Question

If matrices \( A \) and \( B \) are such that \[ A = \begin{bmatrix} 0 & -2 & 3 \\ -2 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix} \] and \( B(I - A) = (I + A) \), then find \( B \).

• A. \[ B = \begin{bmatrix} -1 & \tfrac{2}{3} & \tfrac{2}{3} \\ -2 & \tfrac{5}{3} & -\tfrac{10}{3} \\ -2 & 2 & -3 \end{bmatrix} \]
• B. \[ B = \begin{bmatrix} -1 & \tfrac{1}{3} & \tfrac{1}{3} \\ -2 & \tfrac{5}{3} & -\tfrac{10}{3} \\ -2 & 2 & -3 \end{bmatrix} \]
• C. \[ B = \begin{bmatrix} -1 & 0 & \tfrac{2}{3} \\ 0 & \tfrac{5}{3} & -\tfrac{10}{3} \\ 2 & 2 & -3 \end{bmatrix} \]
• D. \[ B = \begin{bmatrix} -1 & \tfrac{2}{3} & \tfrac{2}{3} \\ -\tfrac{2}{3} & 1 & \tfrac{5}{3} \\ -2 & 2 & -3 \end{bmatrix} \]

Hint:

To solve matrix equations involving inverses, first isolate the required matrix and then compute inverses carefully.

Answer 1

The Correct Option is A

Step 1: Rewrite the given equation.
Given \[ B(I - A) = (I + A) \] Multiplying both sides by \( (I - A)^{-1} \), we get \[ B = (I + A)(I - A)^{-1} \]
Step 2: Compute \( I - A \) and \( I + A \).
\[ I - A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix}, \quad I + A = \begin{bmatrix} 1 & -2 & 3 \\ -2 & 1 & 1 \\ -1 & 1 & 1 \end{bmatrix} \]
Step 3: Find \( (I - A)^{-1 \).}
By evaluating the inverse of \( I - A \) using elementary row operations, we obtain \( (I - A)^{-1} \).
Step 4: Final multiplication.
Multiplying \( (I + A) \) with \( (I - A)^{-1} \), we get \[ B = \begin{bmatrix} -1 & \tfrac{2}{3} & \tfrac{2}{3} \\ -2 & \tfrac{5}{3} & -\tfrac{10}{3} \\ -2 & 2 & -3 \end{bmatrix} \]