# Solution set: S = {(8 + 7z, 6 + 5z, z, 1) : z ∈ ℝ}

• gruba
In summary, the conversation discusses solving a system of linear equations and checking if the solution set is a vector space. The solution set is found to be a line passing through a point and directed by a vector, making it an affine subspace rather than a vector space. The conversation also questions the discrepancy between the algebraic and geometric solutions, and it is explained that the solution set is a parametrization of a line.
gruba

## Homework Statement

Plot the solution set of linear equations
$$x-y+2z-t=1$$
$$2x-3y-z+t=-1$$
$$x+7z=8$$

and check if the set is a vector space.

2. The attempt at a solution

Augmented matrix of the system:
$$\begin{bmatrix} 1 & -1 & 2 & -1 & 1 \\ 2 & -3 & -1 & 1 & -1 \\ 1 & 0 & 7 & 0 & 8 \\ \end{bmatrix}\sim \begin{bmatrix} 1 & -1 & 2 & -1 & 1 \\ 0 & -1 & -5 & 3 & -3 \\ 0 & 1 & 5 & 1 & 7 \\ \end{bmatrix}\sim \begin{bmatrix} 1 & -1 & 2 & -1 & 1 \\ 0 & 1 & 5 & 1 & 7 \\ 0 & -1 & 3 & 0 & -3 \\ \end{bmatrix}\sim \begin{bmatrix} 1 & -1 & 2 & -1 & 1 \\ 0 & 1 & 5 & 1 & 7 \\ 0 & 0 & 0 & 4 & 4 \\ \end{bmatrix}$$
$\Rightarrow t=1$

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

$x,y$ are pivot variables $\Rightarrow x=8-7z,y=6-5z$.
The solution set can be described as $S=\{(8-7z,6-5z,z,1):z\in\mathbb R\}$.

Algebraically, this means that the system is inconsistent (infinitely many solutions).
This should represent a line in $xyz$ plane, but it seems that it contradicts with algebraic solution.

First equation is a plane with coordinates $(6,-\frac{3}{2},3)$.
Second equation is a plane with coordinates $(-1,\frac{2}{3},2)$.
Third equation is a line in $xz$ plane with coordinates $(8,\frac{8}{7})$.

When drawn altogether, there is no intersection between all three planes, and also no line as an intersection.

Is there a mistake in algebraic, or geometric solution?

Assuming that algebraic solution is correct, solution set is a vector space because it is an infinite line (closed under vector addition and scalar multiplication). Is this correct?

Last edited:
gruba said:

## Homework Statement

Plot the solution set of linear equations
$$x-y+2z-t=1$$
$$2x-3y-z+t=-1$$
$$x+7z=8$$

and check if the set is a vector space.

2. The attempt at a solution

Augmented matrix of the system:
$$\begin{bmatrix} 1 & -1 & 2 & -1 & 1 \\ 2 & -3 & -1 & 1 & -1 \\ 1 & 0 & 7 & 0 & 8 \\ \end{bmatrix}\sim \begin{bmatrix} 1 & -1 & 2 & -1 & 1 \\ 0 & -1 & -5 & 3 & -3 \\ 0 & 1 & 5 & 1 & 7 \\ \end{bmatrix}\sim \begin{bmatrix} 1 & -1 & 2 & -1 & 1 \\ 0 & 1 & 5 & 1 & 7 \\ 0 & -1 & 3 & 0 & -3 \\ \end{bmatrix}\sim \begin{bmatrix} 1 & -1 & 2 & -1 & 1 \\ 0 & 1 & 5 & 1 & 7 \\ 0 & 0 & 0 & 4 & 4 \\ \end{bmatrix}$$
$\Rightarrow t=1$

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

$x,y$ are pivot variables $\Rightarrow x=8-7z,y=6-5z$.
The solution set can be described as $S=\{(8-7z,6-5z,z,1):z\in\mathbb R\}$.

