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Mathematics Question on Matrices

Compute the following:
(i)[ab−ba]+[ab\ba](i)\begin{bmatrix}a&b\\\\-b&a\end{bmatrix}+\begin{bmatrix}a&b\\\b&a\end{bmatrix}
(ii)[a2+b2b2+c2 a2+c2a2+b2]+[2ab2bc−2ac−2ab](ii)\begin{bmatrix}a^2+b^2& b^2+c^2\\\ a^2+c^2& a^2+b^2\end{bmatrix}+\begin{bmatrix}2ab& 2bc \\\\-2ac& -2ab\end{bmatrix}
(iii)[−14−6 8516 285]+[1276\805 324](iii)\begin{bmatrix}-1&4&-6\\\ 8&5&16\\\ 2&8&5\end{bmatrix}+\begin{bmatrix}12&7&6 \\\8&0&5\\\ 3&2&4\end{bmatrix}
(iv)[cos2xsin2x sin2xcos2x]+[sin2xcos2x cos2xsin2x](iv)\begin{bmatrix}cos^2x& sin^2x\\\ sin^2x& cos^2x\end{bmatrix}+\begin{bmatrix}sin^2x& cos^2x\\\ cos^2x& sin2x\end{bmatrix}

Answer

(i)\begin{bmatrix}a&b\\\\-b&a\end{bmatrix}+\begin{bmatrix}a&b\\\b&a\end{bmatrix}$$=\begin{bmatrix}a+a& b+b\\\ -b+b& a+a\end{bmatrix}$$=\begin{bmatrix}2a& 2b\\\ 0& 2a\end{bmatrix}

(ii)[a2+b2b2+c2 a2+c2a2+b2]+[2ab2bc−2ac−2ab](ii)\begin{bmatrix}a^2+b^2& b^2+c^2\\\ a^2+c^2& a^2+b^2\end{bmatrix}+\begin{bmatrix}2ab& 2bc \\\\-2ac& -2ab\end{bmatrix}
=[a2+b2+2abb2+c2+2bc a2+c2−2aca2+b2−2ab]=\begin{bmatrix}a^2+b^2+2ab& b^2+c^2+2bc\\\ a^2+c^2-2ac& a^2+b^2-2ab\end{bmatrix}
=[(a+b)2(b+c)2 (a−c)2(a−b)2]=\begin{bmatrix}(a+b)^2& (b+c)^2\\\ (a-c)^2& (a-b)^2\end{bmatrix}

(iii)[−14−6 8516 285]+[1276\805 324](iii)\begin{bmatrix}-1&4&-6\\\ 8&5&16\\\ 2&8&5\end{bmatrix}+\begin{bmatrix}12&7&6 \\\8&0&5\\\ 3&2&4\end{bmatrix}
=[−1+124+7−6+6 8+85+016+5 2+38+25+4]=\begin{bmatrix}-1+12& 4+7& -6+6\\\ 8+8& 5+0& 16+5\\\ 2+3& 8+2& 5+4\end{bmatrix}
=[11110 16521 5109]=\begin{bmatrix}11& 11& 0\\\ 16& 5& 21\\\ 5& 10& 9\end{bmatrix}

(iv)[cos2xsin2x sin2xcos2x]+[sin2xcos2x cos2xsin2x](iv)\begin{bmatrix}cos^2x& sin^2x\\\ sin^2x& cos^2x\end{bmatrix}+\begin{bmatrix}sin^2x& cos^2x\\\ cos^2x& sin2x\end{bmatrix}
=[cos2x+sin2xsin2x+cos2x sin2x+cos2xcos2x+sin2x]=\begin{bmatrix}cos^2x+sin^2x& sin^2x+cos^2x\\\ sin^2x+cos^2x& cos^2x+sin^2x\end{bmatrix}
=\begin{bmatrix}1& 1\\\ 1& 1\end{bmatrix}$$(\because sin^2x+cos^2x=1)