53. Find the matrix inverse of the matrix

$$
A=\left(\begin{array}{rrr}
1 & 2 & 3 \\
0 & -1 & 2 \\
3 & 0 & 7
\end{array}\right)
$$
Solution. $1^{0}$. Find the determinant of matrix $A$:

$$
\begin{aligned}
& D=\left|\begin{array}{rrr}
1 & 2 & 3 \\
0 & -1 & 2 \\
3 & 0 & 7
\end{array}\right|=1 \cdot(-1) \cdot 7+2 \cdot 2 \cdot 3+0 \cdot 0 \cdot 3-3 \cdot(-1) \cdot 3- \\
& -2 \cdot 0 \cdot 7-1 \cdot 2 \cdot 0=-7+12+9=14
\end{aligned}
$$

Since $D \neq 0$, matrix $A$ is non-singular and, therefore, we can find the matrix $A^{-1}$.

$2^{0}$. Find the algebraic complements of all elements of matrix $A$:

$$
\begin{gathered}
A_{11}=(-1)^{1+1} \cdot\left|\begin{array}{rr}
-1 & 2 \\
0 & 7
\end{array}\right|=-7 ; \quad A_{12}=(-1)^{1+2} \cdot\left|\begin{array}{ll}
0 & 2 \\
3 & 7
\end{array}\right|=6 \\
A_{13}=(-1)^{+3} \cdot\left|\begin{array}{cc}
0 & -1 \\
3 & 0
\end{array}\right|=3 ; \\
A_{21}=(-1)^{2+1} \cdot\left|\begin{array}{ll}
2 & 3 \\
0 & 7
\end{array}\right|=-14 ; A_{22}=(-1)^{2+2} \cdot\left|\begin{array}{ll}
1 & 3 \\
3 & 7
\end{array}\right|=-2 ; \\
A_{23}=(-1)^{2+3}\left|\begin{array}{cc}
1 & 2 \\
3 & 0
\end{array}\right|=6 ; \\
A_{31}=(-1)^{3+1} \cdot\left|\begin{array}{rr}
2 & 3 \\
-1 & 2
\end{array}\right|=7 ; \quad A_{32}=(-1)^{3+2} \cdot\left|\begin{array}{ll}
1 & 3 \\
0 & 2
\end{array}\right|=-2 \\
A_{33}=(-1)^{3+3}\left|\begin{array}{rr}
1 & 2 \\
0 & -1
\end{array}\right|=-1 .
\end{gathered}
$$

Write down the new matrix:

$$
\left(\begin{array}{rrr}
-7 & 6 & 3 \\
-14 & -2 & 6 \\
7 & -2 & -1
\end{array}\right)
$$

$3^{0}$. Transpose the obtained matrix:

$$
\left(\begin{array}{rrr}
-7 & -14 & 7 \\
6 & -2 & -2 \\
3 & 6 & -1
\end{array}\right)
$$