Question

Let \(A\) be an \(n\times n\) real matrix. Consider the following statements.

• A. \(P\) is correct and \(Q\) is NOT correct
• B. \(P\) is NOT correct and \(Q\) is correct
• C. Both \(P\) and \(Q\) are correct
• D. Neither \(P\) nor \(Q\) is correct

Hint:

For a skew-symmetric matrix \(A\), always remember \(X^TAX=0\) for real column vectors \(X\). Also, a skew-symmetric orthogonal matrix can exist only in even order.

Answer 1

The Correct Option is C

Step 1: Use the scalar trick for $P$.
For a column $X$, $S=X^TAX$ is a single number, so $S^T=S$.

Step 2: Flip it with skew symmetry.
But $S^T=X^TA^TX=-X^TAX=-S$. So $S=-S$, forcing $S=0$. Thus $X^TAX=0$ always, and $P$ is correct.

Step 3: Use determinants for $Q$.
From $A^T=-A$, taking determinants gives $\det A=(-1)^n\det A$.

Step 4: Use orthogonality.
Since $A$ is orthogonal, $(\det A)^2=1$, so $\det A\neq0$. Cancelling it leaves $1=(-1)^n$, which needs $n$ even. So such a matrix exists only in even order, and $Q$ is correct.

Step 5: Conclude.
Both hold, option (C).
\[ \boxed{(C)} \]