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(C) First,we calculate $BB^T$. Given $B = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$,its tra

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$\begin{bmatrix} 1 & 5 \ 0 & 1 \end{bmatrix}$

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(C) First,we calculate $BB^T$. Given $B = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$,its transpose is $B^T = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$.$BB^T = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} \frac{3}{4} + \frac{1}{4} & -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \ -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} & \frac{1}{4} + \frac{3}{4} \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Now,$(BB^TA)^5 = (IA)^5 = A^5$.We have $A = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$.$A^2 = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix}$.$A^3 = A^2 A = \begin{bmatrix} 1 & 2 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \ 0 & 1 \end{bmatrix}$.By induction,$A^n = \begin{bmatrix} 1 & n \ 0 & 1 \end{bmatrix}$.Therefore,$A^5 = \begin{bmatrix} 1 & 5 \ 0 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$,then $(BB^TA)^5$ is equal to

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