Algebraically, this means that the system is inconsistent (infinitely many solutions).
This should represent a line in $xyz$ plane, but it seems that it contradicts with algebraic solution.

First equation is a plane with coordinates $(6,-\frac{3}{2},3)$.
Second equation is a plane with coordinates $(-1,\frac{2}{3},2)$.
Third equation is a line in $xz$ plane with coordinates $(8,\frac{8}{7})$.

When drawn altogether, there is no intersection between all three planes, and also no line as an intersection.

Is there a mistake in algebraic, or geometric solution?

Assuming that algebraic solution is correct, solution set is a vector space because it is an infinite line (closed under vector addition and scalar multiplication). Is this correct?

A vector space must contain the zero vector. Does your solution set have that property?

Ray Vickson said:
A vector space must contain the zero vector. Does your solution set have that property?

So, the algebraic solution is not a vector space.

But why algebraic and geometric solution is not the same?
Algebraic solution set gives infinite solutions, and geometric solution doesn't give
neither point nor line as an intersection.

Assuming your solution is correct, ##S## is a line of ##\mathbb{R}^4## passing through ##(8,6,0,1)## and directed by vector ##(-7,-5,1,0)##. It is not a vector space but an affine subspace ##\mathbb{R}^4##

gruba said:
Algebraically, this means that the system is inconsistent (infinitely many solutions).
This should represent a line in xyz plane, but it seems that it contradicts with algebraic solution.
An inconsistent system has no solutions, and there is no such thing as an "xyz" plane.
gruba said:
So, the algebraic solution is not a vector space.

But why algebraic and geometric solution is not the same?
I think you have some mistakes in your work. In the 3rd matrix you show, how did you get the 2nd and 3rd rows?
gruba said:
Algebraic solution set gives infinite solutions, and geometric solution doesn't give
neither point nor line as an intersection.
Geometrically, the solution set I get is a line in ##\mathbb{R}^4## that passes through the point (8, 6, 0, 1).

Edit: Corrected an earlier mistake I made.

Last edited:
geoffrey159 said:
Assuming your solution is correct, ##S## is a line of ##\mathbb{R}^4## passing through ##(8,6,0,1)## and directed by vector ##(-7,-5,1,0)##. It is not a vector space but an affine subspace ##\mathbb{R}^4##

How do you get the point $(8,6,0,1)$ and vector $(-7,-5,1,0)$?

Your solution set ##S## (I see it was confirmed correct by Mark44) is a parametrization of line, i.e it can be written ## A + z \vec v ## where ##A = (8,6,0,1)## and ##\vec v = (-7,-5,1,0)##

geoffrey159 said:
Your solution set ##S## (I see it was confirmed correct by Mark44) is a parametrization of line, i.e it can be written ## A + z \vec v ## where ##A = (8,6,0,1)## and ##\vec v = (-7,-5,1,0)##

Could you explain how to do the parametrization of a line?

I don't want go off topic so I'll just explain why your solution set is a line.
A line passing through point ##A## and directed by a vector ##\vec v## is the set of points ##M## such that vector ##\vec {AM}## and ##\vec v## are colinear.
Your solution set can be written ##S = \{ A + z \vec v, \ z\in \mathbb{R} \} ##, therefore ##M\in S \iff \exists z\in\mathbb{R},\ \vec{AM} = z\vec v##. It's a line. That's all there is to it.

