Question 8.6.6: 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}-9 x_{2}=15 \\ & 3 x_{1}+6 x_{2}=16. \end{aligned}
The given system can be written as
\left(\begin{array}{rr} 2 & -9 \\ 3 & 6 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{l} 15 \\ 16 \end{array}\right).
Since \left|\begin{array}{rr}2 & -9 \\ 3 & 6\end{array}\right|=39 \neq 0, the coefficient matrix is nonsingular. Consequently, from (4) we get
A ^{-1}=\frac{1}{\operatorname{det} A }\left(\begin{array}{rr}a_{22} & -a_{12} \\-a_{21} & a_{11}\end{array}\right) . (4)
\left(\begin{array}{rr} 2 & -9 \\ 3 & 6 \end{array}\right)^{-1}=\frac{1}{39}\left(\begin{array}{rr} 6 & 9 \\ -3 & 2 \end{array}\right).
Using (8) it follows that
X = A ^{-1} B (8)
\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\frac{1}{39}\left(\begin{array}{rr} 6 & 9 \\ -3 & 2 \end{array}\right)\left(\begin{array}{l} 15 \\ 16 \end{array}\right)=\frac{1}{39}\left(\begin{array}{r} 234 \\ -13 \end{array}\right)=\left(\begin{array}{r} 6 \\ -\frac{1}{3} \end{array}\right),
and so x_{1}=6 and x_{2}=-\frac{1}{3}.