Let z=-1+3i2, where i=-1, and r,s∈1,2,3. Let P=(-z)rz2sz2szr and I be the identity matrix of order 2. Then the total number of ordered pairs (r,s) for which P2=-I is
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answer is 1.
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Detailed Solution
z=ω, given that P2=-I ⇒P2=(−ω)rω2sω2sωr(−ω)rw2sω2sωr=−100−1-ω2r+ω4s=-1, (-ω)rω2s+(ω)r(ω)2s=01+ω2r+ω4s=1 and ⇒ωr+(-ω)r=0 ⇒r must be add Clearly r=1, s=1 is the only solutions