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[SOLVED] 311.1.5.12 Ax=0 in parametric vector form

karush

Well-known member
Jan 31, 2012
3,068
$\tiny{1.5.12}$
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
Jan 30, 2018
821
??This problem has nothing to do with "the origin" or "planes"!
 

karush

Well-known member
Jan 31, 2012
3,068
ok i presume it is about parallel planes
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
Why? What was the exact statement of this problem and what makes you "presume" it is about parallel planes?
 

karush

Well-known member
Jan 31, 2012
3,068

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
I see nothing there that says anything about "planes" or "parallel planes"!
 

karush

Well-known member
Jan 31, 2012
3,068
ok so there is no possible graph of this
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
821
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 [tex]R^6[/tex].

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 [tex]R^3[/tex], x= 13t, y= 5t, z= t.