Let $$A$$ and $$B$$ be real matrices of the form $$\begin{bmatrix}\alpha & 0 \\ 0 & \beta\end{bmatrix}$$ and $$\begin{bmatrix}0 & \gamma \\ \delta & 0\end{bmatrix}$$, respectively. Statement 1: $$AB - BA$$ is always an invertible matrix. Statement 2: $$AB - BA$$ is never an identity matrix.

We first compute $$AB$$ and $$BA.$$

$$AB=\begin{bmatrix}\alpha & 0 \\ 0 & \beta\end{bmatrix}\begin{bmatrix}0 & \gamma \\ \delta & 0\end{bmatrix}=\begin{bmatrix}0 & \alpha\gamma \\ \beta\delta & 0\end{bmatrix}.$$

Similarly,

$$BA=\begin{bmatrix}0 & \gamma \\ \delta & 0\end{bmatrix}\begin{bmatrix}\alpha & 0 \\ 0 & \beta\end{bmatrix}=\begin{bmatrix}0 & \beta\gamma \\ \alpha\delta & 0\end{bmatrix}.$$

Therefore,

$$AB-BA=\begin{bmatrix}0 & \gamma(\alpha-\beta) \\ \delta(\beta-\alpha) & 0\end{bmatrix}.$$

Equivalently,

$$AB-BA=(\alpha-\beta)\begin{bmatrix}0 & \gamma \\ -\delta & 0\end{bmatrix}.$$

To examine invertibility, compute the determinant:

$$\det(AB-BA)=0\cdot0-\gamma(\alpha-\beta)\delta(\beta-\alpha).$$

Since $$\beta-\alpha=-(\alpha-\beta),$$

$$\det(AB-BA)=\gamma\delta(\alpha-\beta)^2.$$

If $$\alpha=\beta,$$ then

$$AB-BA=\begin{bmatrix}0&0\\0&0\end{bmatrix},$$

which is not invertible.

Hence Statement 1 is false.

Now consider Statement 2.

The identity matrix is

$$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$$

But every matrix of the form

$$AB-BA=\begin{bmatrix}0 & \gamma(\alpha-\beta) \\ \delta(\beta-\alpha) & 0\end{bmatrix}$$

has diagonal entries equal to $$0.$$

Therefore $$AB-BA$$ can never be equal to $$I,$$ whose diagonal entries are $$1.$$

Hence Statement 2 is true.

Therefore, Statement 1 is false and Statement 2 is true.