In each of the following problems, find the $A=L U$ factorization of the coefficient matrix, and then use Forward and Back Substitution to solve the corresponding linear systems $A \mathbf{x}=\mathbf{b}_j$ for each of the indicated right-hand sides:
(a) $A=\left(\begin{array}{rr}-1 & 3 \\ 3 & 2\end{array}\right), \mathbf{b}_1=\left(\begin{array}{r}1 \\ -1\end{array}\right), \mathbf{b}_2=\left(\begin{array}{l}2 \\ 5\end{array}\right), \mathbf{b}_3=\left(\begin{array}{l}0 \\ 3\end{array}\right)$.
(b) $A=\left(\begin{array}{rrr}-1 & 1 & -1 \\ 1 & 1 & 1 \\ -1 & 1 & 2\end{array}\right), \mathbf{b}_1=\left(\begin{array}{r}1 \\ -1 \\ 1\end{array}\right), \mathbf{b}_2=\left(\begin{array}{r}-3 \\ 0 \\ 2\end{array}\right)$.
(c) $A=\left(\begin{array}{rrr}9 & -2 & -1 \\ -6 & 1 & 1 \\ 2 & -1 & 0\end{array}\right), \mathbf{b}_1=\left(\begin{array}{r}2 \\ -1 \\ 0\end{array}\right), \mathbf{b}_2=\left(\begin{array}{l}1 \\ 2 \\ 5\end{array}\right)$.
(d) $A=\left(\begin{array}{rrr}2.0 & .3 & .4 \\ .3 & 4.0 & .5 \\ .4 & .5 & 6.0\end{array}\right), \mathbf{b}_1=\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right), \mathbf{b}_2=\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right), \mathbf{b}_3=\left(\begin{array}{l}0 \\ 0 \\ 1\end{array}\right)$.
(e) $A=\left(\begin{array}{rrrr}1 & 0 & -1 & 0 \\ 0 & 2 & 3 & -1 \\ -1 & 3 & 2 & 2 \\ 0 & -1 & 2 & 1\end{array}\right), \mathbf{b}_1=\left(\begin{array}{r}1 \\ 0 \\ -1 \\ 1\end{array}\right), \mathbf{b}_2=\left(\begin{array}{r}0 \\ -1 \\ 0 \\ 1\end{array}\right)$.
(f) $A=\left(\begin{array}{rrrr}1 & -2 & 0 & 2 \\ 4 & 1 & -1 & -1 \\ -8 & -1 & 2 & 1 \\ -4 & -1 & 1 & 2\end{array}\right), \mathbf{b}_1=\left(\begin{array}{l}1 \\ 0 \\ 0 \\ 0\end{array}\right), \mathbf{b}_2=\left(\begin{array}{r}3 \\ 0 \\ -1 \\ 2\end{array}\right), \quad \mathbf{b}_3=\left(\begin{array}{r}2 \\ 3 \\ -2 \\ 1\end{array}\right)$.