Linear Algebra and Its Applications, exercise 1.3.6

Exercise 1.3.6. Given the following system of equations:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcr}au&+&v&=&1 \\ 4u&+&av&=&2 \end{array}$

for what values of $a$ 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 $u$ in the first equation to be 0, thereby forcing a row exchange as the very first step in elimination. This corresponds to the case $a = 0$ and the following set of equations:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcr}&&v&=&1 \\ 4u&&&=&2 \end{array}$

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 $a$ 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:

$2(au + v) = 4u + av \quad\rightarrow\quad 2au + 2v = 4u + av \quad\rightarrow\quad a = 2$

So if $a = 2$ then we have the following set of equations:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcr}2u&+&v&=&1 \\ 4u&+&2v&=&2 \end{array}$

which after elimination produces the single equation

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcr}2u&+&v&=&1\end{array}$

for which no unique solution exists.

Another approach is to assume that $a \ne 0$ and to perform elimination. In order to eliminate the first term of the second equation we multiply the first equation by $4/a$ and subtract it from the second. This gives us

$4u+av - 4/a(au+v) = 2 - (4/a)\cdot 1$

$\rightarrow 4u - 4u + av - (4/a)v = 2 - 4/a$

$\rightarrow (a - 4/a)v = 2 - 4/a$

Elimination can fail at this point if the term involving $v$ is zero, which would be the case if $a - 4/a = 0$ . Multiplying this equation by $a$ on both sides gives us $a^2 - 4 = 0$ or $a^2 = 4$ with solutions $a = 2$ and $a = -2$ .

We’ve already considered the case $a = 2$ . When $a = -2$ the original equations become

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcr}-2u&+&v&=&1 \\ 4u&-&2v&=&2 \end{array}$

which after elimination becomes

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcr}-2u&+&v&=&1 \\ &&0&=&4 \end{array}$

This system has no solution.

To sum up, there are three values of $a$ for which elimination could temporarily or permanently break down:

When $a = 0$ elimination temporarily breaks down and requires a row exchange.
When $a = 2$ elimination permanently breaks down and the system has an infinite number of solutions.
When $a = -2$ 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.

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5 Responses to Linear Algebra and Its Applications, exercise 1.3.6

freaky4you says:

November 11, 2012 at 12:14 pm

Doubt
——–
case 1: we get unique solutions u=1/2 and v=1, then how could u say that elimination breaks down ?
Case 2: Can u explain it with an example..?
case 3 : Agreeing that there is no solution , but how could you say that the elimination process get broke down ???

Plz explain … Including conditions at which the elimination process breaks down

- hecker says:
  
  November 11, 2012 at 3:39 pm
  
  In the first case, elimination breaks down temporarily, because we have to do a row exchange before we can do the next step in elimination.
  
  There are only two cases. The second case is where one equation is a multiple of the other equation. My original answer was confusing because I didn’t connect the paragraph where I referenced that possibility with the following paragraph where I tried to find a value of a for which this was the case. I’ve added a new sentence to provide a better transition between the two paragraphs.
  
- hecker says:
  
  November 11, 2012 at 3:43 pm
  
  I forgot to add: In the second case I say that the elimination process breaks down because elimination would produce the result 0 = 0 for the second equation.
  
coruja38 says:

January 23, 2013 at 1:10 am

There are 2 values of “a” such that the elimination breakdown in a permanent way. a =2 and a=-2. Note that the breakdown would be permanent if the result of the elimination is the elimination of both terms of the second equation, i.e. (assuming a!=0) if a – (4/a) = 0. Where 4/a is the multiplier. Solving this equation we have the result that if a = +/- 2 the breakdown would be permanent.

- hecker says:
  
  January 23, 2013 at 2:24 pm
  
  Thanks for pointing this out; you are (once again!) correct. I will correct the post to reflect your comment.