An n × n square matrix A satisfies AT = A-1. The determinant of this matrix may
take which of the following value(s)?
Options:
(A) +1
(B) -1
(C) n
(D) 0
Official Answer Key:
A, B
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.22)**
**1. Intuitive Physical Setup & Given Parameters**:
We are given an $n \times n$ real square matrix $A$ satisfying:
$A^T = A^{-1}$
This is the defining condition of an **orthogonal matrix** ($A^T A = I$).
We are asked to identify all possible values that the determinant $\det(A)$ can take (MSQ).
**2. Step-by-Step Mathematical Derivation**:
1. **Determinant Properties of Transpose and Inverse**:
- For any square matrix $A$:
$\det(A^T) = \det(A)$
- For any invertible matrix $A$:
$\det(A^{-1}) = \frac{1}{\det(A)}$
2. **Equating Determinants**:
- Taking the determinant on both sides of $A^T = A^{-1}$:
$\det(A^T) = \det(A^{-1})$
$\det(A) = \frac{1}{\det(A)}$
$[\det(A)]^2 = 1$
3. **Solving for Determinant**:
$\det(A) = \pm 1$
- Thus, the determinant can take the values **$+1$** (proper orthogonal matrix, e.g., pure rotations) or **$-1$** (improper orthogonal matrix, e.g., reflections).
**3. Comprehensive Options & Distractor Analysis**:
- **(A) +1**: ✅ **CORRECT**. Represents rotation matrices in $SO(n)$ where $\det(A) = +1$.
- **(B) -1**: ✅ **CORRECT**. Represents reflection/inversion matrices in $O(n) \setminus SO(n)$ where $\det(A) = -1$.
- **(C) n**: ❌ **INCORRECT**. The determinant of an orthogonal matrix cannot equal $n$ (unless $n = 1$ and $A = [1]$, but in general it cannot take the value $n$).
- **(D) 0**: ❌ **INCORRECT**. Since $A^{-1}$ exists, $A$ is non-singular, so $\det(A) \neq 0$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **A, B**
- **Exam Takeaway**: The determinant of any real orthogonal matrix is strictly constrained to $\det(A) \in \{+1, -1\}$ because orthogonal transformations preserve Euclidean lengths and volumes.
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