Find inverse of A=\(\begin{bmatrix}5&3\\4&1\end{bmatrix}\).

Consider the matrix A=\(\begin{bmatrix}5&3\\4&1\end{bmatrix}\)
By using the elementary row operations, we write A=IA
\(\begin{bmatrix}5&3\\4&1\end{bmatrix}\)=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)A
Applying R₁→R₁-3R₂ and R₁→R₁/(-7), we get
\(\begin{bmatrix}1&0\\4&1\end{bmatrix}\)=\(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\0&1\end{bmatrix}\)
Applying R₂→R₂-4R₁, we get
\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=\(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\)A
?A^-1=\(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\).