Find x,0

1.	Find x,0
Answer» 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, $\boxed{ \bf \: {(A + B)}^{'} = A' + B' }$ $\boxed{ \bf \: {(A - B)}^{'} = A' - B' }$ $\boxed{ \bf \: {(kA)}^{'} = k \: A' } \: \sf \: where \: k \in \: R$ $\boxed{ \bf \: {(A' )}^{'} = A}$ $\boxed{ \bf \: {(AB)}^{'} = B' A' }$

Answer»

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,

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

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