Solving for $x_{4}$ in the last equation yields
$$
x_{4}=\frac{6}{3}=2
$$
Using $x_{1}=2$ in the third equation, we obtain
$$
x_{3}=\frac{4-5(2)}{6}=-1
$$
Now $x_{3}=-1$ and $x_{4}=2$ are used to find $x_{2}$ in the second equation:
$$
x_{2}=\frac{-7-7(-1)+4(2)}{-2}=-4 .
$$
Finally, $x_{1}$ is obtained using the first equation:
$$
x_{1}=\frac{20+1(-4)-2(-1)-3(2)}{4}=3 .
$$
MATH0086 COURSE NOTES ：
Modify Program so that it will compute $A^{-1}$ by repeatedly solving $N$ linear systems
$$
A C_{J}=E_{J} \quad \text { for } J=1,2, \ldots, N
$$
Then and
$$
\begin{gathered}
A\left[\begin{array}{llll}
C_{1} & C_{2} & \ldots & C_{N}
\end{array}\right]=\left[\begin{array}{llll}
E_{1} & \boldsymbol{E}{2} & \ldots & \boldsymbol{E}{N}
\end{array}\right] \
A^{-l}=\left[\begin{array}{llll}
C_{1} & C_{2} & \ldots & C_{N}
\end{array}\right] .
\end{gathered}
$$
Make sure that you compute the $\boldsymbol{L} \boldsymbol{I}$ factorization only once!