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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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    Nevermind the previous comment, I mixed up columnspace when I saw the words 'all solutions'.2012-04-22

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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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    The span of the vector (-3,1,1) are all the vectors parallel to this one since $span(\vec{u})=c_1 \cdot \vec{u}$. So it's actually the same thing. You're wrong about the columnspace. The columnspace is $span(A\cdot x)$ , where x all vectors in $R^N$. What you're looking for here is the Nullspace, which has dimension 1 (nullity=number of zero rows in row-echelon form)2012-04-22