Question

Let \(A = \left[ \begin{array}{ccc}cosα & -sinα & 0 \\ sinα & cosα & 0 \\ 0 & 0 & 1\end{array} \right]\). If \(B = \text{adj}\,A\), then the matrix \(B^{-1}\) is equal to...

• A. \(I\)
• B. \(A^{-1}\)
• C. \(-A\)
• D. \(A\)

Hint:

For a matrix with determinant 1, the adjoint equals the inverse.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a square matrix, \(A \cdot \text{adj}A = |A| I\). If \(|A| = 1\), then \(\text{adj}A = A^{-1}\).

Step 2: Key Formula or Approach:
Find \(|A|\) by expanding along the third row.

Step 3: Detailed Explanation:
\(|A| = 1 \times (\cos^2\alpha + \sin^2\alpha) = 1\).
So \(B = \text{adj}A = |A| A^{-1} = A^{-1}\).
Then \(B^{-1} = (A^{-1})^{-1} = A\).
Option A would need \(A = I\), option B would mean \(B^{-1} = B\), and option C has the wrong sign.

Final Answer:
\(B^{-1} = A\), option (D). \[ \boxed{A} \]