Answer
$A^{-1}=\begin{bmatrix}
\frac{7}{11} &\frac{-5}{11} \\
\frac{-2}{11} & \frac{3}{11}
\end{bmatrix}$
Work Step by Step
Given: $A=\begin{bmatrix}
3 & 5 \\
2 & 7
\end{bmatrix} \rightarrow adj(A)=\begin{bmatrix}
7 & -5 \\
-2 & 3
\end{bmatrix}$
$\det(A)=3.7-2.5=11$
Since $D=11 \ne0$, the matrix is invertible.
Use the adjoint method to find $A^{-1}:$
$A^{-1}=\frac{1}{\det(A)}adj(A)$
$A^{-1}=\frac{1}{11}\begin{bmatrix}
7 & -5 \\
-2 & 3
\end{bmatrix}$
$A^{-1}=\begin{bmatrix}
\frac{7}{11} &\frac{-5}{11} \\
\frac{-2}{11} & \frac{3}{11}
\end{bmatrix}$
