Review exercise 2.15. For each of the following, find a matrix for which

i) there is either no solution or one solution to depending on

ii) there are an infinite number of solutions to for all

iii) there is either no solution or an infinite number of solutions to depending on

iv) there is exactly one solution for for any

Answer: i) The following system of two equations in two unknowns

obviously has a solution no matter the value of . However if we add a third equation as follows

then the resulting system may or may not have a solution depending on the value of . In particular, in order for the system to have a solution we must have in which case the single solution is .

The system above corresponds to where

ii) In order for a system to have an infinite number of solutions we can specify more unknowns than there are equations, so that some variables are free variables that can take on any value. For example, if we start again with the system of two equations in two unknowns

we can add another unknown to obtain the following system of two equations in three unknowns:

This system has the general solution for any with being an arbitrary value. Since can take on any value the number of solutions to the system is thus infinite.

The system above corresponds to where

iii) The system of two equations in three unknowns in (ii) above has an infinite number of solutions. To allow for the possibility of having either an infinite number of solutions or no solution at all, we can follow the example of (i) above and add a third equation which might or might not be satisfiable depending on the value of (and in particular the value of ).

For example, consider the following system of equations:

From (ii) above we know that (where is an arbitrary constant) is a solution to the first two equations. For this to also be a solution to the third equation we must have

So the system has an infinite number of solutions if but has no solution otherwise.

The system above corresponds to where

iv) Consider taking the system of three equations in three unknowns from (iii) above and changing the third equation as follows:

From the third equation we have . From the second equation we have or . From the first equation we have or . So the system has the single solution .

This corresponds to where

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.