Question

12 24 8N If trace of the matrix 22 0 is given by TrA) = lim Tr1 then possible value(s) 1 of I is(are) A 2 B4 C -2 D-4​

Answer 1

Answer:

Step-by-step explanation:

Given matrix is,

$A=\left[\begin{array}{ccc}2x&2&2x\\0&x^2&-1\\1&3x&-4\end{array}\right]$

Now, the trace of matrix A

$Tr(A)=2x+x^2-4$

Also, given

$\displaystyle\,Tr(A)=\lim_{n\to0}\,\left(\dfrac{2^n+4^n+8^n}{3}\right)^{\frac{1}{n} }$

So,

$\displaystyle\,\implies\,x^2+2x-4=\lim_{n\to0}\,\left(\dfrac{2^n+4^n+8^n}{3}\right)^{\frac{1}{n} }$

$\displaystyle\,\implies\,x^2+2x-4=\lim_{n\to0}\,e^{\displaystyle\ln\left(\dfrac{2^n+4^n+8^n}{3}\right)^{\frac{1}{n}} }$

$\displaystyle\,\implies\,x^2+2x-4=\lim_{n\to0}\,e^{\displaystyle\dfrac{1}{n}\ln\left(\dfrac{2^n+4^n+8^n}{3}\right) }$

$\displaystyle\,\implies\,x^2+2x-4=\lim_{n\to0}\,e^{\displaystyle\dfrac{1}{n}\left\{\ln\left(2^n+4^n+8^n\right)-\ln(3)\right\} }$

$\displaystyle\,\implies\,x^2+2x-4=\lim_{n\to0}\,e^{\displaystyle\dfrac{\ln\left(2^n+4^n+8^n\right)-\ln(3)}{n} }$

$\displaystyle\,\implies\,x^2+2x-4=\lim_{n\to0}\,e^{\displaystyle\dfrac{\ln\left(2^n+4^n+8^n\right)}{n}-\dfrac{\ln(3)}{n} }$

$\displaystyle\,\implies\,x^2+2x-4=e^{\displaystyle\,\,\lim_{n\to0}\dfrac{\ln\left(2^n+4^n+8^n\right)}{n}-\lim_{n\to0}\dfrac{\ln(3)}{n} }$

Applying L'hospital's rule,

$\displaystyle\,\implies\,x^2+2x-4=e^{\displaystyle\,\,\lim_{n\to0}\dfrac{2^n\cdot\ln(2)+4^n\cdot\ln(4)+8^n\cdot\ln(8)}{\left(2^n+4^n+8^n\right)\cdot1}-\lim_{n\to0}\dfrac{0}{1}}$

$\displaystyle\,\implies\,x^2+2x-4=e^{\displaystyle\,\,\lim_{n\to0}\dfrac{2^n\cdot\ln(2)+\left(2^n\right)^2\cdot2\ln(2)+\left(2^n\right)^3\cdot3\ln(2)}{2^n+4^n+8^n}-0}$

$\displaystyle\,\implies\,x^2+2x-4=e^{\displaystyle\,\,\lim_{n\to0}\dfrac{2^n\cdot\ln(2)\{1+2^n\cdot2+\left(2^n\right)^2\cdot3\}}{2^n\{1+2^n+4^n\}}}$

$\displaystyle\,\implies\,x^2+2x-4=e^{\displaystyle\,\,\lim_{n\to0}\dfrac{\ln(2)\{1+2^n\cdot2+4^n\cdot3\}}{1+2^n+4^n}}$

$\displaystyle\,\implies\,x^2+2x-4=e^{\displaystyle\,\,\dfrac{\ln(2)\{1+2+3\}}{1+1+1}}$

$\displaystyle\,\implies\,x^2+2x-4=e^{\displaystyle2\,\ln(2)}$

$\displaystyle\,\implies\,x^2+2x-4=e^{\displaystyle\ln\left(2^2\right)}$

$\displaystyle\,\implies\,x^2+2x-4=e^{\displaystyle\ln\left(4\right)}$

$\displaystyle\,\implies\,x^2+2x-4=4$

$\displaystyle\,\implies\,x^2+2x-4-4=0$

$\displaystyle\,\implies\,x^2+2x-8=0$

$\displaystyle\,\implies\,x^2+4x-2x-8=0$

$\displaystyle\,\implies\,x(x+4)-2(x+4)=0$

$\displaystyle\,\implies\,(x-2)(x+4)=0$

$\displaystyle\,\implies\,x=2\,\,\,\,or\,\,\,\,x=-4$

Answer 2

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm :\longmapsto\:A = \begin{gathered}\left[\begin{array}{ccc}2x&2&2x\\0&x^2&-1\\1&3x&-4\end{array}\right]\end{gathered}$

