To determine \( B^{-1} A^{-1} \), the inverses of matrices \( A \) and \( B \) must be calculated first. For a 2x2 matrix \( M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the inverse is \( M^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \), provided that \( ad-bc eq 0 \).
Step 1: Calculate \( A^{-1} \)
Given \( A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \).
The determinant of \( A \) is: \[ \text{det}(A) = (2)(-4) - (3)(1) = -8 - 3 = -11 \]
Therefore, \( A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 4 & 3 \\ 1 & -2 \end{bmatrix} \).
Step 2: Calculate \( B^{-1} \)
Given \( B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} \).
The determinant of \( B \) is: \[ \text{det}(B) = (1)(3) - (-2)(-1) = 3 - 2 = 1 \]
Thus, \( B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \).
Step 3: Compute \( B^{-1} A^{-1} \)
Multiply \( B^{-1} \) by \( A^{-1} \): \[ B^{-1} A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \frac{4}{11} & \frac{3}{11} \\ \frac{1}{11} & -\frac{2}{11} \end{bmatrix} \]
Performing matrix multiplication:
\[ B^{-1} A^{-1} = \begin{bmatrix} (3 \times \frac{4}{11}) + (2 \times \frac{1}{11}) & (3 \times \frac{3}{11}) + (2 \times -\frac{2}{11}) \\ (1 \times \frac{4}{11}) + (1 \times \frac{1}{11}) & (1 \times \frac{3}{11}) + (1 \times -\frac{2}{11}) \end{bmatrix} \]
\[ = \begin{bmatrix} \frac{12}{11} + \frac{2}{11} & \frac{9}{11} - \frac{4}{11} \\ \frac{4}{11} + \frac{1}{11} & \frac{3}{11} - \frac{2}{11} \end{bmatrix} = \begin{bmatrix} \frac{14}{11} & \frac{5}{11} \\ \frac{5}{11} & \frac{1}{11} \end{bmatrix} \]
This result simplifies to: \(\displaystyle \frac{1}{11} \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \).
The final answer is: \( \frac{1}{11} \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \).