The problem asks to show that matrices $A$ and $B$ are invertible and that they are inverses of each other. The given matrices are:

$A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}$

To show that $A$ and $B$ are inverses of each other, we need to verify that:

$A \cdot B = B \cdot A = I$

Where $I$ is the identity matrix:

$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Step 1: Compute $A \cdot B$

$A \cdot B = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \cdot \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}$

The multiplication is done as follows:

$A \cdot B = \begin{pmatrix} (2 \cdot 3 + 5 \cdot (-1)) & (2 \cdot (-5) + 5 \cdot 2) \\ (1 \cdot 3 + 3 \cdot (-1)) & (1 \cdot (-5) + 3 \cdot 2) \end{pmatrix}$

Simplifying each element:

$A \cdot B = \begin{pmatrix} (6 - 5) & (-10 + 10) \\ (3 - 3) & (-5 + 6) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

So, $A \cdot B = I$.

Step 2: Compute $B \cdot A$

$B \cdot A = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$

Similarly, the multiplication is:

$B \cdot A = \begin{pmatrix} (3 \cdot 2 + (-5) \cdot 1) & (3 \cdot 5 + (-5) \cdot 3) \\ (-1 \cdot 2 + 2 \cdot 1) & (-1 \cdot 5 + 2 \cdot 3) \end{pmatrix}$

Simplifying each element:

$B \cdot A = \begin{pmatrix} (6 - 5) & (15 - 15) \\ (-2 + 2) & (-5 + 6) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

So, $B \cdot A = I$.

Conclusion

Since $A \cdot B = I$ and $B \cdot A = I$, we have shown that $A$ and $B$ are inverses of each other. Therefore, both $A$ and $B$ are invertible matrices, and they are each other's inverses.

Would you like more details or explanations on any part of this?

Here are 5 related questions for further practice:

1. How do you determine if a matrix is invertible?
2. What is the general formula for the inverse of a 2x2 matrix?
3. Can a non-square matrix be invertible? Why or why not?
4. How does matrix multiplication differ from scalar multiplication?
5. Why is the identity matrix crucial in matrix operations?

Tip: The inverse of a matrix $A$, when it exists, satisfies $A \cdot A^{-1} = I$, where $I$ is the identity matrix.