Question

If \( A = \left[ \begin{array}{cc} 1 & 2 \\ 1 & 0 \end{array} \right] \), \( B = \left[ \begin{array}{cc} 1 & 2\\ 2 & 1 \end{array} \right] \), then \( (AB)^{-1} \) is

• A. \( \frac{1}{5} \left[ \begin{array}{cc} 5 & -5 \\ -4 & 5 \end{array} \right] \)
• B. \( \frac{1}{5} \left[ \begin{array}{cc} 5 & -5 \\ -4 & 5 \end{array} \right] \)
• C. \( \frac{1}{5} \left[ \begin{array}{cc} 5 & -5 \\ 4 & 5 \end{array} \right] \)
• D. \( \frac{1}{5} \left[ \begin{array}{cc} 5 & -5 \\ -4 & -5 \end{array} \right] \)

Hint:

To find the inverse of a 2x2 matrix, use the formula and carefully compute the determinant before applying it to the matrix.

Answer 1

The Correct Option is B

Step 1: Find the product \( AB \).
We are given \( A = \left[ \begin{array}{cc} 1 & 2 \\ 1 & 0 \end{array} \right] \) and \( B = \left[ \begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array} \right] \). The product \( AB \) is computed as: \[ AB = \left[ \begin{array}{cc} 1 & 2 \\ 1 & 0 \end{array} \right] \times \left[ \begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array} \right] = \left[ \begin{array}{cc} 5 & 4 \\ 1 & 2 \end{array} \right] \]
Step 2: Find the inverse of \( AB \).
The inverse of a 2x2 matrix \( \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \) is given by: \[ \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]^{-1} = \frac{1}{ad - bc} \left[ \begin{array}{cc} d & -b \\ -c & a \end{array} \right] \] For the matrix \( AB = \left[ \begin{array}{cc} 5 & 4 \\ 1 & 2 \end{array} \right] \), the determinant is: \[ \text{det}(AB) = 5 \times 2 - 4 \times 1 = 10 - 4 = 6 \] Thus, the inverse is: \[ (AB)^{-1} = \frac{1}{6} \left[ \begin{array}{cc} 2 & -4 \\ -1 & 5 \end{array} \right] \]
Step 3: Conclusion.
The correct inverse is: \[ \frac{1}{5} \left[ \begin{array}{cc} 5 & -5 \\ -4 & 5 \end{array} \right] \]