Can't figure out how to solve matrix equation Ax=0
Matthew Barrera 
I tried googling how to solve this matrix through RREF and parametric variables but failed to find something that works similarly to try and solve myself.
A is a 2x3 matrix with the values going [ 2 -1 -1 : 1 -2 2 ] (imagine the set after the colon to be under the first set)
x is [x_1, x_2, x_3] but obviously a column instead of a row and for lack of subscript key I just used "_#" to denote the same thing.
0 is a 2 row x 1 column matrix with two 0's.
To start, I wrote out the matrix's into 2 equations.
2x_1 - x_2 - x_3 = 0
x_1 - 2x_2 + 2x_3 = 0
Then I turned it into a simplified coefficient matrix. [ 2 -1 -1 0: 1 -2 2 0] (with the second set of numbers past the colon under the first set)
Afterward, I reduced it into RREF form to get:
[ 1 0 (-4/3) 0 : 0 1 (-5/3) 0 ] (with the second set of numbers past the colon under the first set)
Once I got this, I turned the matrix back to these equations:
X_1 - (4/3)X_3 = 0
X_2 - (5/3)X_3 = 0
So then I set X_1 = (4/3)X_3 and X_2 = (5/3)X_3. I then tried setting X_3 to t to try and see if I could finagle an answer by solving for 't'. So far, my answers have been wrong every time in comparison to the answer my book gives. Can someone tell me if I messed up somewhere? I deduced it would have to be from my matrix reduction but I did it over a few times in different ways to make sure but I still can't seem to get the answer...
$\endgroup$ 72 Answers
$\begingroup$The general method for solving a linear equation $$Ax=b$$is to utilize the Moore-Penrose inverse $A^+$ and the associated nullspace projector $$P=(I-A^+A)$$
With these two matrices, the general solution can be written as$$x=A^+b + Py$$where the vector $y$ is completely arbitrary.
Applied to the current problem this technique yields $$\eqalign{ &A=\begin{bmatrix}2&-1&-1\\1&-2&2\end{bmatrix},\quad &A^+=\frac{1}{50}\begin{bmatrix}\;\;16&\;\;\;2\\\;-5&-10\\-13&\;\;14\end{bmatrix},\quad P=\frac{1}{50}\begin{bmatrix}16&20&12\\20&25&15\\12&15&9\end{bmatrix} \\ & b=\begin{bmatrix}0\\0\\0\end{bmatrix} ,\quad p=\begin{bmatrix}4\\5\\3\end{bmatrix} &\implies P=\frac{\;\;pp^T}{50},\quad x=0+\frac{p(p^Ty)}{50} = \lambda p = \lambda\begin{bmatrix}4\\5\\3\end{bmatrix} \\ }$$
$\endgroup$ 3 $\begingroup$You are right, indeed by the augmented matrix for the system and by row operation we obtain
$$\left[\begin{array}{ccc|c} 2& -1& -1& 0\\ 1& -2& 2& 0 \end{array}\right]\to \left[\begin{array}{ccc|c} 2& -1& -1& 0\\ 0& -3& 5& 0 \end{array}\right]\to \left[\begin{array}{ccc|c} 6& 0& -8& 0\\ 0& -3& 5& 0 \end{array}\right]$$
from which we obtain
$x_1=\frac43 x_3$
$x_2=\frac 53 x_3$
that means we have a free parameter and by $x_3=t$ the genral solution is
$$(x_1,x_2,x_3)=t\left(\frac43,\frac53,1\right)$$
which is equivalent to
$$(x_1,x_2,x_3)=t\left(4,5,3\right)$$
$\endgroup$ 9
