For every real number x, \(\begin{bmatrix}\cos^2 x&\sin^2 x\\\sin^2 x&\cos^2 x\end{bmatrix} + \begin{bmatrix}\sin^2 x&\cos^2 x\\\cos^2 x&\sin^2 x\end{bmatrix}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}1&1\\1&1\end{bmatrix}\)
Matrices of the same order are added element by element:
\( \displaystyle \begin{bmatrix}\cos^2 x + \sin^2 x&\sin^2 x + \cos^2 x\\\sin^2 x + \cos^2 x&\cos^2 x + \sin^2 x\end{bmatrix} \)
Since \(\sin^2 x + \cos^2 x = 1\), every element is 1, and the sum is
\( \displaystyle \begin{bmatrix}1&1\\1&1\end{bmatrix} \)