Linear Algebra and Its Applications, exercise 1.3.2

Exercise 1.3.2. Perform Gaussian elimination on the following system of equations:

\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}u&+&v&+&w&=&2 \\ u&+&3v&+&3w&=&0 \\ u&+&3v&+&5w&=&2 \end{array}

and find the resulting triangular system and the solution.

Answer: The first pivot is 1. We subtract the first equation from the second and the third:

\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}u&+&v&+&w&=&2  \\ u&+&3v&+&3w&=&0 \\  u&+&3v&+&5w&=&2 \end{array} \Rightarrow \begin{array}{rcrcrcl}u&+&v&+&w&=&2  \\ &&2v&+&2w&=&-2 \\ &&2v&+&4w&=&0 \end{array}

The second pivot is 2. We substract the second equation from the third, giving us the final triangular system:

\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}u&+&v&+&w&=&2  \\  &&2v&+&2w&=&-2 \\  &&2v&+&4w&=&0 \end{array} \Rightarrow \begin{array}{rcrcrcl}u&+&v&+&w&=&2  \\  &&2v&+&2w&=&-2 \\  &&&&2w&=&2 \end{array}

We can then back-substitute, starting with solving for w:

\begin{array}{l}2w = 2 \Rightarrow w = 1 \\ 2v + 2w = -2 \Rightarrow 2v + 2 = -2 \Rightarrow v = -2 \\ u + v + w = 2 \Rightarrow u -2 + 1 = 2 \Rightarrow u = 3\end{array}

So the solution is u = 3, v = -2, w = 0 w = 1.

UPDATE: Corrected statement of solution for w in last paragraph; 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 1.3.2

  1. freaky4you says:

    written wrongly in solution, w=1

  2. hecker says:

    Good catch, thanks! Fixed.

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