Question

Match LIST-I with LIST-II

Choose the correct answer from the options given below:

• A. A-II, B-III, C-I, D-IV
• B. A-III, B-II, C-I, D-IV
• C. A-III, B-II, C-IV, D-I
• D. A-II, B-III, C-IV, D-I

Hint:

Determinant summary: - Real Orthogonal matrix $\implies \det A = \pm 1$. - Complex Unitary matrix $\implies |\det A| = 1$ (on the complex unit circle).

Answer 1

The Correct Option is D

Step 1 : Concept:
This matching question tests definitions of special matrices: skew-symmetric, orthogonal, and unitary matrices along with their determinant properties.

Step 2 : Key Formulas and Approach:
1. Skew-symmetric: $A^T = -A$. 2. Orthogonal: $A^T A = I \implies (\det A)^2 = 1 \implies \det A = \pm 1$. 3. Unitary: $A^\dagger A = I \implies |\det A|^2 = 1 \implies |\det A| = 1$.

Step 3 : Step-by-step Explanation:

• Item A:
Let $A = \begin{bmatrix} 0 & -1 & -2
1 & 0 & 5
2 & -5 & 0 \end{bmatrix}$. Taking transpose: \[ A^T = \begin{bmatrix} 0 & 1 & 2 -1 & 0 & -5 -2 & 5 & 0 \end{bmatrix} = -A \] Thus, $A$ is a skew-symmetric matrix. Matches with II.

• Item B:
Let $B = \begin{bmatrix} \cos\theta & 0 & -\sin\theta
0 & 1 & 0
\sin\theta & 0 & \cos\theta \end{bmatrix}$. Computing $B^T B$: \[ B^T B = \begin{bmatrix} \cos\theta & 0 & \sin\theta 0 & 1 & 0 -\sin\theta & 0 & \cos\theta \end{bmatrix} \begin{bmatrix} \cos\theta & 0 & -\sin\theta 0 & 1 & 0 \sin\theta & 0 & \cos\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{bmatrix} = I \] Thus, $B$ is an orthogonal matrix. Matches with III.

• Item C:
If $A$ is an $n \times n$ orthogonal matrix, $A^T A = I \implies \det(A^T A) = \det(I) \implies (\det A)^2 = 1 \implies \det A = \pm 1$. Matches with IV.

• Item D:
If $A$ is an $n \times n$ unitary matrix, $A^\dagger A = I \implies \det(A^\dagger A) = 1 \implies \overline{\det A} \cdot \det A = 1 \implies |\det A|^2 = 1 \implies |\det A| = 1$. Matches with I.

Step 4 : Final Answer:
The correct matching is A-II, B-III, C-IV, D-I, which corresponds to option (D).