# Can I further reduce this matrix?

Initial matrix was $$\begin{pmatrix}3& 7& 4\\2&5&2 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}= \begin{pmatrix}178\\122\end{pmatrix}$$

I managed to reduce it to $$\begin{pmatrix}1&0&6\\0&1&-2 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}= \begin{pmatrix}36\\10 \end{pmatrix}$$ when I try solving for the variables themselves I can't get a solid number, would this mean that one variable is dependent on the other? Forgive the simpleness of the question, first semester in linear.

Last question, just to make sure, did I set up the problem correctly? A potter is making cups, vases and plates. Each cup uses 3lb. of clay and 2 oz. of glaze, each vase uses 7lb. of clay and 5 oz. of glaze and each plate uses 4lb. of clay and 2 oz. of glaze. She has 178 lb. of clay and 122 oz. of glaze on hand. How many whole cups, vases and plates can she make if she uses all available clay and glaze?

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I guess you want a solution over $\mathbb{N}$, so with a bit of luck you only get one solution :) – Dominic Michaelis Feb 21 '13 at 6:52
Your interpretation looks right to me. Are you sure you reduced it right? If you multiply the original equation on the left by $\left(\begin{array}{cc} 5 & -7 \\ -2 & 3\end{array}\right)$, you get the reduced matrix you wrote, but you don't get $(36,10)^T$ on the right hand side. I'd start there. – Will Nelson Feb 21 '13 at 7:05
Ah, I see the problem. In the original equation, it should be $(178,122)^T$, not $(122,178)^T$. Please correct this. The reduced form you wrote is indeed correct, $(36,10)^T$. – Will Nelson Feb 21 '13 at 7:11
@WillNelson, I see, thank you for pointing that out! – Ceelos Feb 21 '13 at 7:15

You've reduced the problem nicely. Staring at your reduced form, you can see that all solutions have the form $$\left(\begin{array}{c} x \\ y\end{array}\right) = \left(\begin{array}{c} 36 - 6z \\ 10 + 2 z\end{array}\right).$$ Now find all integers $z\ge 0$ such that $x$ and $y$ are positive. For example, $z=0$ works. But so does $z=1$. So do a few more values of $z$.
Hint: First equation (after you reduced) depends on $x$ and $z$. Thus, you should be able to write $x$ in terms of $z$. Similarly for second equation, $y$ should be written in terms of $z$. Now, look for natural numbers as $z$ which will give positive $x$ and $y$.