JEE Advanced 2013 · Paper 2 · Mathematics Q43

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MathematicsMultiple correct+3 / -1Question 43 of 60

Let $\omega$ be a complex cube root of unity with $\omega\ne1$ and $P=[p_{ij}]$ be a $n\times n$ matrix with $p_{ij}=\omega^{i+j}$ . Then $P^2\ne0$ , when $n=$

Answer and worked solution

Official answer: B, C, D

Entries of P².
$(P^2)_{ij}=\sum_{k=1}^n\omega^{i+k}\omega^{k+j}=\omega^{i+j}\sum_{k=1}^n\omega^{2k}.$

The sum vanishes exactly when $3\mid n$ (each block of three terms is $\omega^2+\omega^4+\omega^6=0$ ). So $P^2\ne0$ unless $n$ is a multiple of 3: true for 55, 58, 56, false for 57.

(B), (C), (D)

Topics:Matrix algebra Roots of unity

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JEE Advanced 2013 · Paper 2, Mathematics question 43