Chapter 1.4, Problem 36E

To determine

If linear system in three variables has no solution. Your friend concludes that it is not possible for two of the three equations to have any points in common. Is your friend correct? Explain your reasoning.

Answer to Problem 36E

My friend is correct.

Explanation of Solution

Given information: A linear system in three variable has no solution.

Calculation: The system of equations:

  $\begin{array}{l}-x+y-2z=-6(i)\\ 2x-2y+4z=6(ii)\\ x-y+2z=3(iii)\end{array}$

From first equation we find $x$ and put into $(i)\mathrm{and}(iii)$ :

  $\begin{array}{l}x=6+y-2z\\ 2(6+y-2z)-2y+4z=6\\ 12+2y-4z-2y+4z=6\\ 12=6\\ 6+y-2z-y+2z=3\\ 6=3\end{array}$

Therefore my friend is correct.