I have this system of equations that forms a homogeneous system like this:
$$
\left[\begin{array}{rrr|r}
2 & -1 & -1 & 0 \\
1 & -2 & -1 & 0 \\
1 & -1 & -2 & 0
\end{array}\right]
$$
I tried to do elimination process to try to get the infinite solutions case, but every time I try I keep getting the trivial solution of (0,0,0).
I know this system should get me a parametrized solution of a line.
Doing this purely algebraic one gets : y =-x and z=3x
, where the free parameter is x, so hence you can say let x=t, then the solution is:
(x,y,z) = (t, -t,3t)
I tried out many online systems of equations solvers, and found out that many are not equipped to handle the 'infinitely many solutions' case, for the Homogeneous systems they would only give the Trivial solution.
BUT, Wolfram alpha was able to get this non-trivial infinite solutions, parameterized solution.
But I want to know how can I get this via doing elementary row operations in this augmented matrix.
I know I should get a row of zeros across this matrix in order to get to the infinite solutions case.
Hope someone can explain why I keep encountering the trivial solution when there is this infinite case.
Regards,
=======================
I see I have made an Error, I do apologize. I derived a third plane, which is the third row, which was actually unjustified. I actually have a system of equations of the first 2 rows in that matrix, the third row does not exist.
SO since we have these 2 planes, and hence these planes actually intersect in a line.
Once again my apologies.
Best Answer
Then you are row reducing wrong! If you show what you did we might be able to point out an error.
This problem can be written as 2x- y- z= 0, x- 2y- z= 0, and x- y- 2z= 0.
If we subtract the second equation from the third, we eleminate x leaving y+ z= 0. If we subtract twice the second equation from the first, we also eleminate x leaving 3y+ z= 0.
Subtracting y+ z= 0 from 3y+ z= 0 eleminates z leaving 2y= 0 so y=0. Then y+ z= z= 0 and x- 2y- z= x= 0. The "trivial solurion, x= y= z= 0, is the only solution.
A quicker but more "formulaic" way of seeing that is to take the determinant of the coefficient matrix: $\left|\begin{array}{ccc} 2 & -1 & -1 \\ 1 & -2 & -1 \\ 1 & -1 & -2\end{array}\right|= 2\left|\begin{array}{cc}-2 & -2 \\ -1 & -2\end{array}\right|+ \left|\begin{array}{cc} 1 & -1 \\ 1 & -2 \end{array}\right|- \left|\begin{array}{cc}1 & -2 \\ 1& -1\end{array}\right|= 2(4- 2)+ (-2+ 1)- (-1+ 2)= 4- 1+ 1= 4$.
Since that is not 0, the matrix is invertible and there exist a unique solution which is, of course, x= y= z= 0.