Let S = ${\sqrt[]{n} : 1 \leq n \leq 50 a n d n i s o d d}$ .

Let $a \in S a n d A = [\begin{matrix} 1 & 0 & a \\ - 1 & 1 & 0 \\ - a & 0 & 1 \end{matrix}] .$ If $\sum_{a \in S}^{} d e t (a d j A)$ = 100l, then l is equal to :

Option 1 -

218

Option 2 -

221

Option 3 -

663

Option 4 -

1717

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

S = ${\sqrt[]{n} : 1 \leq n \leq 50 & n = o d d}$

= ${\sqrt[]{1}, \sqrt[]{3}, \sqrt[]{5}, . . . . . ., \sqrt[]{49} (25 t e r m s)}$

$| A | = | \begin{matrix} 1 & 0 & 4 \\ - 1 & 1 & 0 \\ - a & 0 & 1 \end{matrix} |$ = 1 + a²

^{$∴ \sum_{a \in S}^{} d e t (a d j A) = 100 λ \Rightarrow \sum {(1 + a^{2})}^{2} = 100 λ \Rightarrow λ = 221$}

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Let S = ${\sqrt[]{n} : 1 \leq n \leq 50 a n d n i s o d d}$ .

Let $a \in S a n d A = [\begin{matrix} 1 & 0 & a \\ - 1 & 1 & 0 \\ - a & 0 & 1 \end{matrix}] .$ If $\sum_{a \in S}^{} d e t (a d j A)$ = 100l, then l is equal to :

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Let S = { n : 1 ≤ n ≤ 5 0 a n d n i s o d d } . Let a ∈ S a n d A = [ 1 0 a − 1 1 0 − a 0 1 ] . If ∑ a ∈ S d e t ( a d j A ) = 100l, then l is equal to :

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Let S = ${\sqrt[]{n} : 1 \leq n \leq 50 a n d n i s o d d}$ .

Let $a \in S a n d A = [\begin{matrix} 1 & 0 & a \\ - 1 & 1 & 0 \\ - a & 0 & 1 \end{matrix}] .$ If $\sum_{a \in S}^{} d e t (a d j A)$ = 100l, then l is equal to :