Question

If \( A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} \), then \( B^{-1} A^{-1} \) is equal to:

• A. \( -\frac{1}{11} \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \)
• B. \( \frac{1}{11} \begin{bmatrix} 15 & 11 \\ 1 & 0 \end{bmatrix} \)
• C. \( \frac{1}{11} \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \)
• D. \( -\frac{1}{11} \begin{bmatrix} 15 & 11 \\ 1 & 0 \end{bmatrix} \)

Answer 1

The Correct Option is C

To find \( B^{-1} A^{-1} \), we need to first find the inverses of matrices \( A \) and \( B \). The inverse of a 2x2 matrix \( M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by: \[ M^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \], provided that \( ad-bc \neq 0 \). 

Step 1: Find \( A^{-1} \)

Matrix \( A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix} \).

The determinant of \( A \) is: \[ \text{det}(A) = (2)(-4) - (3)(1) = -8 - 3 = -11 \]

\( A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 4 & 3 \\ 1 & -2 \end{bmatrix} \).

Step 2: Find \( B^{-1} \)

Matrix \( B = \begin{bmatrix} 1 & -2 \\ -1 & 3 \end{bmatrix} \).

The determinant of \( B \) is: \[ \text{det}(B) = (1)(3) - (-2)(-1) = 3 - 2 = 1 \]

\( B^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \).

Step 3: Calculate \( B^{-1} A^{-1} \)

Now multiply \( B^{-1} \) and \( A^{-1} \): \[ B^{-1} A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \frac{4}{11} & \frac{3}{11} \\ \frac{1}{11} & -\frac{2}{11} \end{bmatrix} \]

Using 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 simplifies to: \(\displaystyle \frac{1}{11} \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \).

The correct answer is: \( \frac{1}{11} \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix} \).

Answer 2

First, compute the inverses of A and B:

The determinant of A is:

\[ \text{det}(A) = (2)(-4) - (3)(1) = -8 - 3 = -11. \]

The inverse of A is:

\[ A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 4 & 3 \\ 1 & -2 \end{bmatrix}. \]

The determinant of B is:

\[ \text{det}(B) = (1)(3) - (-2)(-1) = 3 - 2 = 1. \]

The inverse of B is:

\[ B^{-1} = \frac{1}{1} \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}. \]

Now compute \( B^{-1}A^{-1} \):

\[ B^{-1}A^{-1} = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \cdot \frac{1}{11} \begin{bmatrix} 4 & 3 \\ 1 & -2 \end{bmatrix}. \]

Perform the matrix multiplication:

\[ \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 4 & 3 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} (3)(4) + (2)(1) & (3)(3) + (2)(-2) \\ (1)(4) + (1)(1) & (1)(3) + (1)(-2) \end{bmatrix}. \]

\[ = \begin{bmatrix} 12 + 2 & 9 - 4 \\ 4 + 1 & 3 - 2 \end{bmatrix} = \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix}. \]

Thus:

\[ B^{-1}A^{-1} = \frac{1}{11} \begin{bmatrix} 14 & 5 \\ 5 & 1 \end{bmatrix}. \]