I'm working through an example and my answer is not coming out right. Two equations are given and then the solution is shown.
Equations:
$$\begin{aligned}20q_{1}+15q_{2}+7.5q_{3}&=10\\ q_{1}+q_{2}+q_{3}&=1\end{aligned}$$
Solution Given:
$(X, (1/3)(1-5X), (1/3)(2+2X))$ for arbitrary $X$.
I set up a matrix and solved and my solution is
$((3/2)X+1,(-5/2)X-2,X)$
What's going on here?
$\endgroup$ 13 Answers
$\begingroup$The matrix representing the system is $$(A|B) = \left(\begin{array}{ccc|c} 4 & 3 & 1.5 & 2\\ 1 & 1 & 1 & 1 \end{array}\right) \longrightarrow \left(\begin{array}{ccc|c} 4 & 3 & 1.5 & 2\\ -1.5 & -1.5 & -1.5 & -1.5 \end{array}\right) \longrightarrow \left(\begin{array}{ccc|c} 4 & 3 & 1.5 & 2\\ 2.5 & 1.5 & 0 & 0.5 \end{array}\right)$$
where we see that $$2.5q_1 + 1.5q_2 = 0.5 \implies q_2 = \frac13 - \frac53q_1.$$
Substituting into the first equation we get $$4q_1 + 1 - 5q_1 + 1.5q_3 = 2 \implies q_3 = \frac23 + \frac23q_1.$$
Now let $q_1 = X$ and the set of solutions is $$\left\{\left(X, \frac13 - \frac53X, \frac23 + \frac23X\right)\colon X \in \mathbb R\right\}.$$
$\endgroup$ $\begingroup$HINT: we write $$4q_1+3q_2+1.5q_3=2$$ $$q_1+q_2+q_3=1$$ multiplying the second equation by $$-1.5$$ and adding this to the first one we get $$2.5q_1+1.5q_2=0.5$$ or $$5q_1+3q_2=1$$ we get $$q_1=\frac{1}{5}-\frac{3}{5}q_2$$ setting $q_2=5x$ we get $q_1=1/5-3x$ and $$q_3=1+3x-1/5-5x=4/5-2x$$
$\endgroup$ $\begingroup$Did you set up the matrix equation correctly?
$$\pmatrix{20 & 15 &7.5\\ 1 & 1 & 1}\pmatrix{q_1 \\ q_2\\ q_3} = \pmatrix{10\\1}$$
When I solve it, I get
$q=\pmatrix{-1+\frac{3}{2}X\\2-\frac{5}{2}X\\X}$
Note that you can confirm you did your math wrong somewhere because in your answer $q_1+q_2+q_3 \neq 1$ for every $X$. It does for both mine and the solution (which are in fact the same).
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