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Let P=[(1,0,0),(4,1,0),(16,4,1)] and I b...

Let `P=[(1,0,0),(4,1,0),(16,4,1)]` and `I` be the identity matrix of order 3. If `Q=[q_("ij")]` is a matrix such that `P^(50)-Q=I`, then `(q_(31)+q_(32))/q_(21)` equals

A

52

B

201

C

103

D

205

Text Solution

Verified by Experts

The correct Answer is:
B
Here, `P = [{:(1, 0, 0), (4, 1, 0), (16, 4, 1):}]`
`therefore P^(2) = [{:(1, 0, 0), (4, 1, 0), (16, 4, 1):}][{:(1, 0, 0), (4, 1, 0), (16, 4, 1):}]= [{:(1, 0, 0), (4+4, 1, 0), (16+32, 4+4, 1):}]`
`= [{:(1, 0, 0), (4 xx 2, 1, 0), (16 (1+2), 4 xx 2, 1):}] " "....(i)`
and `P^(2)= [{:(1, 0, 0), (4 xx 2, 1, 0), (16 (1+2), 4 xx 2, 1):}][{:(1, 0, 0), (4, 1, 0), (16, 4, 1):}]`
`= [{:(1, 0, 0), (4 xx 3, 1, 0), (16 (1+2 +3), 4 xx 3, 1):}]" "...(ii)`
From symmetry,
`P^(50)= [{:(1, 0, 0), (4 xx 50, 1, 0), (16 (1+2 +3+... +50), 4 xx 50, 1):}]`
`because P^(50)-Q = I " " ["given"]`
`therefore [{:(1-q_(11), -q_(12), -q_(13)), (200-q_(21), 1-q_(22), -q_(23)), (16 xx (50)/(2) (51)-q_(31), 200-q_(32), 1-q_(33)):}] = [{:(1, 0, 0), (0, 1, 0), (0,0, 1):}]`
`rArr 200-q_(21) = 0, (16 xx 50 xx 51)/(2)-q_(31) = 0,`
`200-q_(32) = 0`
`therefore q_(21) = 200, q_(32) = 200, q_(31) = 20400`
Thus, `(9_(31) + q_(32))/(q_(21)) = (20400 + 200)/(200) = (20600)/(200) = 103`
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