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?

$\mathit{B}{\mathit{B}}^{\mathbf{T}}$.

Short answer

The matrix $B{B}^T$ is 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,

Let the matrix is $B=\left(\begin{array}{ccc}1& 1& -1\\ 1& 2& 0\\ -1& 0& 5\end{array}\right)$.

Thus, it gives

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

Thus, If A and B are arbitrary$n\times n$ matrices then the matrices$B{B}^T$ is symmetric.

Hence, the matrices is symmetric.