info@clay6.com
+91-9566306857
(9am to 6pm)
logo

Ask Questions, Get Answers

X
Want help in doing your homework? We will solve it for you. Click to know more.
 
Home  >>  CBSE XII  >>  Math  >>  Matrices

If $A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix} $ , Evaluate $A^TA^{-1}$


$(A)\; \begin{bmatrix} -\cos 2x & \sin 2x \\ \sin 2x & -\cos 2x \end{bmatrix}$
$(B)\; \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}$
$(C)\; \begin{bmatrix} \cos 2x & -\sin 2x \\ -\sin 2x & \cos 2x \end{bmatrix}$
$(D)\; \begin{bmatrix} -\cos 2x & \sin 2x \\ -\sin 2x & \cos 2x \end{bmatrix}$

1 Answer

Need homework help? Click here.
$A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix} \; \; \; \;\; \therefore A^T = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$
$adj\; A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$
$|A| = 1+ \tan^2x$
$\therefore A^{-1} =\frac{1}{1+ \tan^2x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$
$\therefore A^T.A^{-1} = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} . \frac{1}{1+ \tan^2x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$
$= \frac{1}{\tan ^2x} \begin{bmatrix} 1 -\tan^2x & -\tan x -\tan x \\ \tan x + \tan x & -\tan^2x +1\end{bmatrix}$
$= \begin{bmatrix} \frac{1-\tan^2x}{1+\tan^2x} & \frac{-2\tan x}{1+\tan^2x} \\ \frac{2\tan x}{1+\tan^2x} & \frac{1-tan^2x}{1+tan^2x} \end{bmatrix} $
$= \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}$
answered Dec 1, 2016 by priyanka.c
 

Related questions

...