Question

Consider the matrix $A=\left[\begin{array}{cc}\sqrt{3}& -2\\ 0& 1\end{array}\right]$ and $B$ be a square matrix of order $2$ such that $B{B}^T=B^{T}B=I$ . Let $C=BA{B}^T$ and $D={\left[d_{ij}\right]}_{2\times 2}=B^T{C}^{6}B$ , then $d_{11}$ is equal to

• A. $25$
• B. $27$
• C. $30$
• D. $81$

Answer 1

Answer: (B)

$D=B^T\left(BA{B}^T\right)\left(BA{B}^T\right)\left(BA{B}^T\right)\left(BA{B}^T\right)\left(BA{B}^T\right)\left(BA{B}^T\right)B$
$D=A^6$
Now, $A^2=\left[\begin{array}{cc}\sqrt{3}& -2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\sqrt{3}& -2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}3& -2\sqrt{3}-2\\ 0& 1\end{array}\right]$
$A^3=A^2\cdot A=\left[\begin{array}{cc}3& -2\sqrt{3}-2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\sqrt{3}& -2\\ 0& 1\end{array}\right]$ $=\left[\begin{array}{cc}3\sqrt{3}& -2\sqrt{3}-8\\ 0& 1\end{array}\right]$
$A^6=A^3\cdot A^3=\left[\begin{array}{cc}3\sqrt{3}& -2\sqrt{3}-8\\ 0& 1\end{array}\right]\left[\begin{array}{cc}3\sqrt{3}& -2\sqrt{3}-8\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}27& -26-26\sqrt{3}\\ 0& 1\end{array}\right]=D$
$\Rightarrow d_{11}=27$