Introduction to Determinants

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Question

$\mathrm{A}=\left[\begin{array}{cc}1& \mathrm{tan}\mathrm{x}\\ -\mathrm{tan}\mathrm{x}& 1\end{array}\right]$ and f(x) is defined as f(x) = det. $\left({\mathrm{A}}^{\mathrm{T}}{\mathrm{A}}^{-1}\right)$ then the value of   is $\left(\mathrm{n}\ge 2\right)$

Moderate
Solution

$\mathrm{A}=\left[\begin{array}{cc}1& \mathrm{tan}\mathrm{x}\\ -\mathrm{tan}\mathrm{x}& 1\end{array}\right]$Hence, detA = sec2x$\therefore \mathrm{det}{\mathrm{A}}^{\mathrm{T}}={\mathrm{sec}}^{2}\mathrm{x}$now $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{det}\cdot \left({\mathrm{A}}^{\mathrm{T}}{\mathrm{A}}^{-1}\right)$$\begin{array}{c}=\left(\mathrm{det}\cdot {\mathrm{A}}^{\mathrm{T}}\right)\left(\mathrm{det}\cdot {\mathrm{A}}^{-1}\right)=\left(\mathrm{det}\cdot {\mathrm{A}}^{\mathrm{T}}\right)\left(\mathrm{det}\cdot \mathrm{A}{\right)}^{-1}\\ =\frac{\mathrm{det}\cdot \left({\mathrm{A}}^{\mathrm{T}}\right)}{\mathrm{det}\cdot \left(\mathrm{A}\right)}=1\end{array}$Hence, f(x) =1

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