If $A$=$\left[ {\begin{array}{*{20}{c}}2&0&0\\0&2&0\\0&0&2\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}1&2&3\\0&1&3\\0&0&2\end{array}} \right],$ then $|AB|$ is equal to

Consider the matrix $f(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$

Given below are two statements:

Statement I: $f(-x)$ is the inverse of the matrix $f(x)$.

Statement II: $f(x) f(y)=f(x+y)$.

In the light of the above statements, choose the correct answer from the options given below

Let $\mathrm{p}$ be an odd prime number and $\mathrm{T}_{\mathrm{p}}$ be the following set of $2 \times 2$ matrices :

$T_p=\left\{A=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{a}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1, \ldots ., \mathrm{p}-1\}\right\}$

$1.$ The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and $\operatorname{det}(\mathrm{A})$ divisible by $\mathrm{p}$ is

$(A)$ $(\mathrm{p}-1)^2$  $(B)$ $2(\mathrm{p}-1)$

$(C)$ $(\mathrm{p}-1)^2+1$  $(D)$ $2 \mathrm{p}-1$

$2.$ The number of $A$ in $T_p$ such that the trace of $A$ is not divisible by $p$ but det $(A)$ is divisible by $p$ is [Note: The trace of a matrix is the sum of its diagonal entries.]

$(A)$ $(\mathrm{p}-1)\left(\mathrm{p}^2-\mathrm{p}+1\right)$ $(B)$ $\mathrm{p}^3-(\mathrm{p}-1)^2$

$(C)$ $(\mathrm{p}-1)^2$ $(D)$ $(p-1)\left(p^2-2\right)$

$3.$ The number of $A$ in $T_p$ such that det $(A)$ is not divisible by $p$ is

$(A)$ $2 \mathrm{p}^2$ $(B)$ $p^3-5 p$ $(C)$ $p^3-3 p$ $(D)$ $p^3-p^2$

Give the answer question $1,2$ and $3.$

If $A = \left[ {\begin{array}{*{20}{c}}0&1\\1&0\end{array}} \right],B = \left[ {\begin{array}{*{20}{c}}0&{ – i}\\i&0\end{array}} \right]$ then ${(A + B)^2}$ equals

Show that

$\left[ {\begin{array}{*{20}{c}}
  5&{ – 1} \\ 
  6&7 
\end{array}} \right]\left[ {\begin{array}{*{20}{l}}
  2&1 \\ 
  3&4 
\end{array}} \right]$ $ \ne \left[ {\begin{array}{*{20}{l}}
  2&1 \\ 
  3&4 
\end{array}} \right]\left[ {\begin{array}{*{20}{l}}
  5&{ – 1} \\ 
  6&7 
\end{array}} \right]$