# [SOLVED]311.1.5.12 Ax=0 in parametric vector form

#### karush

##### Well-known member
$\tiny{1.5.12}$
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

#### Country Boy

##### Well-known member
MHB Math Helper
??This problem has nothing to do with "the origin" or "planes"!

#### karush

##### Well-known member
ok i presume it is about parallel planes

#### Country Boy

##### Well-known member
MHB Math Helper
Why? What was the exact statement of this problem and what makes you "presume" it is about parallel planes?

#### karush

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

#12

#### Country Boy

##### Well-known member
MHB Math Helper
I see nothing there that says anything about "planes" or "parallel planes"!

#### karush

##### Well-known member
ok so there is no possible graph of this

#### Country Boy

##### Well-known member
MHB Math Helper
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.