91影视

Consider the matrix of the given transformation be T

Let $T\left(f\right)=f"+2f\text{'}+3f$be the transformation.

Assume$f\left(x\right)=a\cos \left(x\right)+b\sin \left(x\right)$be the function.

Hence$B=\left[{\left[T\left(\cos x\right)\right]}_B\text{鈥夆赌夆赌�}{\left[T\left(\sin x\right)\right]}_B\text{鈥�}\right]$

Then, the transformation be as follows.

role="math" localid="1660309390137" $\begin{array}{c}T\cos \left(x\right)=鈭�\cos \left(x\right)鈭�2\sin \left(x\right)+3\cos \left(x\right)\\ \text{鈥夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆��}=鈭�2\sin \left(x\right)+2\cos \left(x\right)\\ \text{鈥夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆��}=2\cos \left(x\right)鈭�2\sin \left(x\right)\end{array}$

Similarly further simplification is as follows

$\begin{array}{c}T\left(\sin x\right)=鈭�\sin \left(x\right)+2\cos \left(x\right)+3\sin \left(x\right)\\ \text{鈥夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆��}=2\sin \left(x\right)+2\cos \left(x\right)\\ \text{鈥夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆�夆赌夆赌夆��}=2\cos \left(x\right)+2\sin \left(x\right)\end{array}$

Therefore the matrix be B as follows

$B=\left(\begin{array}{cc}2& 2\\ 鈭�2& 2\end{array}\right)$

Hence the solution.

Consider $B=\left(\begin{array}{cc}2& 2\\ 鈭�2& 2\end{array}\right)$ be the matrix of the given transformation T.

$\begin{array}{c}\left[B\right]=\det \left(B\right)\\ \text{鈥夆赌夆赌夆�夆赌夆赌夆�夆��}=ad鈭�bc\\ \text{鈥夆赌夆赌夆�夆赌夆赌夆�夆��}=4鈭�(鈭�4)\\ \text{鈥夆赌夆赌夆�夆赌夆赌夆��}=8鈮�0\end{array}$

Thus the transformation T is invertible as well.So T is an isomorphism.

Hence the solution.