A) \[\left[ \begin{matrix} 2015 & 1 \\ 0 & 2015 \\ \end{matrix} \right]\]
B) \[\left[ \begin{matrix} 1 & 2015 \\ 0 & 1 \\ \end{matrix} \right]\]
C) \[\left[ \begin{matrix} 0 & 2015 \\ 0 & 0 \\ \end{matrix} \right]\]
D) \[\left[ \begin{matrix} 2015 & 0 \\ 1 & 2015 \\ \end{matrix} \right]\]
Correct Answer: B
Solution :
\[P{{P}^{T}}=\left[ \begin{matrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \end{matrix} \right]\left[ \begin{matrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]={{P}^{T}}P\] Now,\[{{P}^{T}}{{Q}^{2015}}P={{P}^{T}}\underbrace{\begin{matrix} PA{{P}^{T}} & PA{{P}^{T}} & ......PA{{P}^{T}}P \\ \end{matrix}}_{2015\,\text{times}}\] because\[={{A}^{2015}}\] Now\[{{A}^{2}}-2A+I=0\] \[\Rightarrow \]\[{{A}^{n}}=nA-(n-1)|\] \[\Rightarrow \]\[{{A}^{2015}}=2015\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]-(2014)\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 2015 \\ 0 & 1 \\ \end{matrix} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
