴. You don't need to sketch the space the plane is in, just the plane itself.

 

[Mark44]

Looking ahead, it's helpful to be systematic in how you present the solutions to these kinds of problems. In reduced row-echelon form, the augmented matrix looks like this:
$\begin{bmatrix} 1&0 & -4/7 & 0 & | & -5/7 \\ 0 & 1 & -11/7 & 1 & | & 9/7 \\\end{bmatrix}$
This means that

x = 4/7 z +  0 u - 5/7
y = 11/7 z - 1 u + 9/7
z =      z
u =           u

It's very important to understand the system of equations that a matrix or augmented matrix represents. From the above, a solution vector $\vec x$ ( = <x, y, z, u>^T )can be written by merely picking off the coordinates of z and u + the constant vector.
I.e.
$\vec x = \begin{bmatrix} 4/7 \\ 11/7 \\ 1 \\ 0 \end{bmatrix}z + \begin{bmatrix} 0 \\ -1 \\ 0 \\ 1\end{bmatrix}u + \begin{bmatrix} -5/7 \\ 9/7 \\ 0 \\ 0\end{bmatrix}$

This arrangement shows the two vectors that define the plane, plus a vector from the origin up to a point in the plane.

 

[FactChecker]

In the future, please state the problem completely, especially if you are going to ask about a part of the problem like "all the solutions". You did not even include that part in your problem statement above (although it was mentioned in the title).

 

[FactChecker]

Examples of equation sets, solutions, and dimensions of solution sets:
A) x=6 One equation with 1 variable, one solution (single point), solution set has 0 dimension
B) y = x+5 One equation with 2 variable, infinite solutions (a line), solution set has 1 dimension
C) y=x+5, y=2x Two equations with 2 variables, one solution (single point), solution set has 0 dimension
In general, N independent equations with M variables (unknowns) will have a solution set with M-N dimensions.

 

[Mark44]

C) y=x+5, y=2x Two equations with 2 variables, one solution (single point), solution set has 0 dimension
In general, N independent equations with M variables (unknowns) will have a solution set with M-N dimensions.

To elaborate on the above, and omitting knowledge about whether the equations are independent, if we have three equations in three unknowns, we could have:
No solutions
The system of equations is inconsistent. Geometrically, the planes are all mutually parallel.

Exactly one solution
The system of equations is consistent, meaning that the reduced row echelon form will have a leading entry for each row. Geometrically, the three planes intersect in a single point.

Infinite number of solutions - can occur in two ways
The reduced row echelon form will have one row of zeroes. Geometrically, the three planes intersect in a line.
The reduced row echelon form will have two rows of zeroes. Geometrically, the three planes all intersect everywhere (the three equations all represent the same plane).

 

[sysprog]

An illustrated list of the possible geometric arrangements (found on stackexchange):