Exercise 1.3.6. Given the following system of equations:
for what values of would Gaussian elimination break down, either in a fixable way (i.e., via row exchange) or in a non-fixable way?
Answer: One way for elimination to break down temporarily would be for the coefficient of in the first equation to be 0, thereby forcing a row exchange as the very first step in elimination. This corresponds to the case and the following set of equations:
A second way for elimination to fail would be for one equation to be equal to the other equation times some constant c. This would correspond to the equations representing the same line in the u-v plane, with no unique solution (singular system). We now try to find a value of for which this situation would occur.
Note that we can multiply the right-hand side of the first equation by 2 to get the right hand side of the second equation. Multiplying the left-hand side of the first equation by 2 and setting it equal to the left-hand side of the second equation gives us:
So if then we have the following set of equations:
which after elimination produces the single equation
for which no unique solution exists.
Another approach is to assume that and to perform elimination. In order to eliminate the first term of the second equation we multiply the first equation by and subtract it from the second. This gives us
Elimination can fail at this point if the term involving is zero, which would be the case if . Multiplying this equation by on both sides gives us or with solutions and .
We’ve already considered the case . When the original equations become
which after elimination becomes
This system has no solution.
To sum up, there are three values of for which elimination could temporarily or permanently break down:
- When elimination temporarily breaks down and requires a row exchange.
- When elimination permanently breaks down and the system has an infinite number of solutions.
- When elimination permanently breaks down and the system has no solution.
UPDATE: Tried to clarify the explanation of the second way in which elimination could break down, and included a third way elimination could fail based on comments from coruja38.
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.