Determinants Matrices Question 67
Question: If $ A= \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \\ \end{bmatrix} $ , then $ A^{T}{A^{-1}} $ is
Options:
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 & \cos 2x \\ \sin 2x & \sin 2x \\ \end{bmatrix} $
D) none of these
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Answer:
Correct Answer: B
Solution:
- [b] $ | A |= \begin{vmatrix} 1 & \tan x \\ -\tan x & 1 \\ \end{vmatrix}=1+{{\tan }^{2}}x\ne 0 $ So A is invertible. Also, Adj $ A={{ \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \\ \end{bmatrix} }^{T}}= \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \\ \end{bmatrix} $ Now, $ {A^{-1}}=\frac{1}{| A |}adjA $
$ =\frac{1}{(1+{{\tan }^{2}}x)} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \\ \end{bmatrix} $
$ = \begin{bmatrix} \frac{1}{1+{{\tan }^{2}}x} & \frac{-\tan x}{1+{{\tan }^{2}}x} \\ \frac{\tan x}{1+{{\tan }^{2}}} & \frac{1}{1+{{\tan }^{2}}x} \\ \end{bmatrix} $
$ \therefore A^{T}{A^{-1}}= \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \\ \end{bmatrix} $
$ \begin{bmatrix} \frac{1}{1+{{\tan }^{2}}x} & \frac{-\tan x}{1+{{\tan }^{2}}x} \\ \frac{\tan x}{1+{{\tan }^{2}}} & \frac{1}{1+{{\tan }^{2}}x} \\ \end{bmatrix} $ = $ \begin{bmatrix} \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} & \frac{-2\tan x}{1+{{\tan }^{2}}x} \\ \frac{2\tan x}{1+{{\tan }^{2}}x} & \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \\ \end{bmatrix} $
$ = \begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \\ \end{bmatrix} $
BETA
