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.]

B2=IAB=[axpyqbrcz][001010100]=[pxabqyzcf]AB=AB2=……………=AB19=[pxabqyzcr]Tr.(AB+AB2+……..+AB19)=210⇒10(p+q+r)=210⇒p+q+r=21,p,q,r∈Np′+q′+r′=18,p′,q′⋅r′∈W∴Number of ordered triplets (p,q,r)=20Cr=20×192=190

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Let $A = [\begin{array}{ccc} a & x & p \\ y & q & b \\ r & c & z \end{array}] and B = [\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}]$ where a, b, c, x, y, z, p, q, r are natural numbers.
If Tr $(AB + {AB}^{3} + {AB}^{5} + \dots \dots + {AB}^{19}) = 210,$ then find number of ordered triplets $(p, q, r)$ [Note : tr(P) denotes the trace of matrix P.]

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answer is 190.

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Detailed Solution

$B^{2} = I$

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