Exercise 1.2.11. Assume that we have the following system of two equations in two unknowns:

Under what circumstances would this system have a set of solutions constituting an entire line in the x-y plane?

Answer: Each of the equations corresponds to a line in the x-y plane. There are thus three possibilities:

- The lines intersect in a point. The point corresponds to a unique solution to the system of equations.
- The lines are parallel and do not intersect. There is no solution to the system of equations.
- The lines are identical, i.e., they form a single line. The set of solutions consists of all points on the line.

For this exercise we want case (3), i.e., the lines are identical. In order for the lines to be identical the two equations need to be identical or one needs to be equal to the other times some constant. In this case the equations can be made identical simply by setting a = 2. The resulting line is then described by the equation 2x + 2y = 0, or y = -x.

Another solution is found by setting a = -2. In that case the second equation is equal to -1 times the first. The resulting line is then described by the equation 2x – 2y = 0, or y = x.

UPDATE: Added the solution a = -2. Thanks to coruja38 for pointing this out.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

x= – 2 is also a solution.

a=-2 and not x=-2.

Fixed. Thanks!

This blog will be very useful to anyone interested in learning Linear Algebra.. Thanks for your valuable time and efforts !