1.10. Using the method of the adjugate matrix, find $A^{-1}$ if

$$
A=\left(\begin{array}{lll}
2 & 5 & 6 \\
1 & 2 & 5 \\
1 & 2 & 3
\end{array}\right)
$$
Solution. First, we calculate the determinant of the given matrix to ensure the existence of the inverse matrix. We have

$$
\operatorname{det} A=\left|\begin{array}{ccc}
2 & 5 & 6 \\
1 & 2 & 5 \\
1 & 2 & 3
\end{array}\right|=\left|\begin{array}{lll}
2 & 5 & 6 \\
0 & 0 & 2 \\
1 & 2 & 3
\end{array}\right|=(-2) \cdot\left|\begin{array}{ll}
2 & 5 \\
1 & 2
\end{array}\right|=2
$$

Here, we added to the elements of the second row the elements of the third row, multiplied by (-1) beforehand, and then expanded the determinant along the second row. Since the determinant of the given matrix is not zero, the inverse matrix exists. To construct the adjoint matrix, we find the algebraic complements of the elements of the given matrix. We have

$$
\begin{array}{lll}
A_{11}=(-1)^{1+1}\left|\begin{array}{ll}
2 & 5 \\
2 & 3
\end{array}\right|=-4, & A_{12}=(-1)^{1+2}\left|\begin{array}{ll}
1 & 5 \\
1 & 3
\end{array}\right|=2, & A_{13}=(-1)^{1+3}\left|\begin{array}{ll}
1 & 2 \\
1 & 2
\end{array}\right|=0, \\
A_{21}=(-1)^{2+1}\left|\begin{array}{ll}
5 & 6 \\
2 & 3
\end{array}\right|=-3, & A_{22}=(-1)^{2+2}\left|\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right|=0, & A_{23}=(-1)^{2+3} \left\lvert\,\left[\left.\begin{array}{ll}
2 & 5 \\
1 & 2
\end{array} \right\rvert\,=1,\right.\right. \\
A_{31}=(-1)^{3+1}\left|\begin{array}{ll}
5 & 6 \\
2 & 5
\end{array}\right|=13, & A_{32}=(-1)^{3+2}\left|\begin{array}{ll}
2 & 6 \\
1 & 5
\end{array}\right|=-4, & A_{33}=(-1)^{3+3}\left|\begin{array}{ll}
2 & 5 \\
1 & 2
\end{array}\right|=-1
\end{array}
$$

In accordance with formula (1.15), we transpose the matrix $A^{*}:$

$$
A^{*}=\left(\begin{array}{ccc}
-4 & 2 & 0 \\
-3 & 0 & 1 \\
13 & -4 & -1
\end{array}\right),\left(A^{*}\right)^{\mathrm{T}}=\left(\begin{array}{ccc}
-4 & -3 & 13 \\
2 & 0 & -4 \\
0 & 1 & -1
\end{array}\right)
$$

Then, by formula (1.15), we have

$$
A^{-1}=\left(\begin{array}{ccc}
-2 & -3 / 2 & 13 / 2 \\
1 & 0 & -2 \\
0 & 1 / 2 & -1 / 2
\end{array}\right)
$$