## Linear Algebra and Its Applications, exercise 1.3.1

Exercise 1.3.1. Solve the following equation using Gaussian elimination: $\setlength\arraycolsep{0.2em}\begin{array}{rrrrrrr}2u&-&3v&&&=&3 \\ 4u&-&5v&+&w&=&7 \\ 2u&-&v&-&3w&=&5 \end{array}$

Answer: The first pivot is 2 (the coefficient of u in the first equation). We multiply the first equation by 2 (the coefficient of u in the second equation divided by the pivot) and subtract it from the second. We also multiply the first equation by 1 (the coefficient of u in the third equation divided by the pivot) and subtract it from the third: $\setlength\arraycolsep{0.2em}\begin{array}{rrrrrrr}2u&-&3v&&&=&3 \\ 4u&-&5v&+&w&=&7 \\ 2u&-&v&-&3w&=&5 \end{array} \Rightarrow \begin{array}{rrrrrrr}2u&-&3v&&&=&3 \\ &&v&+&w&=&1 \\ &&2v&-&3w&=&2 \end{array}$

The second pivot is 1 (the coefficient of v in the second equation). We multiply the second equation by 2 (the coefficient of v in the third equation divided by the pivot) and subtract it from the third: $\setlength\arraycolsep{0.2em}\begin{array}{rrrrrrr}2u&-&3v&&&=&3 \\ &&v&+&w&=&1 \\ &&2v&-&3w&=&2 \end{array} \Rightarrow \begin{array}{rrrrrrr}2u&-&3v&&&=&3 \\ &&v&+&w&=&1 \\ &&&&-5w&=&0 \end{array}$

We can then back-substitute, starting with solving for w: $-5w = 0 \Rightarrow w = 0, v + w = 1 \Rightarrow v = 1, 2u - 3v = 3 \Rightarrow u = 3$

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

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.