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 March 1st, 2017, 08:53 AM #1 Newbie   Joined: Mar 2017 From: Florida Posts: 1 Thanks: 0 putting a word problem into a matrix Help! I have a word problem that I need to solve using a matrix. I don't know how to put it into a matrix though. If I could get that, I would be able to solve the problem. The problem is this: A company produces computer chips, resistors, and transistors. Each computer chip requires 2 units of copper, 2 units of zinc, and 1 unit of glass. Each resistor requires 1 unit of copper, 3 units of zinc, and 2 units of glass. Each transistor requires 3 units of copper, 2 units of zinc, and 2 units of glass. There are 70 units of copper, 80 units of zinc, and 55 units of glass available for use. Find the numbers of computer chips, resistors, and transistors the company can produce. I found the answer in the book and it said that 15 computer chips, 10 resistors, and 10 transistors were made. If someone could tell me how to put this into a matrix, that would be great. Last edited by skipjack; March 1st, 2017 at 09:00 AM.
 March 1st, 2017, 09:08 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198 $\begin{pmatrix}copper \\ zinc \\ glass \end{pmatrix} = \begin{pmatrix}2 &1 &3 \\ 2 &3 &2 \\1 &2 &2 \end{pmatrix} \begin{pmatrix}chip \\ resistor \\ transistor \end{pmatrix} \leq \begin{pmatrix} 70 \\ 80 \\ 55 \end{pmatrix}$ $\begin{pmatrix}chip \\ resistor \\ transistor \end{pmatrix} \leq \begin{pmatrix}2 &1 &3 \\ 2 &3 &2 \\1 &2 &2 \end{pmatrix}^{-1} \begin{pmatrix} 70 \\ 80 \\ 55 \end{pmatrix}= \begin{pmatrix}15 \\10\\10\end{pmatrix}$ Do you know how to find the inverse of that matrix? If not do you know how to form the augmented matrix and Gaussian reduce it?
 March 5th, 2017, 08:09 AM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 630 Thanks: 85 I spent time on it as a system of equations without matrices because I hate matrices. My solution may not be the fewest steps that could have been taken, but it matches the book's answers. a = amount of computer chips, b = amount of resistors, and c = amount of transistors. 1. 2a + b + 3c = 70 2. 2a + 3b + 2c = 80 3. a + 2b + 2c = 55 4. Line 1 + Line 3: 3a + 3b + 5c = 125 5. Line 4 - Line 2: a + 3c = 45 6. Line 1 - Line 5: a + b = 25 7. Line 6 multiplied by 3: 3a + 3b = 75 8. Line 4 - Line 7: 5c = 50 9. Line 8 divided by 5: c = 10 10. Line 9 multiplied by 2: 2c = 20 11. Line 10 substituted into Line 2: 2a + 3b + 20 = 80 12. Line 11 -20 on both sides: 2a + 3b = 60 13. Line 7 - Line 12: a = 15 14. Substitute Line 9 and Line 13 into Line 3: 15 + 2b + 20 = 55 15. Subtract constants from both sides of Line 14: 2b = 20 16. Line 15 divided by 2: b = 10
March 5th, 2017, 09:22 AM   #4
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Quote:
 Originally Posted by EvanJ I spent time on it as a system of equations without matrices because I hate matrices.
The way to deal with a math topic you aren't comfortable with is not to ignore it. It is to practice at it so hard that you are no longer uncomfortable with it.

March 7th, 2017, 03:29 PM   #5
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 Originally Posted by romsek The way to deal with a math topic you aren't comfortable with is not to ignore it. It is to practice at it so hard that you are no longer uncomfortable with it.
I'm not in school or doing anything that requires me to learn more math. If I have one way of doing a problem, I don't care about other ways.

March 7th, 2017, 04:28 PM   #6
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Quote:
 Originally Posted by EvanJ I'm not in school or doing anything that requires me to learn more math. If I have one way of doing a problem, I don't care about other ways.
So, you limit your knowledge of problem solving for the sake of utility and/or convenience ... makes me wonder why you even bother to frequent this site.

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