Question

If A and B are arbitrary$\mathit{n}\mathbf{\times }\mathit{n}$ matrices, which of the matrices in Exercise 21 through 26 must be symmetric?

${\mathit{A}}^{\mathbf{\Gamma }}\mathit{B}\mathit{A}$.

Short answer

The matrix$A^{T}BA$ is not symmetric.

Step-by-step solution

Condition to be a symmetric.

A matrix is symmetric if and only if it is equal to its transpose.

All entries above the main diagonal of a symmetric matrix are reflected into equal entries below the diagonal. A matrix is skew-symmetric if and only if it is the opposite of its transpose.

For example,

$\left(\begin{array}{ccc}1& 1& -1\\ 1& 2& 0\\ -1& 0& 5\end{array}\right)$

Thus, it gives

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

Therefore, if A and B are arbitrary $n\times n$matrices then the matrices $A^{T}BA$is not symmetric.

Hence, the matrix is not necessarily symmetric.