If $A$ and $B$ are symmetric matrices of the same order, then show that $AB$ is symmetric if and only if $A$ and $B$ commute, that is $AB = BA$.

Show that the matrix $B ^{\prime}A B$ is symmetric or skew symmetric according as $A$ is symmetric or skew symmetric.

Let the matrix

 $A = \left[ {\begin{array}{*{20}{c}}
{{{10}^{30}} + 5}&{{{10}^{20}} + 4}&{{{10}^{20}} + 6}\\
{{{10}^4} + 2}&{{{10}^8} + 7}&{{{10}^{10}} + 2n}\\
{{{10}^4} + 8}&{{{10}^6} + 4}&{{{10}^{15}} + 9}
\end{array}} \right]$ ,

 $n \in N$, then

If $P$ is a $3 \times 3$ matrix such that $P^{\top}=2 P+I$, where $P^{\top}$ is the transpose of $P$ and $I$ is the $3 \times 3$ identity matrix, then there exists a column matrix $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \neq\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ such that 

Let $A=\left\{X=(x, y, z)^{T}: P X=0\right.$ and $\left.\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=1\right\}$ where $\mathrm{P}=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{array}\right]$ then the set $\mathrm{A}$