Question

If trace of the matrix $\left[\begin{array}{ccc}2x& 2& 2x\\ 0& x^2& -1\\ 1& 3x& -4\end{array}\right]$ is given by $Tr\left(A\right)=\underset{n\to 0}{lim}{\left(\frac{2^n+4^n+8^n}{3}\right)}^{1/n},$ then possible value(s) of $x$ is(are)

• A. $2$
• B. $4$
• C. $-2$
• D. $-4$

Answer 1

Answer: (A), (D)

$-4$

Let $A=\left[\begin{array}{ccc}2x& 2& 2x\\ 0& x^2& -1\\ 1& 3x& -4\end{array}\right]$
Trace of the matrix $Tr\left(A\right)=2x+x^2-4$
Now,
$Tr\left(A\right)=\underset{n\to 0}{lim}{\left(\frac{2^n+4^n+8^n}{3}\right)}^{1/n}=e^{\underset{n\to 0}{lim}\left((\frac{2^n+4^n+8^n}{3}-1)\frac{1}{n}\right)}=e^{\underset{n\to 0}{lim}\left(\frac{2^n+4^n+8^n-3}{3n}\right)}=e^{\underset{n\to 0}{lim}\left((\frac{2^n-1}{n}+\frac{4^n-1}{n}+\frac{8^n-1}{n})\frac{1}{3}\right)}=e^{\underset{n\to 0}{lim}\left(\frac{\ln 2+\ln 4+\ln 8}{3}\right)}=e^{\underset{n\to 0}{lim}\left(\frac{1}{3}\ln \right(2\cdot 4\cdot 8\left)\right)}=\sqrt[3]{32\cdot 4\cdot 8}=4$
$\therefore x^2+2x-4=4$
$\Rightarrow x^2+2x-8=0$
$\Rightarrow (x+4)(x-2)=0$
$\Rightarrow x=2,-4$