Let A=[axpyqbrcz] and B=[001010100] where a, b, c, x, y, z, p, q, r are natural numbers. If Tr (AB+AB3+AB5+……+AB19)=210, then find number of ordered triplets (p, q, r) [Note : tr(P) denotes the trace of matrix P.]

Solution:

$B^{2} = I$

$A B = [\begin{array}{lll} a & x & p \\ y & q & b \\ r & c & z \end{array}] [\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}] = [\begin{array}{ccc} p & x & a \\ b & q & y \\ z & c & f \end{array}]$

$A B = {A B}^{2} = \dots \dots \dots \dots \dots = {A B}^{19} = [\begin{array}{lll} p & x & a \\ b & q & y \\ z & c & r \end{array}]$

$Tr. (A B + {A B}^{2} + \dots \dots . . + {A B}^{19}) = 210$

$\Rightarrow 10 (p + q + r) = 210 \Rightarrow p + q + r = 21, p, q, r \in N$

$p^{'} + q^{'} + r^{'} = 18, p^{'}, q^{'} \cdot r^{'} \in W$

$∴$ Number of ordered triplets $(p, q, r) =^{20} C_{r} = \frac{20 \times 19}{2} = 190$

Solution:

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