Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 1 (a)

MATH152 April 2015

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Question B 1 (a)

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. (a) Let $\mathbf {x} ={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}$ be the vector of unknowns, where $x_{1},x_{2}$ and $x_{3}$ is the amount of money Uno, Duo, and Traea owe, respectively. Describe the information above as a linear system in the form $A\mathbf {x} =\mathbf {b}$ (write $A$ and $\mathbf {b}$ with specific values).

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?

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!

Hint
Express the three statements (1) "All together they owe $600", (2) "Duo owes $200 more than Uno", and (3) "Uno and Duo combined owe as much as Traea" as equations in $x_{1}$, $x_{2}$, and $x_{3}$ first. For instance, (1) may be expressed as $x_{1}+x_{2}+x_{3}=600$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Note that $x_{1}$, $x_{2}$, and $x_{3}$ denote the amount of money Uno, Duo, and Traea owe, respectively. We first express the three statements (1) "All together they owe $600", (2) "Duo owes $200 more than Uno", and (3) "Uno and Duo combined owe as much as Traea" as linear equations in $x_{1}$, $x_{2}$, and $x_{3}$: $${\begin{cases}x_{1}+x_{2}+x_{3}=600&(1)\\x_{2}-x_{1}=200&(2)\\x_{1}+x_{2}=x_{3}&(3)\end{cases}}.$$ This system can be rewritten as $${\begin{cases}x_{1}+x_{2}+x_{3}=600\\-x_{1}+x_{2}+0x_{3}=200\\x_{1}+x_{2}-x_{3}=0\end{cases}}.$$ Then, the coefficient matrix $A$ can be formed by reading off the coefficients of each variable in each row. For the constant vector $\mathbf {b}$, we keep the same values: $${\begin{bmatrix}1&1&1\\-1&1&0\\1&1&-1\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}600\\200\\0\end{bmatrix}}.$$ Therefore, we find that $$\color {blue}A={\begin{bmatrix}1&1&1\\-1&1&0\\1&1&-1\end{bmatrix}},\quad \mathbf {b} ={\begin{bmatrix}600\\200\\0\end{bmatrix}}.$$ Note that the rows of the matrix $A$ and vector $\mathbf {b}$ may be freely permuted without affecting the system; thus, for instance, $$A={\begin{bmatrix}-1&1&0\\1&1&1\\1&1&-1\end{bmatrix}},\quad \mathbf {b} ={\begin{bmatrix}200\\600\\0\end{bmatrix}}.$$ is also a valid answer.