Linear Algebra and Its Applications, exercise 1.3.3

Exercise 1.3.3. Given the following system of equations:

\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcrcl}2u&-&v&&&&&=&0  \\ -u&+&2v&-&w&&&=&0 \\ &&-v&+&2w&-&z&=&0 \\ &&&&-w&+&2z&=&5 \end{array}

find a solution to the system, and give the pivots. You can use a matrix to represent the system (including the right-hand side).

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

\left[ \begin{array}{rrrrr} 2&-1&0&0&0 \\ -1&2&-1&0&0 \\ 0&-1&2&-1&0 \\ 0&0&-1&2&5 \end{array} \right]

The first pivot is 2. The first step in elimination gives the following matrix:

\left[ \begin{array}{rrrrr} 2&-1&0&0&0 \\ 0&\frac{3}{2}&-1&0&0 \\ 0&-1&2&-1&0 \\  0&0&-1&2&5 \end{array} \right]

The second pivot is 3/2. The second step in elimination gives the following:

\left[ \begin{array}{rrrrr} 2&-1&0&0&0 \\  0&\frac{3}{2}&-1&0&0 \\ 0&0&\frac{4}{3}&-1&0 \\   0&0&-1&2&5 \end{array} \right]

The third pivot is 4/3. The third step in elimination gives the following:

\left[ \begin{array}{rrrrr}  2&-1&0&0&0 \\  0&\frac{3}{2}&-1&0&0 \\  0&0&\frac{4}{3}&-1&0 \\   0&0&0&\frac{5}{4}&5  \end{array} \right]

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

\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcrcl}2u&-&v&&&&&=&0   \\ &&\frac{3}{2}v&-&w&&&=&0 \\  &&&&\frac{4}{3}w&-&z&=&0 \\  &&&&&&\frac{5}{4}z&=&5 \end{array}

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

\begin{array}{rcrcr}\frac{5}{4}z = 5&\Rightarrow&z = 4&& \\  \frac{4}{3}w - z = 0&\Rightarrow&\frac{4}{3}w = 4&\Rightarrow&w = 3 \\ \frac{3}{2}v - w = 0&\Rightarrow&\frac{3}{2}v = 3&\Rightarrow&v = 2 \\ 2u  - v = 0&\Rightarrow&2u = 2&\Rightarrow&u =  1\end{array}

So the solution is u = 1, v = 2, w = 3, z = 4.

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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