Question

$\color{red}\rm If \: A= \left[\begin{array}{ccc} \rm{2}^x &0&0\\ \rm0& \rm{2}^{ {2}^{x} } &0\\0&0& \rm {2}^{ {2}^{ {2}^{x} } } \end{array}\right] , \: then \: find \\ \color{blue}\displaystyle \rm \int det(A)dx$​

Answer 1

Step-by-step explanation:

We have,

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

Now,

$\det(A)= \left| \begin{array}{ccc}{2}^x &0&0\\ 0& {2}^{ {2}^{x} } &0\\ 0& 0& {2}^{ {2}^{ {2}^{x} } } \end{array}\right|$

$\implies \det(A)= {2}^{x} \left| \begin{array}{cc} {2}^{ {2}^{x} } &0\\ 0& {2}^{ {2}^{ {2}^{x} } } \end{array}\right|$

$\implies \det(A)= {2}^{x} \left( {2}^{ {2}^{x} } \cdot {2}^{ {2}^{ {2}^{x} } } - 0\right)$

$\implies \det(A)= {2}^{x} \cdot {2}^{ {2}^{x} } \cdot {2}^{ {2}^{ {2}^{x} } }$

Now,

$\displaystyle \int \det(A) \: dx$

$\displaystyle = \int {2}^{x} \cdot {2}^{ {2}^{x} } \cdot {2}^{ {2}^{ {2}^{x} } } \: dx$

$\displaystyle = \int {2}^{ {2}^{ {2}^{x} } } \cdot{2}^{ {2}^{x} } \cdot {2}^{x} \: dx$

$\bf{Put \: \: {2}^{ {2}^{ {2}^{x} } } = t }$

$\bf{ \mapsto \: \: {2}^{ {2}^{ {2}^{x} } } \cdot ln(2) \cdot {2}^{ {2}^{x} } \cdot ln(2) \cdot {2}^{x} \cdot ln(2) dx =d t }$

$\bf{ \mapsto \: \: {2}^{ {2}^{ {2}^{x} } } \cdot {2}^{ {2}^{x} } \cdot {2}^{x} \cdot \big \{ ln(2) \big \} ^{3} dx =d t }$

$\bf{ \mapsto \: \: {2}^{ {2}^{ {2}^{x} } } \cdot {2}^{ {2}^{x} } \cdot {2}^{x} dx = \dfrac{d t}{\big \{ ln(2) \big \} ^{3} } }$

So,

$\displaystyle = \int \dfrac{dt}{ \big \{ \ln(2) \big \}^{3} }$

$= \dfrac{t}{ \big \{ \ln(2) \big \}^{3} } + c$

$= \dfrac{ {2}^{ {2}^{ {2}^{x} } } }{ \big \{ \ln(2) \big \}^{3} } + c$