An \(n \times n\) square matrix \(A\) satisfies \(A^{T} = A^{-1}\). The determinant of this matrix may take which of the following value(s)?
Show Hint
- \(+1\)
- \(-1\)
- \(n\)
- \(0\)
The Correct Option is A, B
Solution and Explanation
The matrix \(A\) satisfies \(A^{T} = A^{-1}\). A matrix with this property is called an orthogonal matrix. It means the rows (and columns) of \(A\) form an orthonormal set, so \(A^{T}A = AA^{T} = I\), where \(I\) is the identity matrix.
Step 2: Take the Determinant of Both Sides.
Starting from \(A^{T} = A^{-1}\), take the determinant of each side.
\[ \det(A^{T}) = \det(A^{-1}) \]
Use two standard determinant facts here: \(\det(A^{T}) = \det(A)\) for any square matrix, and \(\det(A^{-1}) = \dfrac{1}{\det(A)}\) for any invertible matrix.
Step 3: Solve for the Determinant.
Substituting these facts gives
\[ \det(A) = \frac{1}{\det(A)} \]
Multiply both sides by \(\det(A)\):
\[ [\det(A)]^{2} = 1 \]
So \(\det(A) = +1\) or \(\det(A) = -1\). Both values are possible: \(\det(A) = +1\) for a proper rotation (a matrix in \(SO(n)\)) and \(\det(A) = -1\) for an improper rotation that includes a reflection.
Step 4: Check the Other Options.
Option (C), \(n\), cannot be a general answer. The value \(n\) has no connection to the orthogonality condition, and it does not satisfy \([\det(A)]^{2}=1\) for a general \(n\). Option (D), \(0\), is also wrong: if \(\det(A) = 0\) then \(A\) would be singular and \(A^{-1}\) would not even exist, contradicting the given condition \(A^{T}=A^{-1}\).
Final Answer:
The determinant can only be \(+1\) or \(-1\). \[ \boxed{\det(A) = \pm 1 \implies \text{options (A) and (B)}} \]
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