66. Solve the system of equations using the matrix method

$$
\left\{\begin{array}{cc}
x_{1}+2 x_{2} & =10 \\
3 x_{1}+2 x_{2}+x_{3} & =23 \\
x_{2}+2 x_{3} & =13
\end{array}\right.
$$
Solution. Let's form the matrix equation $A X=B$, where

$$
A=\left(\begin{array}{lll}
1 & 2 & 0 \\
3 & 2 & 1 \\
0 & 1 & 2
\end{array}\right), X=\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right), B=\left(\begin{array}{l}
10 \\
23 \\
13
\end{array}\right)
$$

and solve it using the specified method. We find

$$
D=\left|\begin{array}{lll}
1 & 2 & 0 \\
3 & 2 & 1 \\
0 & 1 & 2
\end{array}\right|=-9 \neq 0
$$

$$
\begin{aligned}
& A_{11}=3 ; A_{12}=-6 ; A_{13}=3 ; A_{21}=-4 ; A_{22}=2 ; A_{23}=-1 \\
& A_{31}=2 ; A_{32}=-1 ; A_{33}=-4
\end{aligned}
$$

Let's form the matrix

$$
\left(\begin{array}{rrr}
3 & -6 & 3 \\
-4 & 2 & -1 \\
2 & -1 & -4
\end{array}\right)
$$

and transpose it:

$$
\left(\begin{array}{rrr}
3 & -4 & 2 \\
-6 & 2 & -1 \\
3 & -1 & -4
\end{array}\right)
$$

Let's write the inverse matrix:

$$
A^{-1}=\left(\begin{array}{rrr}
-\frac{1}{3} & \frac{4}{9} & -\frac{2}{9} \\
\frac{2}{3} & -\frac{2}{9} & \frac{1}{9} \\
-\frac{1}{3} & \frac{1}{9} & \frac{4}{9}
\end{array}\right)
$$

Therefore,

$$
X=A^{-1} B=\left(\begin{array}{rrr}
-\frac{1}{3} & \frac{4}{9} & -\frac{2}{9} \\
\frac{2}{3} & -\frac{2}{9} & \frac{1}{9} \\
-\frac{1}{3} & \frac{1}{9} & \frac{4}{9}
\end{array}\right)\left(\begin{array}{l}
10 \\
23 \\
13
\end{array}\right)=\left(\begin{array}{l}
4 \\
3 \\
5
\end{array}\right)
$$

Thus, the solution to the system of equations is $x_{1}=4, x_{2}=3, x_{3}=5$. 67-70. Solve the systems of linear equations by the matrix method: