Let S = [n] : 1 [] n [] 50 and n is odd.

Question

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

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

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

Accepted Answer

Given, $$A = {\left[ {\begin{pmatrix} 1 & 0 & a  \
   { - 1} & 1 & 0  \
   { - a} & 0 & 1  \

\end{pmatrix} } \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{pmatrix} 1 & 0 & a  \
   { - 1} & 1 & 0  \
   { - a} & 0 & 1  \

\end{pmatrix} } \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, $$\sum\limits_{a\, \in \,S}^{} {\det (adj\,A)} $$

$$ = \sum\limits_{a\, \in \,S}^{} {{{({a^2} + 1)}^2}} $$

= $${\left( {{1^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 3 } \right)}^2} + 1} \right)^2} + {\left( {{{\left( {\sqrt 5 } \right)}^2} + 1} \right)^2} + .... + {\left( {{{\left( {\sqrt {49} } \right)}^2} + 1} \right)^2}$$

= $${\left( {{1^2} + 1} \right)^2} + {\left( {3 + 1} \right)^2} + {\left( {5 + 1} \right)^2} + .... + {\left( {49 + 1} \right)^2}$$

= ${2^2} + {4^2} + {6^2} + .... + {50^2}$

= $${2^2}\left( {{1^2} + {2^2} + {3^2} + .... + {{25}^2}} \right)$$

= $$4.{{25.26.51} \over 6} = 100.221$$

$$\therefore$$ $100K = 100.221$

$$ \Rightarrow K = 221$$

Solutions · 1