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Let S = {$$\sqrt{n}$$ : 1 $$\le$$ n $$\le$$ 50 and n is odd}. Let a $$\in$$ S and $$A = \left[ {\begin{matrix} 1 & 0 & a \\ { - 1} & 1 & 0 \\ { - a} & 0 & 1 \\ \end{matrix} } \right]$$. If $$\sum\limits_{a\, \in \,S}^{} {\det (adj\,A) = 100\lambda } $$, then $$\lambda$$ is equal to :

JEE · Math · previous-year question

  1. A.218
  2. B.221correct
  3. C.663
  4. D.1717

Answer

B. 221

Explanation

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

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