Find the value of b for which the following system has a non-trivial solution and find all the solutions in this case
$$2x + 6z = 0$$ $$4x + y + bz = 0$$ $$y - z = 0$$
I put this in a matrix and row reduced and got
$\begin{bmatrix}2 & 0 & 6 \\ 0 & 1 & b-12 \\ 0 & 0 & -b+11\end{bmatrix}$
So this has a non-trivial solution when b = 11.
Now what is the way to find all solutions?
Edit:
Had a 6 instead of a b in the second row, third column of the original matrix.
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$\begingroup$Let your system be (assuming b=11, whatever b is): $$\begin{bmatrix}2 & 0 & 6 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{bmatrix}\cdot\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$then, by multiplication you get: $$2x+6z=0 \ and \ y-z=0$$Transforming this system of equations you get: $$x=-3z \ and \ y=z$$ The general solution of this system is the vector $\mathbf{x}$, given by: $$\mathbf{x}=\begin{bmatrix}x \\ y \\ z \end{bmatrix}=\begin{bmatrix}-3z \\ z \\ z \end{bmatrix}=\begin{bmatrix}-3 \\ 1 \\ 1 \end{bmatrix}\cdot z$$Therefore, the solutions to this system are infinite, but all of them are vectors parallel to the vector $$\begin{bmatrix}-3 \\ 1 \\ 1 \end{bmatrix}$$and their length is $|z|$ times the length of the vector above.
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