60. Solve the matrix equation

$$
\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right) x=\binom{7}{17}
$$
Solution. $1^{0}$. We will find the inverse matrix $A^{-1}$.

Let's find the determinant of matrix $A$:

$$
D=\left|\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right|=1 \cdot 4-2 \cdot 3=4-6=-2 \neq 0
$$

Calculate the algebraic complements of each element of matrix $A$:
$A_{11}=(-1)^{1+1} \cdot 4=4, A_{12}=(-1)^{1+2} \cdot 3=-3 ; A_{21}=(-1)^{2+1} \cdot 2=$ $=-2 ; A_{22}=(-1)^{2+2} \cdot 1=1$.

Write the matrix $\left(\begin{array}{rr}4 & -3 \\ -2 & 1\end{array}\right)$ and transpose it: $\left(\begin{array}{rr}4 & -2 \\ -3 & 1\end{array}\right)$. Considering that $1 / D=-1 / 2$, write the inverse matrix:

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

$2^{0}$. Multiply matrix $A^{-1}$ by matrix $B$:

$$
X=A^{-1} B=\left(\begin{array}{rr}
-2 & 1 \\
\frac{3}{2} & -\frac{1}{2}
\end{array}\right) \cdot\binom{7}{17}=\binom{(-2) \cdot 7+1 \cdot 17}{\frac{3}{2} \cdot 7+\left(-\frac{1}{2}\right) \cdot 17}=\binom{3}{2}
$$

$3^{0}$. Since $\binom{x_{1}}{x_{2}}=\binom{3}{2}$, by the definition of equal matrices, we get $x_{1}=3, x_{2}=2$.