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.]

# Let  where a, b, c, x, y, z, p, q, r are natural numbers. If Tr $\left(\mathrm{AB}+{\mathrm{AB}}^{3}+{\mathrm{AB}}^{5}+\dots \dots +{\mathrm{AB}}^{19}\right)=210,$ then find number of ordered triplets  [Note : tr(P) denotes the trace of matrix P.]

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### Solution:

${B}^{2}=I$

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

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

$\text{Tr.}\left(AB+{AB}^{2}+\dots \dots ..+{AB}^{19}\right)=210$

$⇒10\left(p+q+r\right)=210⇒p+q+r=21,p,q,r\in N$

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

$\therefore$Number of ordered triplets $\left(p,q,r\right){=}^{20}{C}_{r}=\frac{20×19}{2}=190$

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