Question

For x > 0, Let $A=\left[\begin{array}{ccc}x+\frac{1}{x}& 0& 0\\ 0& x& 0\\ 0& 0& 16\end{array}\right]B=\left[\begin{array}{ccc}\frac{5x}{x^2+1}& 0& 0\\ 0& \frac{3}{x}& 0\\ 0& 0& \frac{1}{4}\end{array}\right]$
$X=(AB{)}^{-1}+(AB{)}^{-2}+(AB{)}^{-3}+...\infty$
$Z=X^{-1}-2I$ (I is identity matrix of order 3)
$\begin{array}{cc}\text{(P) minimum value of [Tr(Ax)]}is& \left(1\right)24\\ \text{(when [.])}\to \text{represent integer function}& \\ \text{(Q) det}\left(X^{-1}\right)is& \left(2\right)12\\ \text{(R) If}Tr(z+z^2+---+z^{10})=2^a+b,& \\ (a,b\in N)\text{then a + b is}& \left(3\right)6\\ \text{(S) If value of}|adj\left(\sqrt{5}{X}^{-1}\right)|=k& \\ \text{then number of positive divisors}& \left(4\right)19\\ \text{of k which are odd is}& \end{array}$

• A. $P\to 3,Q\to 1,R\to 2,S\to 4$
• B. $P\to 3,Q\to 1,R\to 4,S\to 2$
• C. $P\to 2,Q\to 4,R\to 3,S\to 1$
• D. $P\to 2,Q\to 4,R\to 1,S\to 3$

Answer 1

The correct option is B $P\to 3,Q\to 1,R\to 4,S\to 2$

$AB=\left[\begin{array}{ccc}5& 0& 0\\ 0& 3& 0\\ 0& 0& 4\end{array}\right]$
$A{B}^{-1}=\left[\begin{array}{ccc}\frac{1}{5}& 0& 0\\ 0& \frac{1}{3}& 0\\ 0& 0& \frac{1}{4}\end{array}\right](AB{)}^{-2}=\left[\begin{array}{ccc}\frac{1}{5^2}& 0& 0\\ 0& \frac{1}{3^2}& 0\\ 0& 0& \frac{1}{4^2}\end{array}\right]$
$x=\left[\begin{array}{ccc}\frac{1}{5}+\frac{1}{5^2}+---\infty & 0& 0\\ 0& \frac{1}{3}+\frac{1}{3^2}+-\infty & 0\\ 0& 0& \frac{1}{4}+\frac{1}{4^2}+---\infty \end{array}\right]$
$X=\left[\begin{array}{ccc}\frac{1}{4}& 0& 0\\ 0& \frac{1}{2}& 0\\ 0& 0& \frac{1}{3}\end{array}\right]$
$X^{-1}=\left[\begin{array}{ccc}4& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right]|$
Q)
$|X{|}^{-1}=4\times 2\times 3=24$

S)
$adj\left(\sqrt{5}{x}^{-1}\right)|=|5adj\left(X^{-1}\right)|$
$=5^3\times |adj\left(X^{-1}\right)|$
$=5^3\times |X^{-1}{|}^2=5^3\times (24{)}^2=5\times (120{)}^2=k$
$k=5\times 2^2\times 3^2\times 4^2\times {\sigma }^2=2^6\times 3^2\times 5^3$
odd divisors $\to 3\times 4=\left(12\right)$

P)
$AX=\left[\begin{array}{ccc}\frac{1}{4}(x+\frac{1}{x})& 0& 0\\ 0& \frac{x}{2}& 0\\ 0& 0& \frac{16}{3}\end{array}\right]$
$Tr\left(AX\right)=\frac{1}{4}(x+\frac{1}{x})+\frac{x}{2}+\frac{16}{3}=f\left(x\right)$
$f\left(x\right)=\frac{3x}{4}+\frac{1}{4x}+\frac{16}{3}$
For min value apply $A.M\ge G.M$ for the expression
$\frac{3x}{4}+\frac{1}{4x}inf\left(x\right)$
$\to \frac{3x}{4}+\frac{1}{4x}\ge \frac{\sqrt{3}}{2}$
$\to f\left(x\right)\ge \frac{\sqrt{3}}{2}+\frac{16}{3}$
minimum value of [Tr(Ax)] = $[\frac{\sqrt{3}}{2}+\frac{16}{3}]$
Therefore the minimum value of [Tr(Ax)] = 6

R)
$Z=X^{-1}-2I=\left[\begin{array}{ccc}2& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right],Z^2=\left[\begin{array}{ccc}{2}^2& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]...$
$Z+Z^2+---Z^{10}=\left[\begin{array}{ccc}2+2^2+--+2^{10}& 0& 0\\ 0& 0& 0\\ 0& 0& 1+---+1\end{array}\right]$
$Tr(z+----+z^{10})=(2^1+-+2^{10}+10)$
$=2.(2^{10}-1)+10=2^{11}+8$
a + b = 19