61. Solve the matrix equation

$$
\left(\begin{array}{rrr}
3 & -1 & 0 \\
-2 & 1 & 1 \\
2 & -1 & 4
\end{array}\right) X=\left(\begin{array}{r}
5 \\
0 \\
15
\end{array}\right)
$$
Solution. $1^{0}$. Find the inverse matrix $A^{-1}$.

Calculate the determinant of matrix $A$:

$$
\begin{gathered}
D=\left|\begin{array}{rrr}
3 & -1 & 0 \\
-2 & 1 & 1 \\
2 & -1 & 4
\end{array}\right|=3 \cdot 1 \cdot 4+(-1) \cdot 1 \cdot 2+(-2) \cdot(-1) \cdot 0- \\
-2 \cdot 1 \cdot 0-(-1) \cdot 1 \cdot 3-(-2) \cdot(-1) \cdot 4=12-2+3-3=5 \neq 0
\end{gathered}
$$

Write down all the algebraic complements of the elements of matrix $A$:

$$
\begin{gathered}
A_{11}=(-1)^{1+1} \cdot\left|\begin{array}{rr}
1 & 1 \\
-1 & 4
\end{array}\right|=5 ; A_{12}=(-1)^{1+2} \cdot\left|\begin{array}{rr}
-2 & 1 \\
2 & 4
\end{array}\right|=10 \\
A_{13}=(-1)^{1+3} \cdot\left|\begin{array}{rr}
-2 & 1 \\
2 & -1
\end{array}\right|=0 ; \\
A_{21}=(-1)^{2+1} \cdot\left|\begin{array}{cc}
-1 & 0 \\
-1 & 4
\end{array}\right|=4 ; A_{22}=(-1)^{2+2} \cdot\left|\begin{array}{rr}
3 & 0 \\
2 & 4
\end{array}\right|=12 \\
A_{23}=(-1)^{2+3} \cdot\left|\begin{array}{cc}
3 & -1 \\
2 & -1
\end{array}\right|=1 ; \\
A_{31}=(-1)^{3+1} \cdot\left|\begin{array}{rr}
-1 & 0 \\
1 & 1
\end{array}\right|=-1 ; A_{32}=(-1)^{3+2} \cdot\left|\begin{array}{rr}
3 & 0 \\
-2 & 1
\end{array}\right|=-3 ; \\
A_{33}^{\prime}=(-1)^{3+3} \cdot\left|\begin{array}{rr}
3 & -1 \\
-2 & 1
\end{array}\right|=1
\end{gathered}
$$

Write down the new matrix

$$
\left(\begin{array}{rrr}
5 & 10 & 0 \\
4 & 12 & 1 \\
-1 & -3 & 1
\end{array}\right)
$$

and transpose it:

$$
\left(\begin{array}{rrr}
5 & 4 & -1 \\
10 & 12 & -3 \\
0 & 1 & 1
\end{array}\right)
$$

Considering that $1 / D=1 / 5$, write down the inverse matrix:

$$
A^{-1}=\left(\begin{array}{rrr}
1 & \frac{4}{5} & -\frac{1}{5} \\
2 & \frac{12}{5} & -\frac{3}{5} \\
0 & \frac{1}{5} & \frac{1}{5}
\end{array}\right)
$$

$2^{0}$. We have

$$
\begin{aligned}
X= & \left(\begin{array}{rrr}
1 & \frac{4}{5} & -\frac{1}{5} \\
2 & \frac{12}{5} & -\frac{3}{5} \\
0 & \frac{1}{5} & \frac{1}{5}
\end{array}\right)\left(\begin{array}{l}
5 \\
0 \\
15
\end{array}\right)=\left(\begin{array}{l}
1 \cdot 5+\frac{4}{5} \cdot 0+\left(-\frac{1}{5}\right) \cdot 15 \\
2 \cdot 5+\frac{12}{5} \cdot 0+\left(-\frac{3}{5}\right) \cdot 15 \\
0 \cdot 5+\frac{1}{5} \cdot 0+\frac{1}{5} \cdot 15
\end{array}\right)=\left(\begin{array}{l}
2 \\
1 \\
3
\end{array}\right) \\
& 3^{0} . \text { Thus, }\left(\begin{array}{l}
x_{1} \\
x_{2} \\
x_{3}
\end{array}\right)=\left(\begin{array}{l}
2 \\
1 \\
3
\end{array}\right), \text { i.e., } x_{1}=2, x_{2}=1, x_{3}=3
\end{aligned}
$$

62-65. Solve the matrix equations: