LIST -1 LIST -2A)If r>1,Mr=r-11r11r-12 then limn→∞(|M2|+|M3|+|M4|+......+|Mn|)logen=P)2B)If 1cosαcosβcosα1cosγcosβcosγ1=0cosαcosβcosα0cosγcosβcosγ0 then sin2α+sin2β+sin2γ=Q)4C)If A=[31−11]and C=(BAB−1)(B−1ATB) then det(C) can beR)1D)If A=[11−11]and A4=−λI then λ=S)3 T)0

A) |Mr|=1r−1−1r∴|M2|+|M3|+......+|Mn|=1−1nLtn→∞(|M2|+|M3|+........+|Mn|)=Ltn→∞(1−1n)logen=1B) 2C) |C|=|A|2=16D) A2=(02−20) and A=(−400−4)

LIST -1 LIST -2A)If r>1,Mr=r-11r11r-12 then limn→∞(|M2|+|M3|+|M4|+......+|Mn|)logen=P)2B)If 1cosαcosβcosα1cosγcosβcosγ1=0cosαcosβcosα0cosγcosβcosγ0 then sin2α+sin2β+sin2γ=Q)4C)If A=[31−11]and C=(BAB−1)(B−1ATB) then det(C) can beR)1D)If A=[11−11]and A4=−λI then λ=S)3 T)0

A) |Mr|=1r−1−1r∴|M2|+|M3|+......+|Mn|=1−1nLtn→∞(|M2|+|M3|+........+|Mn|)=Ltn→∞(1−1n)logen=1B) 2C) |C|=|A|2=16D) A2=(02−20) and A=(−400−4)

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LIST -1 LIST -2
A) If $r > 1, M_{r} = |\begin{matrix} r - 1 & \frac{1}{r} \\ 1 & \frac{1}{{(r - 1)}^{2}} \end{matrix}|$ then $\lim_{n \to \infty} {(| M_{2} | + | M_{3} | + | M_{4} | + . ..... + | M_{n} |)}^{\log_{e}^{n}} =$ P) 2
B) If $|\begin{matrix} 1 & cosα & cosβ \\ cosα & 1 & cosγ \\ cosβ & cosγ & 1 \end{matrix}| = |\begin{matrix} 0 & cosα & cosβ \\ cosα & 0 & cosγ \\ cosβ & cosγ & 0 \end{matrix}|$ then $\sin^{2} α + \sin^{2} β + \sin^{2} γ =$ Q) 4
C) $If A = [\begin{matrix} 3 & 1 \\ - 1 & 1 \end{matrix}] and C = ({BAB}^{- 1}) (B^{- 1} A^{T} B) then \sqrt{\det (C)} can be$ R) 1
D) $If A = [\begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix}] and A^{4} = - λI then λ =$ S) 3
T) 0

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answer is B.

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Detailed Solution

$\begin{array}{l} A) | M_{r} | = \frac{1}{r - 1} - \frac{1}{r} \\ ∴ | M_{2} | + | M_{3} | + ...... + | M_{n} | = 1 - \frac{1}{n} \\ \underset{n \to \infty}{L t} (| M_{2} | + | M_{3} | + ........ + | M_{n} |) = \underset{n \to \infty}{L t} {(1 - \frac{1}{n})}^{\log_{e}^{n}} = 1 \end{array}$

B) 2

C) $| C | = {| A |}^{2} = 16$

D) $A^{2} = (\begin{matrix} 0 & 2 \\ - 2 & 0 \end{matrix}) a n d A = (\begin{matrix} - 4 & 0 \\ 0 & - 4 \end{matrix})$

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