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: Recall the meaning of skew-symmetric matrix.
A real square matrix \(A\) is called skew-symmetric if
\[ A^T=-A \] We need to check both statements \(P\) and \(Q\).

Step 2: Check statement \(P\).
Let
\[ S=X^TAX \] Since \(X\) is an \(n\times 1\) column matrix, \(X^TAX\) is a \(1\times 1\) matrix, that is, a scalar.
So,
\[ S^T=S \] Now,
\[ S^T=(X^TAX)^T \] Using transpose property,
\[ (X^TAX)^T=X^TA^TX \] Since \(A\) is skew-symmetric,
\[ A^T=-A \] Therefore,
\[ S^T=X^T(-A)X \] \[ S^T=-X^TAX \] \[ S^T=-S \] But \(S^T=S\), because \(S\) is a scalar. Hence,
\[ S=-S \] \[ 2S=0 \] \[ S=0 \] Thus,
\[ X^TAX=0 \] for every \(n\times 1\) real matrix \(X\).
Therefore, statement \(P\) is correct.

Step 3: Check statement \(Q\).
Given that \(A\) is skew-symmetric and orthogonal.
Since \(A\) is skew-symmetric,
\[ A^T=-A \] Taking determinant on both sides,
\[ \det(A^T)=\det(-A) \] Now,
\[ \det(A^T)=\det(A) \] and for an \(n\times n\) matrix,
\[ \det(-A)=(-1)^n\det(A) \] Therefore,
\[ \det(A)=(-1)^n\det(A) \] Since \(A\) is orthogonal,
\[ A^TA=I \] So,
\[ \det(A^TA)=\det(I) \] \[ (\det A)^2=1 \] Hence,
\[ \det(A)\neq 0 \] Therefore, from
\[ \det(A)=(-1)^n\det(A) \] we can divide by \(\det(A)\), giving
\[ 1=(-1)^n \] This is possible only when \(n\) is even.
Therefore, statement \(Q\) is correct.

Step 4: Final conclusion.
Both statements \(P\) and \(Q\) are correct.
Hence,
\[ \boxed{(C)} \]