If $A=\left[\begin{array}{cc}a& b\\ b& a\end{array}\right]$ and $A^2=\left[\begin{array}{cc}\alpha & \beta \\ \beta & \alpha \end{array}\right]$, then

If $A=\left[\begin{array}{cc}1& tan\frac{\theta }{2}\\ -tan\frac{\theta }{2}& 1\end{array}\right]$ and AB = I, then B =

Let,
$A=\left[\begin{array}{ccc}4& 6& -1\\ 3& 0& 2\\ 1& -2& 5\end{array}\right],B=\left[\begin{array}{cc}2& 4\\ 0& 1\\ -1& 2\end{array}\right]$
$\text{and}C=\left[\begin{array}{ccc}1& 2& 3\end{array}\right]$
The expression which is not defined is: 

If $A=\left[\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right]$, then $A^n=$

If A is $3\times 4$ matrix and B is a matrix such that A'B and BA' are both defined. Then B is of the type

For each real number x such that $-1<x<1,letA\left(x\right)bethematrix(1-x{)}^{-1}\left[\begin{array}{cc}1& -x\\ -x& 1\end{array}\right]andz=\frac{x+y}{1+xy}Then,$

$IfA=\left[\begin{array}{cc}cos\theta & sin\theta \\ sin\theta & -cos\theta \end{array}\right],B=\left[\begin{array}{cc}1& 0\\ -1& 1\end{array}\right],C=AB{A}^T,then{A}^T{C}^{n}Aequalsto\left(nϵZ^{+}\right)$

If A and B are two non singular matrices and both are symmetric and commute each other then

$A=[a_{ij}{]}_{n\times n}$ and $a_{ij}=i^2-j^2$ then A is necessarily

If A is a non-diagonal involutory matrix, then