Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& M_r\text{is defined as}{M}_r=\left[\begin{array}{cc}(r-1)& \frac{1}{r}\\ 1& \frac{1}{(r-1{)}^2}\end{array}\right]\text{. Then}\underset{n\to \infty }{lim}{(det{M}_2+det{M}_3+\cdots +det{M}_n)}^{\ln n}\text{equals}& \text{(P)}& 16\\ \text{(B)}& \text{If}\left|\begin{array}{ccc}1& \cos \alpha & \cos \beta \\ \cos \alpha & 1& \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0& \cos \alpha & \cos \beta \\ \cos \alpha & 0& \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|,\text{then}{\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma \text{equals}& \text{(Q)}& 4\\ \text{(C)}& \text{If}A=\left[\begin{array}{cc}3& 1\\ -1& 1\end{array}\right]\text{and a matrix}C\text{is defined as}C=\left(BA{B}^{-1}\right)\left(B^{-1}{A}^{\text{T}}B\right),\text{where}|B|\ne 0,& \text{(R)}& 2\\ & \text{then}detC\text{is a square of natural number equal to}& & \\ \text{(D)}& \text{If}A=\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]\text{and}{A}^4=-\lambda I,\text{then}\lambda \text{equals}& \text{(S)}& 1\end{array}$

Which of the following is a CORRECT combination?

• A. $\text{(A)}\to \text{(R)},\text{(B)}\to \text{(P)}$
• B. $\text{(A)}\to \text{(S)},\text{(B)}\to \text{(Q)}$
• C. $\text{(A)}\to \text{(S)},\text{(B)}\to \text{(R)}$
• D. $\text{(A)}\to \text{(Q)},\text{(B)}\to \text{(P)}$

Answer 1

The correct option is C $\text{(A)}\to \text{(S)},\text{(B)}\to \text{(R)}$

$\left(A\right)$
$|M_r|=\frac{1}{r-1}-\frac{1}{r}|M_2|+|M_3|+|M_4|+\cdots +|M_n|=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots +(\frac{1}{n-1}-\frac{1}{n})=1-\frac{1}{n}$
We get $\underset{n\to \infty }{lim}{(det{M}_2+det{M}_3+\cdots +det{M}_n)}^{\ln n}$
$=\underset{n\to \infty }{lim}{(1-\frac{1}{n})}^{\ln n}(1^{\infty }$ form$)$
$=e^{\underset{n\to \infty }{lim}\frac{-\ln n}{n}}=e^0=1$


$\left(B\right)$ Let $A=\left|\begin{array}{ccc}1& \cos \alpha & \cos \beta \\ \cos \alpha & 1& \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=1-{\cos }^2\gamma -{\cos }^2\alpha +2\cos \alpha \cos \beta \cos \gamma -{\cos }^2\beta$

And let $B=\left|\begin{array}{ccc}0& \cos \alpha & \cos \beta \\ \cos \alpha & 0& \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|=2\cos \alpha \cos \beta \cos \gamma$
Given that $A=B$
$\Rightarrow 1={\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma \Rightarrow {\sin }^2\alpha +{\sin }^2\beta +{\sin }^2\gamma =2$