Chapter 2.4, Problem 4E

Solve the system by finding the $\text{PA}=\text{LU}$ factorization and then carrying out the two-step back substitution.

a. $\left[\begin{array}{ccc}4& 2& 0\\ 4& 4& 2\\ 2& 2& 3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}2\\ 4\\ 6\end{array}\right]$

b. $\left[\begin{array}{ccc}-1& 0& 1\\ 2& 1& 1\\ -1& 2& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}-2\\ 17\\ 3\end{array}\right]$