Question

Find x,0<x<π/2, when A+A' = I. If

[cos x sin x]
[sin x cos x] = A

Please tell me the answer
solve on the paper and please upload it ​

Answer 1

Correct Statement is

Find x,

$\sf \: If \: A = \begin{bmatrix} cosx & - sinx\\ sinx & cosx\end{bmatrix}, \: 0 < x < \dfrac{\pi}{2} \: when \: A + A' = I$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:A = \begin{bmatrix} cosx & - sinx\\ sinx & cosx\end{bmatrix}$

So,

$\rm :\longmapsto\:A' = \begin{bmatrix} cosx & sinx\\ - sinx & cosx\end{bmatrix}$

Now,

According to statement,

$\rm :\longmapsto\:A + A' = I$

$\rm :\longmapsto\:\begin{bmatrix} cosx & - sinx\\ sinx & cosx\end{bmatrix} + \begin{bmatrix} cosx & sinx\\ - sinx & cosx\end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$

$\rm :\longmapsto\:\begin{bmatrix} 2cosx & 0\\ 0 & 2cosx\end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$

On comparing, we get

$\rm :\longmapsto\:2cosx = 1$

$\rm :\longmapsto\:cosx = \dfrac{1}{2}$

$\rm :\longmapsto\:cosx = cos60 \degree$

$\bf\implies \:x = 60 \degree \:$

Additional Information :-

For any two matrices A and B,

$\boxed{ \bf \: {(A + B)}^{'} = A' + B' }$

$\boxed{ \bf \: {(A - B)}^{'} = A' - B' }$

$\boxed{ \bf \: {(kA)}^{'} = k \: A' } \: \sf \: where \: k \in \: R$

$\boxed{ \bf \: {(A' )}^{'} = A}$

$\boxed{ \bf \: {(AB)}^{'} = B' A' }$