Question

Let A and B be two real symmetric matrices of order n. Then which of the following is true?

• A. \( AA^T = I \)
• B. \( A = A^{-1} \)
• C. \( AB = BA \)
• D. \( (AB)^T = BA \)

Hint:

Matrix Properties. Symmetric: \(A^T = A\). Transpose of Product: \((AB)^T = B^T A^T\). Combining these for symmetric A and B gives \((AB)^T = BA\).

Answer 1

The Correct Option is D

Given that A and B are real symmetric matrices of order n.
By definition of symmetric matrix: \( A^T = A \) \( B^T = B \) Let's check the options:
(1) \( AA^T = I \): This defines an orthogonal matrix.
A symmetric matrix is not necessarily orthogonal.
Incorrect.

(2) \( A = A^{-1} \): This means \(A^2 = I\).
This defines an involutory matrix.
A symmetric matrix is not necessarily involutory.
Incorrect.

(3) \( AB = BA \): Matrix multiplication is generally not commutative.
AB = BA only if A and B commute, which is not guaranteed even if they are both symmetric.
Incorrect.

(4) \( (AB)^T = BA \): We use the property of transpose of a product: \( (AB)^T = B^T A^T \).
Since A and B are symmetric, \(B^T = B\) and \(A^T = A\).
Substituting these into the property: $$ (AB)^T = B^T A^T = B A $$ Therefore, the statement \( (AB)^T = BA \) is true if A and B are symmetric matrices.