Now, the trace of matrix A is defined as sum of the diagonal elements of matrix and denoted as Tr(A)

So,

$\red{\rm :\longmapsto\:Tr(A) = {x}^{2} + 2x - 4 - - - - (1)}$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to 0} {\bigg[\dfrac{ {2}^{n} + {4}^{n} + {8}^{n} }{3} \bigg]}^{\dfrac{1}{n} }$

can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{n \to 0} {\bigg[1 + \dfrac{{2}^{n} + {4}^{n} + {8}^{n} }{3} - 1 \bigg]}^{\dfrac{1}{n} }$

$\rm \:  =  \: \displaystyle\lim_{n \to 0} {\bigg[1 + \dfrac{{2}^{n} + {4}^{n} + {8}^{n} - 3 }{3} \bigg]}^{\dfrac{1}{n} }$

$\rm \:  =  \: \displaystyle\lim_{n \to 0} {\bigg[1 + \dfrac{{2}^{n} + {4}^{n} + {8}^{n} - 3 }{3} \bigg]}^{\dfrac{3}{ {2}^{n} + {4}^{n} + {8}^{n} - 3} \times \dfrac{{2}^{n} + {4}^{n} + {8}^{n} - 3 }{3n}}$

$\rm \:  =  \: {\bigg[e\bigg]}^{\displaystyle\lim_{n \to 0} \dfrac{{2}^{n} + {4}^{n} + {8}^{n} - 3 }{3n}}$

$\rm \:  =  \: {\bigg[e\bigg]}^{\displaystyle\lim_{n \to 0} \dfrac{{2}^{n} + {4}^{n} + {8}^{n} - 1 - 1 - 1}{3n}}$

$\rm \:  =  \: {\bigg[e\bigg]}^{\displaystyle\lim_{n \to 0} \dfrac{({2}^{n} - 1) + ({4}^{n} - 1) +( {8}^{n} - 1)}{3n}}$

$\rm \:  =  \: {e}^{\displaystyle\lim_{n \to 0}\dfrac{1}{3} \bigg[\dfrac{ {2}^{n} - 1}{n} + \dfrac{ {4}^{n} - 1}{n} + \dfrac{ {8}^{n} - 1}{n} \bigg]}$

$\rm \:  =  \: {e}^{\bigg[\dfrac{1}{3}(log2 \: + \: log4 + \: log8 )\bigg]}$

$\rm \:  =  \: {e}^{\bigg[\dfrac{1}{3}log(2 \times 4 \times 8) \:\bigg]}$

$\rm \:  =  \: {e}^{\bigg[\dfrac{1}{3}log(64) \:\bigg]}$

$\rm \:  =  \: {e}^{\bigg[\dfrac{1}{3}log( {4}^{3} ) \:\bigg]}$

$\rm \:  =  \: {e}^{\bigg[\dfrac{1}{3} \times 3log(4) \:\bigg]}$

$\rm \:  =  \: {e}^{log4}$

$\rm \:  =  \: 4$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{n \to 0} {\bigg[\dfrac{ {2}^{n} + {4}^{n} + {8}^{n} }{3} \bigg]}^{\dfrac{1}{n} } = 4}} - - - (2)$

According to statement,

$\rm :\longmapsto\:\boxed{\tt{Tr(A) = \displaystyle\lim_{n \to 0} {\bigg[\dfrac{ {2}^{n} + {4}^{n} + {8}^{n} }{3} \bigg]}^{\dfrac{1}{n} } }}$

So, on substituting the values, we get

$\rm :\longmapsto\: {x}^{2} + 2x - 4 = 4$

$\rm :\longmapsto\: {x}^{2} + 2x - 4 - 4 = 0$

$\rm :\longmapsto\: {x}^{2} + 2x -8 = 0$

$\rm :\longmapsto\: {x}^{2} + 4x - 2x -8 = 0$

$\rm :\longmapsto\:x(x + 4) - 2(x + 4) = 0$

$\rm :\longmapsto\:(x + 4)(x - 2) = 0$

$\bf\implies \:x = - 4 \: \: or \: \: x = 2$

So, option (a) and (d) is correct.

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Formula Used :-

$\boxed{\tt{ \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1 }{x} = loga \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0} {\bigg[1 + x \bigg]}^{\dfrac{1}{x} } = e \: }}$

$\boxed{\tt{ logx \: + \: logy = log(xy) \: }}$

$\boxed{\tt{ log {x}^{y} = y \: logx \: }}$