Question

Let $A=\left[\begin{array}{cc}2& 0\\ 2& 2\end{array}\right],$ $B=\left[\begin{array}{cc}\sqrt{\frac{7}{8}}& \frac{1}{2\sqrt{2}}\\ \frac{-1}{2\sqrt{2}}& \sqrt{\frac{7}{8}}\end{array}\right]$ and $C=B^{T}AB.$
$IfX=B{C}^{2015}{B}^{T}whereX=\left[X_{ij}\right]$ is a $2\times 2$ matrix then, $\frac{x_{11}+x_{12}+x_{21}+x_{22}}{2^{2016}}$ is

Answer 1

$B{B}^T=\left[\begin{array}{cc}\sqrt{\frac{7}{8}}& \frac{1}{2\sqrt{2}}\\ \frac{-1}{2\sqrt{2}}& \sqrt{\frac{7}{8}}\end{array}\right]\left[\begin{array}{cc}\sqrt{\frac{7}{8}}& \frac{-1}{2\sqrt{2}}\\ \frac{1}{2\sqrt{2}}& \sqrt{\frac{7}{8}}\end{array}\right]$
= I
So $C=B^{T}AB$
$C^2=\left(B^{T}AB\right)\left(B^{T}AB\right)$
$C^2=B^T{A}^{2}B$
Similarly, $C^{2015}=B^T{A}^{2015}B$
Now $X=B{C}^{2015}{B}^T=B\left(B^T{A}^{2015}B\right)B^T$
$X=A^{2015}$
$A^2=\left[\begin{array}{cc}2& 0\\ 2& 2\end{array}\right]\left[\begin{array}{cc}2& 0\\ 2& 2\end{array}\right]=\left[\begin{array}{cc}4& 0\\ 2\times 4& 4\end{array}\right]$
$A^3=\left[\begin{array}{cc}4& 0\\ 2\times 4& 4\end{array}\right]\left[\begin{array}{cc}2& 0\\ 2& 2\end{array}\right]=\left[\begin{array}{cc}8& 0\\ 3\times 8& 8\end{array}\right]$
$A^{2015}=\left[\begin{array}{cc}{2}^{2015}& 0\\ 2015\times 2^{2015}& 2^{2015}\end{array}\right]=X$
So $x_{11}=2^{2015},x_{12}=0,x_{21}=2015\times 2^{2015},x_{22}=2^{2015}$
$\frac{x_{11}+x_{12}+x_{21}+x_{22}}{2^{2016}}=\frac{2^{2015}\left(2017\right)}{2^{2016}}=1008.50$