Question

If \( A \) is a symmetric matrix, then for any matrix \( B \) of order same as \( A \), \( BAB' \) is a/an :

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

Hint:

The expression \( B M B' \) always inherits the symmetry characteristics of the central matrix \( M \). If \( M \) is symmetric, \( B M B' \) is symmetric. If \( M \) is skew-symmetric, \( B M B' \) becomes skew-symmetric.

Answer 1

The Correct Option is C

Concept: A matrix \( M \) is said to be symmetric if it equals its own transpose, meaning \( M' = M \). Conversely, it is skew-symmetric if \( M' = -M \). An essential property of matrix transposes is the reversal rule for products: \( (XYZ)' = Z'Y'X' \).

Step 1: State the given mathematical properties. We are given that \( A \) is a symmetric matrix. By definition: \[ A' = A \] We want to determine the nature of the expression matrix \( X = BAB' \).

Step 2: Take the transpose of the matrix expression \( BAB' \). Let \( X = BAB' \). To find whether it is symmetric or skew-symmetric, let us evaluate its transpose \( X' \): \[ X' = (BAB')' \] Applying the reversal law of transpose operation for multiple matrices, \( (PQR)' = R'Q'P' \), we get: \[ X' = (B')' \cdot A' \cdot B' \]

Step 3: Simplify the expression using transpose rules. We know that the double transpose of any matrix returns the original matrix itself, i.e., \( (B')' = B \). Substituting this and \( A' = A \) back into the expression: \[ X' = B \cdot A \cdot B' \] Notice that this resulting expression is exactly the matrix \( X \) we started with: \[ X' = BAB' = X \]

Step 4: Draw conclusion. Since taking the transpose of \( BAB' \) gives back the original matrix \( BAB' \) unaltered, the matrix \( BAB' \) is conclusively a symmetric matrix.