If `A =[{:(1,2),(1,1):}]and B = [{:(0,-1),(1,2):}]`, then what is `B^(-1)A^(-1)` equal to ?

Options

A. `[{:(1,-3),(-1,2):}]`
B. `[{:(-1,3),(1,-2):}]`
C. `[{:(-1,3),(-1,-2):}]`
D. `[{:(-1,-3),(1,-2):}]`

Explanation

To find \( B^{-1} A^{-1} \), we can use the property of inverses which states that \( A^{-1}B^{-1} = (AB)^{-1} \). Therefore, we can first compute the product \( AB \) and then find its inverse.

### Step 1: Define the matrices
Given:
\[ A = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix} \]

### Step 2: Compute the product \( AB \)
To compute \( AB \):
\[ AB = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix} \]

Calculating the elements of the product:
- First row, first column: \( 1 \cdot 0 + 2 \cdot 1 = 2 \)
- First row, second column: \( 1 \cdot (-1) + 2 \cdot 2 = -1 + 4 = 3 \)
- Second row, first column: \( 1 \cdot 0 + 1 \cdot 1 = 1 \)
- Second row, second column: \( 1 \cdot (-1) + 1 \cdot 2 = -1 + 2 = 1 \)

Thus,
\[ AB = \begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} \]

### Step 3: Compute the determinant of \( AB \)
The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \).
For our matrix \( AB \):
\[ \text{det}(AB) = (2)(1) - (3)(1) = 2 - 3 = -1 \]

### Step 4: Compute the adjoint of \( AB \)
The adjoint of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \).
For our matrix \( AB \):
\[ \text{adj}(AB) = \begin{pmatrix} 1 & -3 \\ -1 & 2 \end{pmatrix} \]

### Step 5: Compute the inverse of \( AB \)
Using the formula for the inverse:
\[ (AB)^{-1} = \frac{1}{\text{det}(AB)} \cdot \text{adj}(AB) \]
Substituting the values we calculated:
\[ (AB)^{-1} = \frac{1}{-1} \begin{pmatrix} 1 & -3 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} -1 & 3 \\ 1 & -2 \end{pmatrix} \]

### Step 6: Conclude \( B^{-1} A^{-1} \)
Thus, we have:
\[ B^{-1} A^{-1} = (AB)^{-1} = \begin{pmatrix} -1 & 3 \\ 1 & -2 \end{pmatrix} \]

### Final Answer:
\[ B^{-1} A^{-1} = \begin{pmatrix} -1 & 3 \\ 1 & -2 \end{pmatrix} \]