Given the inverse matrices $A^{-1}$ and $B^{-1}$, we can use matrix properties to find $(AB)^{-1}$, $(A^T)^{-1}$, and $(2A)^{-1}$.

1. Finding $(AB)^{-1}$:

By the property of the inverse of a product of matrices, we know: $(AB)^{-1} = B^{-1}A^{-1}$ To find $(AB)^{-1}$, we multiply the given matrices $B^{-1}$ and $A^{-1}$.

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

Now, compute $B^{-1}A^{-1}$: $B^{-1}A^{-1} = \begin{pmatrix} 1 & 2 \\ -2 & -3 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix}$ Let's calculate this step by step.

2. Finding $(A^T)^{-1}$:

By the property of transposed inverses: $(A^T)^{-1} = (A^{-1})^T$ So, we just take the transpose of $A^{-1}$.

Given $A^{-1} = \begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix}$, the transpose is: $(A^{-1})^T = \begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix}$ Thus, $(A^T)^{-1} = \begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix}$.

3. Finding $(2A)^{-1}$:

By the property of scalar multiples of matrices, we know: $(2A)^{-1} = \frac{1}{2} A^{-1}$ Thus, multiply $A^{-1}$ by $\frac{1}{2}$: $(2A)^{-1} = \frac{1}{2} \begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -1 \\ -1 & \frac{5}{2} \end{pmatrix}$

Summary:

1. $(AB)^{-1} = B^{-1} A^{-1}$
2. $(A^T)^{-1} = (A^{-1})^T = \begin{pmatrix} 1 & -2 \\ -2 & 5 \end{pmatrix}$
3. $(2A)^{-1} = \begin{pmatrix} \frac{1}{2} & -1 \\ -1 & \frac{5}{2} \end{pmatrix}$

Would you like to go through the detailed calculation of $(AB)^{-1}$? Let me know if you have questions!

Here are some related questions you might consider:

1. How do you compute the inverse of a matrix?
2. What is the significance of the transpose in matrices?
3. How does the determinant affect the inverse of a matrix?
4. What is the property of the inverse of scalar multiples of matrices?
5. How do inverse matrices apply to solving systems of equations?

Tip: When multiplying matrices, pay attention to the order—matrix multiplication is not commutative!