Question:

Compute the following:
(i)[abba]+[abba](i)\begin{bmatrix}a&b\\-b&a\end{bmatrix}+\begin{bmatrix}a&b\\b&a\end{bmatrix}
(ii)[a2+b2b2+c2a2+c2a2+b2]+[2ab2bc2ac2ab](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)[1468516285]+[1276805324](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)[cos2xsin2xsin2xcos2x]+[sin2xcos2xcos2xsin2x](iv)\begin{bmatrix}cos^2x& sin^2x\\ sin^2x& cos^2x\end{bmatrix}+\begin{bmatrix}sin^2x& cos^2x\\ cos^2x& sin2x\end{bmatrix}

Updated On: Sep 23, 2023
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Solution and Explanation

(i)[abba]+[abba](i)\begin{bmatrix}a&b\\-b&a\end{bmatrix}+\begin{bmatrix}a&b\\b&a\end{bmatrix}=[a+ab+bb+ba+a]=\begin{bmatrix}a+a& b+b\\ -b+b& a+a\end{bmatrix}=[2a2b02a]=\begin{bmatrix}2a& 2b\\ 0& 2a\end{bmatrix}


(ii)[a2+b2b2+c2a2+c2a2+b2]+[2ab2bc2ac2ab](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+2bca2+c22aca2+b22ab]=\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(ac)2(ab)2]=\begin{bmatrix}(a+b)^2& (b+c)^2\\ (a-c)^2& (a-b)^2\end{bmatrix}


(iii)[1468516285]+[1276805324](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+76+68+85+016+52+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}
=[11110165215109]=\begin{bmatrix}11& 11& 0\\ 16& 5& 21\\ 5& 10& 9\end{bmatrix}


(iv)[cos2xsin2xsin2xcos2x]+[sin2xcos2xcos2xsin2x](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+cos2xsin2x+cos2xcos2x+sin2x]=\begin{bmatrix}cos^2x+sin^2x& sin^2x+cos^2x\\ sin^2x+cos^2x& cos^2x+sin^2x\end{bmatrix}
=[1111]=\begin{bmatrix}1& 1\\ 1& 1\end{bmatrix}(sin2x+cos2x=1)(\because sin^2x+cos^2x=1)

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