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CGPET 2011 Solved Paper

Section: Mathematics

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Question : 106 of 150

Marks: +1, -0

If

A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}

, then the value of

\left| A^{T} A^{-1} \right|

$\cos 4x$
$\sec^2 x$
$-\cos 4x$
1

Solution:

We have,

A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}

\text{ Now, } A^{T} = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}

\operatorname{adj}(A) = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}

\text{ and } A^{-1} = \frac{1}{|A|}(\operatorname{adj} A)

= \frac{1}{\sec^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}

Now,

\left| A^{T} A^{-1} \right|

= \left| \frac{1}{\sec^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \right|

= \left| \frac{1}{\sec^2 x} \begin{bmatrix} 1 - \tan^2 x & -\tan x - \tan x \\ \tan x + \tan x & 1 - \tan^2 x \end{bmatrix} \right|

= \frac{1}{\sec^2 x} \begin{bmatrix} 1 - \tan^2 x & -2\tan x \\ 2\tan x & 1 - \tan^2 x \end{bmatrix}

= \frac{(1 - \tan^2 x)^2 + 4\tan^2 x}{\sec^2 x}

= \frac{(1 + \tan^2 x)^2}{\sec^2 x}

= \sec^2 x

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