Matrices & Determinants
General
Grade 12
Question:
If $P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$, $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $Q = PAP^T$ and $x = P^T Q^{2005} P$, then $x$ is equal to -
\begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix}
\begin{bmatrix} 4+2005\sqrt{3} & 6015 \\ 2005 & 4-2005\sqrt{3} \end{bmatrix}
\frac{1}{4} \begin{bmatrix} 2+\sqrt{3} & 1 \\ -1 & 2-\sqrt{3} \end{bmatrix}
\frac{1}{4} \begin{bmatrix} 2005 & 2-\sqrt{3} \\ 2+\sqrt{3} & 2005 \end{bmatrix}
Step-by-Step Solution
Key Concept: General
<p>$PP^T = I$</p><p>$\Rightarrow A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \Rightarrow A^2 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$</p><p>and so, on</p><p>$\Rightarrow Q = PAP^T$</p><p>$\Rightarrow Q^2 = (PAP^T)(PAP^T) = PA^2P^T$</p><p>$\Rightarrow Q^{2005} = PA^{2005}P^T$</p><p>$\Rightarrow x = P^T(PA^{2005}P^T)P$</p><p>$\Rightarrow x = A^{2005} = \begin{bmatrix} 1 & 2005 \\ 0 & 1 \end{bmatrix}$</p>
Correct Answer: A