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Question: If A =\(\begin{bmatrix} 1 & \tan x \\ –\tan x & 1 \end{bmatrix}\) then ATA–1 =...

If A =[1tan⁡x–tan⁡x1]\begin{bmatrix} 1 & \tan x \\ –\tan x & 1 \end{bmatrix} then ATA–1 =

A

[cos⁡2x–sin⁡2xsin⁡2xcos⁡2x]\begin{bmatrix} \cos 2x & –\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}

B

[–cos⁡2xsin⁡2x–sin⁡2xcos⁡2x]\begin{bmatrix} –\cos 2x & \sin 2x \\ –\sin 2x & \cos 2x \end{bmatrix}

C

[sin⁡2xcos⁡2xcos⁡2xsin⁡2x]\begin{bmatrix} \sin 2x & \cos 2x \\ \cos 2x & \sin 2x \end{bmatrix}

D

None of these

Answer

[cos⁡2x–sin⁡2xsin⁡2xcos⁡2x]\begin{bmatrix} \cos 2x & –\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}

Explanation

Solution

|A| =∣1tan⁡x–tan⁡x1∣\left| \begin{matrix} 1 & \tan x \\ –\tan x & 1 \end{matrix} \right| = 1 + tan2x ¹ 0.

So A is invertible. Let Cij be the cofactor of aij in A = [aij]

Then C11 = (–1)1 + 1 1 = 1,

C12 = (–1)1+2 (– tanx) = tanx

C21 = (–1)2+1 tanx = – tanx , C22 = (–1)2+2 . 1

\ adj A=[1tan⁡x–tan⁡x1]T\begin{bmatrix} 1 & \tan x \\ –\tan x & 1 \end{bmatrix}^{T}= [1–tan⁡xtan⁡x1]\begin{bmatrix} 1 & –\tan x \\ \tan x & 1 \end{bmatrix}

Now, A–1 = 1∣A∣\frac{1}{|A|} adj A

Ž A–1 = 1(1+tan⁡2x)\frac{1}{(1 + \tan^{2}x)} [1–tan⁡xtan⁡x1]\begin{bmatrix} 1 & –\tan x \\ \tan x & 1 \end{bmatrix}

= [11+tan⁡2x−tan⁡x1+tan⁡2xtan⁡x1+tan⁡2x11+tan⁡2x]\left[ \begin{array} { c c } \frac { 1 } { 1 + \tan ^ { 2 } x } & \frac { - \tan x } { 1 + \tan ^ { 2 } x } \\ \frac { \tan x } { 1 + \tan ^ { 2 } x } & \frac { 1 } { 1 + \tan ^ { 2 } x } \end{array} \right]

\ AT A–1 = [1–tan⁡xtan⁡x1]\begin{bmatrix} 1 & –\tan x \\ \tan x & 1 \end{bmatrix}

= [1–tan⁡2x1+tan⁡2x–2tan⁡x1+tan⁡2x2tan⁡x1+tan⁡2x1–tan⁡2x1+tan⁡2x]\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} =[cos⁡2x–sin⁡2xsin⁡2xcos⁡2x]\begin{bmatrix} \cos 2x & –\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}