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

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:

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