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Consider the following linear system $A\vec{x}=\vec{b}$ where,

$A=\begin{bmatrix} 2&1&4\\ 2&-3&4\\ 3&-2&6 \end{bmatrix}$ and $\vec{b}=\begin{bmatrix} b\\ b\\ b \end{bmatrix}$ for a real number b.

Determine the values of b where so that $A\vec{x}=\vec{b}$ has at least 1 solution and solve the linear system.

I have taken the augmented matrix:

$\begin{bmatrix} 2&1&4&b\\ 2&-3&4&b\\ 3&-2&6&b \end{bmatrix}$ and reduced it to $\Rightarrow$ $\begin{bmatrix} 1&0&2&\frac{3}{7}b\\ 0&1&0&\frac{1}{7}b\\ 0&0&0&\frac{4}{7}b \end{bmatrix}$

From here, do I conclude the that this system will only have a solution for those values of b such that $b=0$?

What else do I have to do to solve the linear system?

Edit this is what i got for my row reduction

enter image description here

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    I don't think that Gauss-Jordan reduction is correct .Check it once? – 2017-02-23
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    I am not sure where my error is. – 2017-02-23
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    In the sixth passage, you do (if I read correctly from the sheet) $\frac{5}{7}\cdot R_2+R_3\to R_3$. Where the $frac{5}{7}$ comes from? – 2017-02-23

1 Answers 1

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Essentially the question is to determine the range of the matrix, that is the space spanned by it's columns. Basically what we do here is to pick a solution candidate $x$ and see what value of $b$ would have that solution.

We can immediately note that the third column is double the first so that vector is redundant in the base. We're then left with the first and second which are linearily independent.

So the values of $b$ for which the equation system has at least one solution is $b = (2s+t, 2s-3t, 3s-2t)$.