To solve the problem, we need to analyze the two statements given about matrices A and B.
Given:
\[ A = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix} \]
### Step 1: Calculate \( AB \)
To find \( AB \), we perform matrix multiplication:
\[
AB = \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 1 \cdot 2 + (-1) \cdot (-1) = 2 + 1 = 3 \)
- First row, second column: \( 1 \cdot 3 + (-1) \cdot (-2) = 3 + 2 = 5 \)
- Second row, first column: \( 2 \cdot 2 + 3 \cdot (-1) = 4 - 3 = 1 \)
- Second row, second column: \( 2 \cdot 3 + 3 \cdot (-2) = 6 - 6 = 0 \)
Thus, we have:
\[
AB = \begin{pmatrix} 3 & 5 \\ 1 & 0 \end{pmatrix}
\]
### Step 2: Calculate \( A^{-1} \)
To find \( A^{-1} \), we need the determinant of \( A \):
\[
\text{det}(A) = (1)(3) - (-1)(2) = 3 + 2 = 5
\]
Now, the adjoint of \( A \) is obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements:
\[
\text{adj}(A) = \begin{pmatrix} 3 & 1 \\ -2 & 1 \end{pmatrix}
\]
Thus,
\[
A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{5} \begin{pmatrix} 3 & 1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} \frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix}
\]
### Step 3: Calculate \( B^{-1} \)
Next, we calculate \( B^{-1} \):
\[
\text{det}(B) = (2)(-2) - (3)(-1) = -4 + 3 = -1
\]
The adjoint of \( B \) is:
\[
\text{adj}(B) = \begin{pmatrix} -2 & -3 \\ 1 & 2 \end{pmatrix}
\]
Thus,
\[
B^{-1} = \frac{1}{\text{det}(B)} \cdot \text{adj}(B) = -1 \begin{pmatrix} -2 & -3 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix}
\]
### Step 4: Calculate \( A^{-1}B^{-1} \)
Now we find \( A^{-1}B^{-1} \):
\[
A^{-1}B^{-1} = \begin{pmatrix} \frac{3}{5} & \frac{1}{5} \\ -\frac{2}{5} & \frac{1}{5} \end{pmatrix} \begin{pmatrix} 2 & 3 \\ -1 & -2 \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( \frac{3}{5} \cdot 2 + \frac{1}{5} \cdot (-1) = \frac{6}{5} - \frac{1}{5} = \frac{5}{5} = 1 \)
- First row, second column: \( \frac{3}{5} \cdot 3 + \frac{1}{5} \cdot (-2) = \frac{9}{5} - \frac{2}{5} = \frac{7}{5} \)
- Second row, first column: \( -\frac{2}{5} \cdot 2 + \frac{1}{5} \cdot (-1) = -\frac{4}{5} - \frac{1}{5} = -1 \)
- Second row, second column: \( -\frac{2}{5} \cdot 3 + \frac{1}{5} \cdot (-2) = -\frac{6}{5} - \frac{2}{5} = -\frac{8}{5} \)
Thus,
\[
A^{-1}B^{-1} = \begin{pmatrix} 1 & \frac{7}{5} \\ -1 & -\frac{8}{5} \end{pmatrix}
\]
### Step 5: Check if \( AB(A^{-1}B^{-1}) \) is a unit matrix
Now we need to calculate \( AB(A^{-1}B^{-1}) \):
\[
AB(A^{-1}B^{-1}) = \begin{pmatrix} 3 & 5 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & \frac{7}{5} \\ -1 & -\frac{8}{5} \end{pmatrix}
\]
Calculating the elements:
- First row, first column: \( 3 \cdot 1 + 5 \cdot (-1) = 3 - 5 = -2 \)
- First row, second column: \( 3 \cdot \frac{7}{5} + 5 \cdot (-\frac{8}{5}) = \frac{21}{5} - \frac{40}{5} = -\frac{19}{5} \)
- Second row, first column: \( 1 \cdot 1 + 0 \cdot (-1) = 1 \)
- Second row, second column: \( 1 \cdot \frac{7}{5} + 0 \cdot (-\frac{8}{5}) = \frac{7}{5} \)
Thus,
\[
AB(A^{-1}B^{-1}) = \begin{pmatrix} -2 & -\frac{19}{5} \\ 1 & \frac{7}{5} \end{pmatrix}
\]
This is not a unit matrix, so the first statement is false.
### Step 6: Check if \( (AB)^{-1} = A^{-1}B^{-1} \)
We already calculated \( AB \) and now we need to find \( (AB)^{-1} \).
\[
\text{det}(AB) = -5 \quad \text{(as calculated earlier)}
\]
The adjoint of \( AB \):
\[
\text{adj}(AB) = \begin{pmatrix} 0 & -1 \\ -5 & 3 \end{pmatrix}
\]
Thus,
\[
(AB)^{-1} = \frac{1}{-5} \begin{pmatrix} 0 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 0 & \frac{1}{5} \\ 1 & -\frac{3}{5} \end{pmatrix}
\]
Now we compare \( (AB)^{-1} \) with \( A^{-1}B^{-1} \):
Since \( A^{-1}B^{-1} \) was calculated as:
\[
A^{-1}B^{-1} = \begin{pmatrix} 1 & \frac{7}{5} \\ -1 & -\frac{8}{5} \end{pmatrix}
\]
Clearly, \( (AB)^{-1} \neq A^{-1}B^{-1} \).
### Conclusion
Both statements are false.
### Final Answer
Neither 1 nor 2 is correct.
