Question 8.6.7: Use the inverse of the coefficient matrix to solve the syste......
Use the inverse of the coefficient matrix to solve the system
\begin{aligned} 2 x_{1}+ \quad \quad \quad x_{3} & =2 \\ -2 x_{1}+3 x_{2}+4 x_{3} & =4 \\ -5 x_{1}+5 x_{2}+6 x_{3} & =-1 \end{aligned}
We found the inverse of the coefficient matrix
\mathrm{A}=\left(\begin{array}{rrr} 2 & 0 & 1 \\ -2 & 3 & 4 \\ -5 & 5 & 6 \end{array}\right)
in Example 4. Thus, (8) gives
\left(\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right)=\left(\begin{array}{rrr}2 & 0 & 1 \\-2 & 3 & 4 \\-5 & 5 & 6\end{array}\right)^{-1}\left(\begin{array}{r}2 \\4 \\-1\end{array}\right)=\left(\begin{array}{rrr}-2 & 5 & -3 \\-8 & 17 & -10 \\5 & -10 & 6\end{array}\right)\left(\begin{array}{r}2 \\4 \\-1\end{array}\right)=\left(\begin{array}{r}19 \\62 \\-36\end{array}\right) .
Consequently, x_{1}=19, x_{2}=62, and x_{3}=-36.