Question

If A and B are square matrices of same order and B is a skew symmetric matrix, then A’BA is

• A. Symmetric matrix
• B. Null matrix
• C. Diagonal matrix
• D. Skew symmetric matrix

Answer 1

The Correct Option is D

If A and B are square matrices of the same order and B is a skew-symmetric matrix, then we need to determine the nature of \(A'BA\).

Since B is a skew-symmetric matrix, we have \(B' = -B\).

Let \(C = A'BA\). To determine if C is symmetric or skew-symmetric, we need to find \(C'\).

\(C' = (A'BA)'\)

Using the property \((ABC)' = C'B'A'\), we have:

\(C' = A'B'(A')' = A'B'A\)

Since B is skew-symmetric, \(B' = -B\), so:

\(C' = A'(-B)A = -A'BA = -C\)

Since \(C' = -C\), C is a skew-symmetric matrix.

Therefore, the correct option is (D) Skew symmetric matrix.

Answer 2

Given that \( A \) and \( B \) are square matrices of the same order, and \( B \) is a skew-symmetric matrix,

Key Property: For a skew-symmetric matrix \( B \), we know that: \[ B^T = -B \] 

Step 1: Consider \( A^T B A \). Now, let's compute \( A^T B A \). 

We use the property of \( B \) being skew-symmetric: \[ (A^T B A)^T = A^T B^T A = A^T (-B) A = - A^T B A \] 

Thus, we have: \[ (A^T B A)^T = - (A^T B A) \] 

This shows that \( A^T B A \) is skew-symmetric because the transpose of a matrix is equal to its negative. 

Conclusion: Since \( A^T B A \) is skew-symmetric, the result of \( A^T B A \) will also be a skew-symmetric matrix.