Exercise 1.3.3. Given the following system of equations:
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]](https://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Barray%7D%7Brrrrr%7D+2%26-1%260%260%260+%5C%5C+-1%262%26-1%260%260+%5C%5C+0%26-1%262%26-1%260+%5C%5C+0%260%26-1%262%265+%5Cend%7Barray%7D+%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
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]](https://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Barray%7D%7Brrrrr%7D+2%26-1%260%260%260+%5C%5C+0%26%5Cfrac%7B3%7D%7B2%7D%26-1%260%260+%5C%5C+0%26-1%262%26-1%260+%5C%5C++0%260%26-1%262%265+%5Cend%7Barray%7D+%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
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]](https://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Barray%7D%7Brrrrr%7D+2%26-1%260%260%260+%5C%5C++0%26%5Cfrac%7B3%7D%7B2%7D%26-1%260%260+%5C%5C+0%260%26%5Cfrac%7B4%7D%7B3%7D%26-1%260+%5C%5C+++0%260%26-1%262%265+%5Cend%7Barray%7D+%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
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]](https://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Barray%7D%7Brrrrr%7D++2%26-1%260%260%260+%5C%5C++0%26%5Cfrac%7B3%7D%7B2%7D%26-1%260%260+%5C%5C++0%260%26%5Cfrac%7B4%7D%7B3%7D%26-1%260+%5C%5C+++0%260%260%26%5Cfrac%7B5%7D%7B4%7D%265++%5Cend%7Barray%7D+%5Cright%5D&bg=ffffff&fg=333333&s=0&c=20201002)
Now that the system is in trangular form we re-express it using u, v, w, and z:
and then back-substitute, starting with solving for z:
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
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition
My Math Homework 