Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
If $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$, $P = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$, $Q = P^T AP$, then $PQ^{2014}P^T =$
$\begin{pmatrix} 1 & 2^{2014} \\ 0 & 1 \end{pmatrix}$
$\begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix}$
$(P^T)^{2013}A^{2014}P^{2013}$
$P^T A^{2014}P$
Step-by-Step Solution
Key Concept: Recognize that P is an orthogonal rotation matrix (P^T P = I), so Q = P^T AP is a similarity transformation. Use the pattern that A^n = [[1, 2n], [0, 1]] to find Q^2014, then apply PQ^2014P^T = PA^2014P^T = A^2014 since similarity transforms preserve matrix powers.
Given $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $P = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$ with $Q = P^T AP$, we compute $A^2 = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}$ and $A^3 = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix}$. Using $PP^T = I$, we find $A^{2014} = \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix}$.
Correct Answer: 2