Chapter 2.2, Problem 4E

Solve the system by finding the LU factorization and then carrying out the two-step back substitution.

$\text{a}.\left[\begin{array}{ccc}3& 1& 2\\ 6& 3& 4\\ 3& 1& 5\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 1\\ 3\end{array}\right]$ $\text{b}.\left[\begin{array}{ccc}4& 2& 0\\ 4& 4& 2\\ 2& 2& 3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}2\\ 4\\ 6\end{array}\right]$