2x2 Matrix Inverse Question Let $a, b, c, d\in \mathbb{R}$. I'm asked to determine the products of the ($2x2$) matrices

$$
\begin{bmatrix}
a&amp;b\\
c&amp;d\\
\end{bmatrix} and
\begin{bmatrix}
d&amp;-b\\
-c&amp;a\\
\end{bmatrix}
$$

and show that the first matrix is invertible if and only if $ad \neq bc$.

Calculating the product is pretty straightforward:

$$
\begin{bmatrix}
a&amp;b\\
c&amp;d\\
\end{bmatrix}
\begin{bmatrix}
d&amp;-b\\
-c&amp;a\\
\end{bmatrix}
= \begin{bmatrix}
ad-bc&amp;-ab+ba\\
cd-dc&amp;-bc+ad\\
\end{bmatrix}
$$

And I know that a matrix A is invertible if I can find a matrix B such that $AB=I$ or $BA = I$. Any hints on how to show that the first matrix is invertible if and only if $ad \neq bc$?
From your step we have

$$\begin{bmatrix}
ad-bc&amp;-ab+ba\\
cd-dc&amp;-bc+ad\\
\end{bmatrix}= \begin{bmatrix}
ad-bc&amp;0\\
0&amp;ad-bc\\
\end{bmatrix}$$

and we need $ad-bc\ne 0$.

Notably form here we can see that

$$\begin{bmatrix}
a&amp;b\\
c&amp;d\\
\end{bmatrix}^{-1}=\frac1{ad-bc}\begin{bmatrix}
d&amp;-b\\
-c&amp;a\\
\end{bmatrix}$$
You can use the fact that a square matrix is invertible if and only if it has a non-zero determinant. If ad = bc then it is easy to show matrix A will have determinant 0.