Let S = {$\sqrt{n}$ : 1 $\le$ n $\le$ 50 and n is odd}.

Let a $\in$ S and $$A = \left[ {\matrix{ 1 & 0 & a \cr { - 1} & 1 & 0 \cr { - a} & 0 & 1 \cr } } \right]$$.

If $\sum\limits_{a\, \in \,S}^{} {\det (adj\,A) = 100\lambda }$, then $\lambda$ is equal to :

<p>Given, $$A = {\left[ {\matrix{ 1 & 0 & a \cr { - 1} & 1 & 0 \cr { - a} & 0 & 1 \cr } } \right]_{3 \times 3}}$$</p> <p>S = {$\sqrt{n}$ : 1 $\le$ n $\le$ 50 and n is odd}</p> <p>$\therefore$ S = $\left\{ {1,\sqrt 3 ,\sqrt 5 ,\sqrt 7 ,....,\sqrt {49} } \right\}$</p> <p>We know,</p> <p>$\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}$</p> <p>Here, n = order of matrix.</p> <p>Here, n = 3</p> <p>$\therefore$ $\left| {adj\,A} \right| = {\left| A \right|^{3 - 1}} = {\left| A \right|^2}$</p> <p>Now, $$\left| A \right| = \left| {\matrix{ 1 & 0 & a \cr { - 1} & 1 & 0 \cr { - a} & 0 & 1 \cr } } \right|$$</p> <p>$= 1(1 - 0) - 0 + a(0 - ( - a))$</p> <p>$= {a^2} + 1$</p> <p>$\therefore$ $\left| {adj\,A} \right| = {\left| A \right|^2} = {({a^2} + 1)^2}$</p> <p>Now, $\sum\limits_{a\, \in \,S}^{} {\det (adj\,A)}$</p> <p>$= \sum\limits_{a\, \in \,S}^{} {{{({a^2} + 1)}^2}}$</p> <p>= $${\left( {{1^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 3 } \right)}^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 5 } \right)}^2} + 1} \right)^2} + .... + {\left( {{{\left( {\sqrt {49} } \right)}^2} + 1} \right)^2}$$</p> <p>= $${\left( {{1^2} + 1} \right)^2} + {\left( {3 + 1} \right)^2} + {\left( {5 + 1} \right)^2} + .... + {\left( {49 + 1} \right)^2}$$</p> <p>= ${2^2} + {4^2} + {6^2} + .... + {50^2}$</p> <p>= ${2^2}\left( {{1^2} + {2^2} + {3^2} + .... + {{25}^2}} \right)$</p> <p>= $4.{{25.26.51} \over 6} = 100.221$</p> <p>$\therefore$ $100K = 100.221$</p> <p>$\Rightarrow K = 221$</p>