"Sketch these three lines and decide if the equations are solvable: $$x+2y=2$$ $$x-y=2$$ $$y=1$$

What happens if all right-hand sides are zero? Is there any nonzero choice of right-hand sides that allows the three lines to intersect at the same point?"

I just started learning Linear Algebra and I am having a hard time solving the Problem Set 1.2 of Gilbert Strang's "Linear Algebra and Its Applications".

Please give some advice on solving problem in Linear Algebra. May be a step by step process through which I can attack the problems.

I have some questions: 1. Do we really need to visualize planes in 2d or 3d when solving equations? 2. This Problem Sets contains problems which are not direct and I don't really know if I have solved the problem correctly.

And a general question, Is there any specific ability needed to solve problems? coz I am really having a hard time!!!

Thanks in advance.

  • $\begingroup$ For this one, have you sketched the lines? $\endgroup$ – Christopher Carl Heckman Apr 17 '16 at 8:06
  • $\begingroup$ Hint: the first two equations imply that $y=0$. $\endgroup$ – Bungo Apr 17 '16 at 8:08
  • $\begingroup$ @CarlHeckman I sketched these lines and they did not intersect at one point. I had two parallel lines and one line[text]x+2y=2[/tex] which cuts the other two equations. The actual problem is I was not sure if I am sovling the equations in the right way.... $\endgroup$ – Rajan Apr 18 '16 at 1:53

You have that $y=1$, hence $x-y=2$ becomes $x-1=2$, thus $x=3$. But then $x+2y=5$, contradicting the first equation. Hence there is no solution to this system of equations. When you draw these lines, you will see that not all three lines intersect in one point. If the right hand side were zero, all three lines would pass the origin, hence there is a solution. You don't need to visualize these lines and such to find solutions to a system of equations, however in some cases it can be useful to understand what you are doing. When you learn the Gaussian algorithm for solving such systems, it might be useful to draw the system in each step to see what the algorithm actually does.

  • $\begingroup$ Thanks for the answer. It really helped me. $\endgroup$ – Rajan Apr 18 '16 at 1:49

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