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Solve the system of equations: $$ \begin{pmatrix} 0 & 1 & 0 & 3 \\ 0 & 2 & 0 & 6 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3\\ x_4 \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2\end{pmatrix}$$

What value of $b_2$ results in this system having a solution?

I did my row reduction and came up with:

$$\begin{pmatrix} 0 & 1 & 0 & 3 & b_1\\ 0 & 0 & 0 & 0 & 2b_1 - b_2\\ \end{pmatrix} $$

therefore, $2b_1=b_2$

Is it really that simple?

Any help is greatly appreciated!

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  • $\begingroup$ Please learn how to present your math in a format that we can read easily. You can refer to your previous question to see how to edit it. At the very least, use lines to space it out. $\endgroup$
    – Calvin Lin
    May 29 '13 at 2:44
  • $\begingroup$ You have only two equations in four unknowns, and the coefficient vectors are not even linearly independent. If $b_1 \ne 2b_2$ then there are no solutions (which you should see even without row reduction; row 2 is just twice row 1) as the system would be inconsistent but if $b_1 = 2b_2$ then you have the equation of a hyperplane of solutions. $\endgroup$
    – Zen
    May 29 '13 at 2:48
  • $\begingroup$ @Calvin~ I'm sorry! Zev sent me the links on how to format. I will screen shot his formatting and learn. $\endgroup$
    – Lanae
    May 29 '13 at 2:49
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    $\begingroup$ @Lanae I also believe that some values are missing. In the first line, you have 11 numbers on one side. I believe there should be 12? The lack of proper formatting makes it impossible for us to figure out what you are trying to say (esp when it's slightly wrong), and people won't be able to help you (even if they wanted to). $\endgroup$
    – Calvin Lin
    May 29 '13 at 2:52
  • $\begingroup$ @Calvin~ the formatting that Zev sent won't work for this system of equations. Ax=b. I have a 2x4 A matrix: [0,1,0,3//0,2,0,6], then the x vector contains [x_1, x_2, x_3, x_4]. The b vector contains [b_1, b_2]. Does that help? $\endgroup$
    – Lanae
    May 29 '13 at 3:26
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Your answer is correct and your reasoning is good (although I think most people would come up with $b_2-2b_1$ rather than $2b_1-b_2$ in the reduced row echelon form). However, in this particular case, there is a conceptually simpler solution (I emphasize "particular" because in general, your way of solving the problem is preferred). Note that $$ \underbrace{\begin{pmatrix}0&1&0&3\\ 0&2&0&6\end{pmatrix}}_{A} \begin{pmatrix}x_1 \\ x_2 \\ x_3\\ x_4\end{pmatrix} =x_1\begin{pmatrix}0\\ 0\end{pmatrix} +x_2\begin{pmatrix}1\\ 2\end{pmatrix} +x_3\begin{pmatrix}0\\ 0\end{pmatrix} +x_4\begin{pmatrix}3\\ 6\end{pmatrix} =(x_2+3x_4)\begin{pmatrix}1\\ 2\end{pmatrix}. $$ Therefore the system of linear equations is solvable if and only if $(b_1,b_2)^T$ is a scalar multiple of $(1,2)^T$, i.e. iff $b_2=2b_1$.

Using the language of linear algebra, $A\mathbf{x}=\mathbf{b}$ is solvable iff $\mathbf{b}$ lies in the column space of $A$. Now the column space of $A$ is the line spanned by $(1,2)^T$. Hence the result.

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  • $\begingroup$ ~ Thank you! I agree that my 2b_1-b_2 is opposite the norm. I'm learning that as I go. I've begun to correct that in my work. It has been at least 15 years since I first learned Linear Algebra and I'm taking a graduate course in it currently. :) $\endgroup$
    – Lanae
    Jun 3 '13 at 0:19

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