Let `P=[(1, 0, 0),(3, 1, 0),(9, 3, 1)]Q=[q_(ij)]` and `Q=P^5+I_3` then `(q_21+q_31)/q_32` is equal to

Options

A. 10
B. 135
C. 9
D. 15

Answer

A

Explanation

Given matrix
`P = [{:(1, 0, 0), (3, 1, 0), (9, 3, 1):}] = [{:(0, 0, 0), (3, 0, 0), (9, 3, 0):}] + [{:(1, 0, 0), (0, 1, 0), (0,0, 1):}]`
`rArr P = X + I ("let")`
Now, `P^(5) = (I + X)^(5)`
` = I + ""^(5)C_(1)(X) + ""^(5)C_(2)(X^(2)) + ""^(5)C_(3)(X^(3)) + ...`
`[because I^(n) = I, I * A " and " (a + x)^(n) = ""^(n)C_(0)alpha^(n) + ""^(n)C_(1)alpha^(n-1)x + ... + ""^(n)C_(n)x^(n)]`
Here, `X^(2) = [{:(0, 0, 0), (3, 0, 0), (9, 3, 0):}][{:(0, 0, 0), (3, 0, 0), (9, 3, 0):}] = [{:(0, 0, 0), (0, 0, 0), (9,0, 0):}]`
and `X^(3) = X^(2), X = [{:(0, 0, 0), (0, 0, 0), (9,0, 0):}][{:(0, 0, 0), (3, 0, 0), (9, 3, 0):}] = [{:(0, 0, 0), (0, 0, 0), (0, 0, 0):}]`
`rArr X^(4) = X^(5) = [{:(0, 0, 0), (0, 0, 0), (0, 0, 0):}] `
So, `P^(5) = I + 5 [{:(0, 0, 0), (3, 0, 0), (9, 3, 0):}] + 10 [{:(0, 0, 0), (0, 0, 0), (9,0, 0):}]`
` = [{:(1, 0, 0), (15, 1, 0), (135,15, 1):}]`
and `Q = I + P^(5) = [{:(2, 0, 0), (15, 2, 0), (135,15, 2):}] = [q_(ij)]`
`rArr q_(21) = 15, q_(31) = 135 " and " q_(32) = 15`
Hence, `(q_(21) + q_(31))/(q_(32)) = (15 + 135) /(15) = (150)/(15) = 10`