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gruba

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## Homework Statement

Plot the solution set of linear equations

[tex]x-y+2z-t=1[/tex]

[tex]2x-3y-z+t=-1[/tex]

[tex]x+7z=8[/tex]

and check if the set is a vector space.

**2. The attempt at a solution**

Augmented matrix of the system:

[tex]

\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}

[/tex]

[itex]\Rightarrow t=1[/itex]

[tex]x-y+2z=2[/tex]

[tex]y+5z=6[/tex]

[itex]x,y[/itex] are pivot variables [itex]\Rightarrow x=8-7z,y=6-5z[/itex].

The solution set can be described as [itex]S=\{(8-7z,6-5z,z,1):z\in\mathbb R\}[/itex].

Algebraically, this means that the system is inconsistent (infinitely many solutions).

This should represent a line in [itex]xyz[/itex] plane, but it seems that it contradicts with algebraic solution.

First equation is a plane with coordinates [itex](6,-\frac{3}{2},3)[/itex].

Second equation is a plane with coordinates [itex](-1,\frac{2}{3},2)[/itex].

Third equation is a line in [itex]xz[/itex] plane with coordinates [itex](8,\frac{8}{7})[/itex].

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?

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