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Matrices & Determinants: Let Pmatrix Pmatrix Let Pmatrix Pmatrix Equal

JEE Maths question with a full step-by-step solution.

Question
Let M=(0ii0)M = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, where i2=−1i^2 = -1, and let I=(1001)I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. Then I+M+M2+M3+⋯+M2010I + M + M^2 + M^3 + \cdots + M^{2010} is equal to:
A(0000)\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
B(0ii0)\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}correct
C(1ii1)\begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}
D(−100−1)\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
Solution
Step 1: Compute powers of MM
M2=(−100−1)=−I,M3=−M,M4=IM^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I, \quad M^3 = -M, \quad M^4 = I
 The powers repeat with period 4. Step 2: Sum over one period
I+M+M2+M3=I+M−I−M=0I + M + M^2 + M^3 = I + M - I - M = 0
 Step 3: Account for all terms The sum has 2011 terms (n=0n = 0 to 20102010). Since 2011=4(502)+32011 = 4(502) + 3, the first 2008 terms cancel in complete groups of four, leaving:
M2008+M2009+M2010=I+M+M2=I+M−I=M=(0ii0)M^{2008} + M^{2009} + M^{2010} = I + M + M^2 = I + M - I = M = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}
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