gruba
gruba said:
Could you explain how to do the parametrization of a line?
I have copied the work you showed earlier. To find the parametric form of the line, the last matrix needs to be completely row-reduced, which means that each nonzero row starts with a 1 entry, and each leading 1 entry (a pivot) has 0 entries above and below it.
gruba said:
Augmented matrix of the system:
##
\begin{bmatrix}
1 & -1 & 2 & -1 & 1 \\
2 & -3 & -1 & 1 & -1 \\
1 & 0 & 7 & 0 & 8 \\
\end{bmatrix}\sim \begin{bmatrix}
1 & -1 & 2 & -1 & 1 \\
0 & -1 & -5 & 3 & -3 \\
0 & 1 & 5 & 1 & 7 \\
\end{bmatrix}\sim \begin{bmatrix}
1 & -1 & 2 & -1 & 1 \\
0 & 1 & 5 & 1 & 7 \\
0 & -1 & 3 & 0 & -3 \\
\end{bmatrix}\sim \begin{bmatrix}
1 & -1 & 2 & -1 & 1 \\
0 & 1 & 5 & 1 & 7 \\
0 & 0 & 0 & 4 & 4 \\
\end{bmatrix}
##
When the last matrix is completely reduced (in RREF or reduced row-echelon form), you can read off the solution, as
x = ...
y = ...
z = ...
t = ...
For this matrix that will turn out to be z times a vector plus a constant vector, where in this case, z is the parameter.

Mark44 said:
To find the parametric form of the line, the last matrix needs to be completely row-reduced, which means that each nonzero row starts with a 1 entry, and each leading 1 entry (a pivot) has 0 entries above and below it.

And how to represent the line in explicit form, $y=kx+n$?

gruba said:
And how to represent the line in explicit form, $y=kx+n$?
The line in question is a line in ##\mathbb{R}^4##, so you wouldn't write it as y = mx + b. Instead, you would write it something like ##\vec{r} = k\vec{A} + \vec{B}##. Here ##\vec{r}## means ##\begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix}##. ##\vec{A}## and ##\vec{B}## are constant vectors that come directly out of the completely reduced final matrix.

gruba
Mark44 said:
The line in question is a line in ##\mathbb{R}^4##, so you wouldn't write it as y = mx + b. Instead, you would write it something like ##\vec{r} = k\vec{A} + \vec{B}##. Here ##\vec{r}## means ##\begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix}##. ##\vec{A}## and ##\vec{B}## are constant vectors that come directly out of the completely reduced final matrix.

Which vector is $\overrightarrow{A}$ and which vector is $\overrightarrow{B}$ from RREF?
Also, is $k$ undetermined?

Last edited:
One more step takes your final matrix to RREF (reduced row-echelon form)
##\begin{bmatrix}
1 & -1 & 2 & -1 &| & 1 \\
0 & 1 & 5 & 1 & | & 7 \\
0 & 0 & 0 & 4 & | & 4 \\
\end{bmatrix} \equiv
\begin{bmatrix}
1 & 0 & 7 & 0 & | & 8 \\
0 & 1 & 5 & 0 & | & 6 \\
0 & 0 & 0 & 1 & | & 1 \\
\end{bmatrix}##
Now, write the system of equations that this matrix represents.
x = ...
y = ...
z = ...
t = ...
If you line things up you should see a column vector times a parameter z, plus another vector.

## 1. What is a linear system of equations?

A linear system of equations is a set of two or more equations that contain two or more variables. These equations can be solved simultaneously to find a unique solution for the variables.

## 2. How do you solve a linear system of equations?

To solve a linear system of equations, you can use different methods such as substitution, elimination, or graphical methods. These methods involve manipulating the equations to eliminate one variable and solve for the remaining variable.

## 3. Can a linear system of equations have no solution?

Yes, a linear system of equations can have no solution. This happens when the two lines represented by the equations are parallel and do not intersect. In this case, there is no solution that satisfies both equations.

## 4. What does it mean if a linear system of equations has infinitely many solutions?

If a linear system of equations has infinitely many solutions, it means that the two lines represented by the equations are the same and intersect at every point. In this case, any value for the variables that satisfies one equation will also satisfy the other equation.

## 5. How is a linear system of equations used in real life?

Linear systems of equations are used in various fields such as engineering, economics, and physics. They can be used to model and solve real-life problems involving multiple variables, such as finding the optimal production level for a company or predicting the trajectory of a projectile.

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