Matrices and Determinants 1 Question 21

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21. Let z=1+3i2, where i=1, and r,s1,2,3. Let P=(z)rz2s z2szr 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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Integer Type Question

####21. Let z=1+3i2, where i=1, and r,s1,2,3. 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

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed

(2016 Adv.)

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Answer:

Correct Answer: 21. (1)

Solution:

  1. Here, z=1+i32=ω

P=[(ω)rω2sω2sωr]P2=[(ω)rω2s(ω)rω2sω2sωrω2sωr]=[ω2r+ω4sωr+2s[(1)r+1]ωr+2s[(1)r+1]ω4s+ω2r]

Given, P2=I

ω2r+ω4s=1 and ωr+2s[(1)r+1]=0

Since, r1,2,3 and (1)r+1=0

r=1,3

Also,

ω2r+ω4s=1

If r=1, then ω2+ω4s=1

which is only possible, when s=1.

As, ω2+ω4=1

r=1,s=1

Again, if r=3, then

[ω6+ω4s=1ω4s=2r3] [never possible] 

(r,s)=(1,1) is the only solution.

Hence, the total number of ordered pairs is 1 .



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