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

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.