Question

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

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

Answer 1

Answer: (B)

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