MHB 311.1.5.12 Ax=0 in parametric vector form

[karush]

$\tiny{1.5.12}$
Describe all solutions of $Ax=0$ in parametric vector form, where $A$ is row equivalent to the given matrix.
RREF
$A=\left[\begin{array}{rrrrrr}
1&5&2&-6&9& 0\\
0&0&1&-7&4&-8\\
0& 0& 0& 0& 0&1\\
0& 0& 0& 0& 0&0
\end{array}\right]
\sim \left[\begin{array}{rrrrrr}
1&5&0&8&1&0\\
0&0&1&-7&4&-8\\
0& 0& 0& 0& 0&1\\
0& 0& 0& 0& 0&0
\end{array}\right]$
$x_1=-5x_2-8x_4-x_5$ $x_2$ free $x_3=7x_4-4x_5$ $x_4$ free $ x_5\ free $x_6=0$
solution\\
$x_2\left[\begin{array}{rrrrrr}
-5\\1\\0\\0\\0\\0
\end{array}\right]
+x_4\left[\begin{array}{rrrrrr}
-8\\0\\7\\1\\0\\0
\end{array}\right]
+x_5\left[\begin{array}{rrrrrr}
-1\\0\\-4\\0\\1\\0
\end{array}\right]$

ok this appears to be the answer but I still don't see how the origin is 0 or we have || planes

 

[HOI]

??This problem has nothing to do with "the origin" or "planes"!

 

[karush]

ok i presume it is about parallel planes

 

[HOI]

Why? What was the exact statement of this problem and what makes you "presume" it is about parallel planes?

 

[karush]

Why? What was the exact statement of this problem and what makes you "presume" it is about parallel planes?

#12

 

[HOI]

I see nothing there that says anything about "planes" or "parallel planes"!

 

[karush]

ok so there is no possible graph of this

 

[HOI]

I have no idea what you are talking about! There is no mention of "planes" or "graphs" in this problem. Where are you getting this from? For problem 12, you have four equations in six unknowns. You could graph it- in 6 dimensions. The solution set is a two dimensional subspace of R^6.

But problem 9 has two equations in three dimensions: 3x- 6y+ 9z= 0 and -x+ 3y- 2z= 0. From the first equation, x= 2y- 3z. From the second equation, x= 3y- 2z. So x= 2y+ 3z= 3y- 2z. Add 2z to both sides and subtract 2y from both sides: 5z= y.
Then x= 3(5z)- 2z= 13z.

It solution space is one dimensional, the line in R^3, x= 13t, y= 5t, z= t.