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# Compute the following $(iv)\;\begin{bmatrix}\cos^2x & \sin^2x\\ \sin^2x & \cos^2x\end{bmatrix}+\begin{bmatrix}\sin^2x & \cos^2x\\ \cos^2x & \sin^2x\end{bmatrix}$

Note: This is the 4th part of a 4 part question, which is split as 4 separate questions here.

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## 1 Answer

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Toolbox:
• The sum $A+B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A + B)_{i,j} = A_{i,j} + B_{i,j}$ where 1 $\leq$ i $\leq$ m and 1$\leq$ j $\leq$ n.
• $\sin^2x+\cos^2x = 1$
$\begin{bmatrix}\cos^2x & \sin^2x\\ \sin^2x & \cos^2x\end{bmatrix}+\begin{bmatrix}\sin^2x & \cos^2x\\ \cos^2x & \sin^2x\end{bmatrix} = \begin{bmatrix}\cos^2x + \sin^2x& \sin^2x+\cos^2x\\ \sin^2x +cos^2x& \cos^2x+\sin^2x\end{bmatrix}$.
Since $\sin^2x+\cos^2x = 1, \begin{bmatrix}\cos^2x & \sin^2x\\ \sin^2x & \cos^2x\end{bmatrix}+\begin{bmatrix}\sin^2x & \cos^2x\\ \cos^2x & \sin^2x\end{bmatrix} = \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}$.

answered Feb 27, 2013
edited Dec 20, 2013

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