Linear Algebra and Its Applications, exercise 1.3.4

Exercise 1.3.4. Given the following system of equations:

\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcr}u&+&v&+&w&=&-2   \\ 3u&+&3v&-&w&=&6 \\ u&-&v&+&w&=&-1 \end{array}

find a solution to the system, exchanging rows when necessary due to a zero pivot. Also, specify a coefficient for v in the third equation that would prevent elimination from being successful.

Answer: We can represent the system as the following matrix:

\left[ \begin{array}{rrrrr} 1&1&1&-2 \\ 3&3&-1&6 \\ 1&-1&1&-1 \end{array} \right]

The first pivot is 1. The first step in elimination is to subtract three times the first equation from the second, and the first equation from the third. This gives the following matrix:

\left[ \begin{array}{rrrrr} 1&1&1&-2 \\ 0&0&-4&12 \\ 0&-2&0&1 \end{array} \right]

Since we have a zero in the pivot position we exchange the second and third rows:

\left[ \begin{array}{rrrrr} 1&1&1&-2 \\ 0&-2&0&1 \\  0&0&-4&12 \end{array} \right]

Now that the system is in trangular form we re-express it using u, v, and w:

\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcrcr}u&+&v&+&w&=&-2    \\ &&-2v&&&=&1 \\   &&&&-4w&=&12 \end{array}

and then back-substitute, starting with solving for w:

\begin{array}{rcrcr}-4w = 12&\Rightarrow&w = -3 \\ -2v = 1&\Rightarrow&v = -\frac{1}{2} \\ u +  v + w = -2&\Rightarrow&u = -2 + \frac{1}{2} +3&\Rightarrow&u = \frac{3}{2}\end{array}

So the solution is u = 3/2, v = -1/2, w = 3 w = -3.

Note that when we exchanged rows the coefficient of v in the third row was -2; if it had been 0 at that point then we would not have been able to exchange rows. The coefficient of -2 was produced by subtracting v in the original first equation from -v in the original second equation. If instead of -v we had v in the original third equation then we would be subtracting v from v to get 0. So elimination would have been impossible if the coefficient of v in the third equation had been 1.

UPDATE: Corrected the statement of the solution for w in the second to last paragraph of the answer. Thanks to freaky4you for the fix!

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.

This entry was posted in linear algebra. Bookmark the permalink.

2 Responses to Linear Algebra and Its Applications, exercise 1.3.4

  1. freaky4you says:

    w= -3 , correction needed in solution

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