MATH152 April 2015
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Question B 1 (c)
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Uno, Duo and Traea are three friends. They all owe money to a loan shark. All together they owe $600. Duo owes $200 more than Uno. Uno and Duo combined owe as much as Traea.
- (c) Solve the system in part (a) using Gaussian elimination on the augmented matrix. How much does each person owe?
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Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
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Hint
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Apply Gaussian elimination: reduce the matrix to reduced row echelon form (wherein all the pivots are 1, and each pivot is the only nonzero entry in its column).
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Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
- If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
- If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
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Solution
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We perform Gaussian elimination to compute the reduced row echelon form of the augmented matrix:
![{\displaystyle {\begin{aligned}\left[{\begin{array}{ccc|c}1&1&1&600\\-1&1&0&200\\1&1&-1&0\end{array}}\right]&\xrightarrow {R_{3}-R_{1}} \left[{\begin{array}{ccc|c}1&1&1&600\\-1&1&0&200\\0&0&-2&-600\end{array}}\right]&\xrightarrow {R_{2}+R_{1}} \left[{\begin{array}{ccc|c}1&1&1&600\\0&2&1&800\\0&0&-2&-600\end{array}}\right]&\xrightarrow {{\frac {R_{2}}{2}},-{\frac {R_{3}}{2}}} \left[{\begin{array}{ccc|c}1&1&1&600\\0&1&1/2&400\\0&0&1&300\end{array}}\right]\\&\xrightarrow {R_{1}-R_{2}} \left[{\begin{array}{ccc|c}1&0&1/2&200\\0&1&1/2&400\\0&0&1&300\end{array}}\right]&\xrightarrow {R_{2}-{\frac {1}{2}}R_{3}} \left[{\begin{array}{ccc|c}1&0&1/2&200\\0&1&0&250\\0&0&1&300\end{array}}\right]&\xrightarrow {R_{1}-{\frac {1}{2}}R_{3}} \left[{\begin{array}{ccc|c}1&0&0&50\\0&1&0&250\\0&0&1&300\end{array}}\right].\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/fda571c0843add682d05a449076ee864d04b7377)
Thus 
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