## 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 .

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### 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.