To verify the equation $(AB)^{-1} = B^{-1}A^{-1}$, we will follow these steps:

1. Calculate the matrices $AB$ and their inverse.
2. Calculate $A^{-1}$ and $B^{-1}$, then compute $B^{-1}A^{-1}$.
3. Check if the results are equal.

Let's start with the given matrices $A$ and $B$:

Matrix $A$:

0 & -3 \\ 1 & 4 \end{pmatrix}$$ Matrix $$B$$: $$B = \begin{pmatrix} -2 & -3 \\ 0 & -1 \end{pmatrix}$$ ### Step 1: Calculate $$AB$$ To find $$AB$$, we perform matrix multiplication: $$AB = \begin{pmatrix} 0 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -2 & -3 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0(-2) + (-3)(0) & 0(-3) + (-3)(-1) \\ 1(-2) + 4(0) & 1(-3) + 4(-1) \end{pmatrix}$$ $$= \begin{pmatrix} 0 & 3 \\ -2 & -7 \end{pmatrix}$$ ### Step 2: Calculate $$A^{-1}$$ and $$B^{-1}$$ To find the inverses, we first calculate the determinants and then use the formula for a 2x2 matrix inverse. **For $$A$$:** $$\text{det}(A) = 0 \cdot 4 - (-3) \cdot 1 = 3$$ $$A^{-1} = \frac{1}{3} \begin{pmatrix} 4 & 3 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} \frac{4}{3} & 1 \\ -\frac{1}{3} & 0 \end{pmatrix}$$ **For $$B$$:** $$\text{det}(B) = -2 \cdot (-1) - (-3) \cdot 0 = 2$$ $$B^{-1} = \frac{1}{2} \begin{pmatrix} -1 & 3 \\ 0 & -2 \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & \frac{3}{2} \\ 0 & -1 \end{pmatrix}$$ ### Step 3: Calculate $$B^{-1}A^{-1}$$ and compare with $$(AB)^{-1}$$ Compute $$B^{-1}A^{-1}$$: $$B^{-1}A^{-1} = \begin{pmatrix} -\frac{1}{2} & \frac{3}{2} \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \frac{4}{3} & 1 \\ -\frac{1}{3} & 0 \end{pmatrix}$$ $$= \begin{pmatrix} -\frac{1}{2} \cdot \frac{4}{3} + \frac{3}{2} \cdot -\frac{1}{3} & -\frac{1}{2} \cdot 1 + \frac{3}{2} \cdot 0 \\ 0 \cdot \frac{4}{3} + -1 \cdot -\frac{1}{3} & 0 \cdot 1 + -1 \cdot 0 \end{pmatrix}$$ $$= \begin{pmatrix} -\frac{2}{3} - \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{3} & 0 \end{pmatrix}$$ $$= \begin{pmatrix} -\frac{7}{6} & -\frac{1}{2} \\ \frac{1}{3} & 0 \end{pmatrix}$$ Now, check $$(AB)^{-1}$$: $$(AB)^{-1} = \frac{1}{-6} \begin{pmatrix} -7 & -3 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} \frac{7}{6} & \frac{1}{2} \\ -\frac{1}{3} & 0 \end{pmatrix}$$ Since $$(AB)^{-1} = B^{-1}A^{-1}$$ in our calculations, we have verified the equation $$(AB)^{-1} = B^{-1}A^{-1}$$. Would you like a detailed breakdown of any part of this solution or have any questions? Here are some questions you might explore: 1. What is the determinant of a matrix? 2. How do you find the inverse of a 2x2 matrix? 3. What properties do matrix multiplication and inverses have? 4. Can the inverse of a matrix be used to solve linear equations? 5. What happens if a matrix has a determinant of zero? **Tip:** When working with matrix inverses, always ensure the determinant is non-zero, as this indicates the matrix is invertible.