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.

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