52. Find the matrix inverse of the matrix

$$
A=\left(\begin{array}{rr}
2 & -1 \\
4 & 3
\end{array}\right)
$$
Solution. $1^{0}$. Find the determinant of matrix $A$:

$$
D=\left|\begin{array}{cc}
2 & -1 \\
4 & 3
\end{array}\right|=2 \cdot 3-(-1) \cdot 4=6+4=10
$$

Since $D \neq 0$, the matrix is non-singular and, therefore, an inverse matrix exists.

$2^{\circ}$. Find the algebraic complements of each element: $A_{11}=(-1)^{1+1} \cdot 3=3, \quad A_{12}=(-1)^{1+2} \cdot 4=-4, \quad A_{21}=(-1)^{2+1} \cdot$ $\cdot(-1)=1, \quad A_{22}=(-1)^{2+2} \cdot 2=2$. Then we get the matrix

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

$3^{0}$. Transpose this matrix:

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

$4^{0}$. Multiply the obtained matrix by $1 / D$, i.e., by $1 / 10$:

$$
A^{-1}=\frac{1}{10}\left(\begin{array}{rr}
3 & 1 \\
-4 & 2
\end{array}\right)=\left(\begin{array}{cc}
3 / 10 & 1 / 10 \\
-2 / 5 & 1 / 5
\end{array}\right)
$$

Check the obtained answer. By performing the multiplication $A A^{-1}$, we find

$$
\begin{gathered}
\left(\begin{array}{cc}
2 & -1 \\
4 & 3
\end{array}\right)\left(\begin{array}{cc}
\frac{3}{10} & \frac{1}{10} \\
-\frac{2}{5} & \frac{1}{5}
\end{array}\right)=\left(\begin{array}{ll}
2 \cdot \frac{3}{10}+(-1) \cdot\left(-\frac{2}{5}\right) & 2 \cdot \frac{1}{10}+(-1) \cdot \frac{1}{5} \\
4 \cdot \frac{3}{10}+3 \cdot\left(-\frac{2}{5}\right) & 4 \cdot \frac{1}{10}+3 \cdot \frac{1}{5}
\end{array}\right)= \\
=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=E
\end{gathered}
$$