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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?


Had a 6 instead of a b in the second row, third column of the original matrix.

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What is $b$? I don't see it anywhere in the equations... – dtldarek Apr 22 '12 at 10:57
When Im asked to find all the solutions, isn't this asking 'what is the columnspace of the matrix'? And in this case of the matrix above, when b = 11 we have all zeros on the last row - so the columnspace is the span of the first two vectors in the original matrix? – Jim_CS Apr 22 '12 at 12:18
Nevermind the previous comment, I mixed up columnspace when I saw the words 'all solutions'. – Jim_CS Apr 22 '12 at 12:29
up vote 2 down vote accepted

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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oopp........... – Jim_CS Apr 22 '12 at 12:22
So the solution of this systems is the span of the vector (-3,1,1). (-9,3,3) is a solution for example...But you say there are other solutions too, parallel to (-3,1,1). Why is this? By parallel do you just mean the span of (-3,1,1)? – Jim_CS Apr 22 '12 at 12:28
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) – chemeng Apr 22 '12 at 12:37

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