In one of my assignments I have one exercise, where I should solve the equations with more than one unknown without performing a single calculation: E.g.: $$x + 15.4y = 12.5$$ or $$a + 4.15b + 7.25c + 22.01d = 2.54$$

My solution for the first one: $$x = 12.5 - 15.14y$$ That means that the equation has infinite number of solutions and the solution are all pairs (x, y).

But what kind of solution is expected from me? How should I write that the solution are all the pairs mentioned above mathematically? This task has appeared in the matrix chapter, so I guess it has to do something with it. Thanks for help.

  • $\begingroup$ What you have tried, mention that. $\endgroup$ – Anurag Nov 5 '16 at 12:02
  • $\begingroup$ If you want to know what is expected of you, you have to ask the person who set the assignment. It's her duty to set clear assignments. $\endgroup$ – Gerry Myerson Nov 5 '16 at 12:39

The question is not very clear, but I'll hazard a guess:

For $x+15.4y=12.5$, the solutions can be written as such: $$\begin{bmatrix}x \\y \end{bmatrix} =\begin{bmatrix}-15.4 \\1 \end{bmatrix}k \ +\ \begin{bmatrix} 12.5\\ 0\end{bmatrix},\quad \text{for }k\in\mathbb{R}$$

For $a+4.15b+7.25c+22.01d=2.54$, the solutions can be written as such: $$\begin{bmatrix}a\\b\\c\\d \end{bmatrix} = \begin{bmatrix}-4.15\\1\\0\\0\end{bmatrix}s + \begin{bmatrix}-7.25\\0\\1\\0\end{bmatrix}t + \begin{bmatrix}-22.01\\0\\0\\1\end{bmatrix}u + \begin{bmatrix}2.54\\0\\0\\0\end{bmatrix}, \quad \text{for } s,t,u\in\mathbb{R} $$

Basically this expresses one of the variables in terms of the others, setting those as free parameters. The set of all the solutions is then obtained by substituting various values of the free parameters ($k,s,t,\text{or }u$). Hopefully this is what your instructor is looking for.

  • $\begingroup$ Shouldn't the numbers that stand before the unknowns be written before the [ ] brackets? E.g. in second example -4.15 would be before the [ ] and then inside those brackets at the place where -4.15 stood would be 0? $\endgroup$ – Chilcone Nov 5 '16 at 13:32
  • $\begingroup$ @Chilcone I'm afraid I'm not sure where you're coming from. Try writing out in component form, i.e. $a=-4.15s-7.25t-22.01u+2.54;\; b=s;\; c=t;\;d=u$, then it should be clear why solutions satisfying these parametric equations are also solutions to the original equation. $\endgroup$ – Troy Nov 5 '16 at 13:56

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