Question

If A= [cos 2x sin 2x -sin2x cos2x] is a matrix of 2 X 2, find A2​

Answer 1

Answer:

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Answer 2

Concept:

This question requires how to solve square of a matrix of a 2X2 matrix

$A^{2} =\left[\begin{array}{ccc}a&b\\c&d\end{array}\right] \left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\\A^{2} =\left[\begin{array}{ccc}a^{2}+bc &ab+bd\\ca+dc&cb+d^{2} \end{array}\right]$

Also, this question requires trigonometric equations like,

$cos^{2}x -sin^{2}x =cos2x ------(1)\\2sinxcosx=sin2x-------(2)$

Given:

$A=\left[\begin{array}{ccc}cos 2x&sin2x\\-sin2x&cos2x\\\end{array}\right]$

To find:

We need to find $A^{2}$

Solution:

Since we know that

$A^{2} =\left[\begin{array}{ccc}a^{2}+bc &ab+bd\\ca+dc&d^{2}+db \end{array}\right]$

where,

a= cos2x

b= sin2x

c= -sin2x

d= cos2x

$A^{2} =\left[\begin{array}{ccc}cos^{2}2x+sin2x(-sin2x) &cos2xsin2x+sinxcos2x\\(-sin2x)cos2x+cos2x(-sin2x)&cos^{2}2x+cos2xsin2x \end{array}\right]\\\\A^{2} =\left[\begin{array}{ccc}cos^{2}2x-sin^{2}2x &2cos2xsin2x\\-2sin2xcos2x&cos^{2}2x-sin^{2}2x \end{array}\right]$

From equations (1) and (2)

We can understand

$A^{2} =\left[\begin{array}{ccc}cos4x &sin4x\\-sin4x&cos4x\end{array}\right]$

Therefore, $A^{2} =\left[\begin{array}{ccc}cos4x &sin4x\\-sin4x&cos4x\end{array}\right]$