Question

Question:TRUE OR FALSE?If A and Bare symmetric$\mathbf{n}\mathbf{\times }\mathbf{n}$ matrices,AB then must be symmetric as well.

Short answer

The given statement is false.

Step-by-step solution

Definition of a symmetric matrix.

Suppose A is an$\mathbf{n}\mathbf{\times }\mathbf{n}$matrix.

Then the matrix is said to be symmetric matrix, if${\mathit{A}}^{\mathbf{T}}\mathbf{=}\mathit{A}$.

By the properties of transpose,${\left(AB\right)}^{\mathbf{T}}\mathbf{=}{\mathit{B}}^{\mathbf{T}}{\mathit{A}}^{\mathbf{T}}$

Check whether the given statement is a true or false.

The given statement is if A and B are symmetric $n\times n$matrices, the AB must be symmetric as well.

It is given that A and B are symmetric $n\times n$matrices.

This means,$A^T=A,B^T=B$

Here, the matrix AB is symmetric, if${\left(AB\right)}^T=AB$

Check:

$$\begin{gathered} {\left(AB\right)}^T=A^T{B}^T \\ =BA \end{gathered}$$

So,${\left(AB\right)}^T=AB$

It is know that the matrix multiplication is not commutative.

So, it is clear that AB = BA is cannot be true for every case.

So,${\left(AB\right)}^T\ne AB$

Then the given statement is false, by the properties of the transpose of matrices.