Matrices & Determinants
Matrix powers and summation
MJAT_TS1_P1
Grade 12
Question:
Let $A = \begin{pmatrix} \cos(\pi/5) & -\sin(\pi/5) \\ \sin(\pi/5) & \cos(\pi/5) \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.
If $\displaystyle\sum_{r=1}^{4}\bigl(B^2 - B\cdot A^{9r} - A^r\cdot B + A^{10r}\bigr) = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then $a^{10} + b^{10} + c^{10} + d^{10} =$
A) $0$
B) $2\left(\cos^{10}\dfrac{\pi}{5} + \sin^{10}\dfrac{\pi}{5}\right)$
C) $4$
D) $2$
Step-by-Step Solution
Key Concept: $A$ is a rotation matrix, so $A^{10} = I$ (rotation by $10\cdot\pi/5 = 2\pi$). $B^2 = I$. Factor the sum: $\sum_r (B^2 - BA^{9r} - A^rB + A^{10r}) = \sum_r (I - BA^{9r} - A^rB + I) = \sum_r (I-A^rB)(I-A^{9r}B^{-1})$... or factor as $(B-A^r)(B-A^{9r})$. Since $A$ and $B$ commute and $A^{10}=I$, the whole sum $= 0$.
$\sum_{r=1}^4 (B-A^r)^2 = 0$ (using $A^{10}=I$, $B^2=I$, commutativity). The matrix is $\mathbf{0}$, so $a=b=c=d=0$ and $a^{10}+b^{10}+c^{10}+d^{10} = 0$.
Correct Answer: A