If A= [ cc 0 & - 2 2 & 0 ] and I is the identity matrix of order 2 , show that I+A=(I-A) [ cc & - & ]

L.H.S. I + A = [ * 20 c 1&0 0&1 ] + [ * 20 c 0& - 2 2 &0 ] = [ * 20 c 1& - 2 2 &1 ] Now, I - A = [ * 20 c 1&0 0&1 ] - [ * 20 c 0& - 2 2 &0 ] = [ * 20 c 1& 2 - 2 &1 ] R.H.S. = (I - A) [ * 20 c & - & ] = [ * 20 c 1& 2 - 2 &1 ] [ * 20 c & - & ] = [ * 20 c + 2 & - + 2 - 2 + & 2 + ] = [ * 20 c + 2 2 & - + 2 2 - 2 2 + & 2 2 + ] = [ * 20 c 2 + 2 2 & - 2 + 2 2 - 2 + 2 2 & 2 + 2 2 ] = [ * 20 c ( - 2 ) 2 & - ( - 2 ) 2 ( - 2 ) 2 & ( - 2 ) 2 ] = [ * 20 c 2 2 & - 2 2 2 2 & 2 2 ] = [ * 20 c 1& - 2 2 &1 ] L.H.S. = R.H.S. Proved.

If A= [ cc 0 & - 2 2 & 0 ] and I is the identity matrix of order …

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Question

If $A=\left[\begin{array}{cc}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{array}\right]$ and I is the identity matrix of order 2 , show that
$I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

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Answer

L.H.S. $I + A = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \frac{\alpha }{2}} \\ {\tan \frac{\alpha }{2}}&0 \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} 1&{ - \tan \frac{\alpha }{2}} \\ {\tan \frac{\alpha }{2}}&1 \end{array}} \right]$
Now, $I - A = \left[ {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&{ - \tan \frac{\alpha }{2}} \\ {\tan \frac{\alpha }{2}}&0 \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} 1&{\tan \frac{\alpha }{2}} \\ { - \tan \frac{\alpha }{2}}&1 \end{array}} \right]$
R.H.S. $ = (I - A)\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}} 1&{\tan \frac{\alpha }{2}} \\ { - \tan \frac{\alpha }{2}}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {\cos \alpha + \sin \alpha \tan \frac{\alpha }{2}}&{ - \sin \alpha + \cos \alpha \tan \frac{\alpha }{2}} \\ { - \cos \alpha \tan \frac{\alpha }{2} + \sin \alpha }&{\sin \alpha \tan \frac{\alpha }{2} + \cos \alpha } \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {\cos \alpha + \sin \alpha \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{ - \sin \alpha + \cos \alpha \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \\ { - \cos \alpha \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}} + \sin \alpha }&{\sin \alpha \frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}} + \cos \alpha } \end{array}} \right]$
$=\left[ {\begin{array}{*{20}{c}} {\frac{{\cos \alpha \cos \frac{\alpha }{2} + \sin \alpha \sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{\frac{{ - \sin \alpha \cos \frac{\alpha }{2} + \cos \alpha \sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \\ {\frac{{ - \cos \alpha \sin \frac{\alpha }{2} + \sin \alpha \cos \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{\frac{{\sin \alpha \sin \frac{\alpha }{2} + \cos \alpha \cos \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \end{array}} \right]$
$ = \left[ {\begin{array}{*{20}{c}} {\frac{{\cos \left( {\alpha - \frac{\alpha }{2}} \right)}}{{\cos \frac{\alpha }{2}}}}&{\frac{{ - \sin \left( {\alpha - \frac{\alpha }{2}} \right)}}{{\cos \frac{\alpha }{2}}}} \\ {\frac{{\sin \left( {\alpha - \frac{\alpha }{2}} \right)}}{{\cos \frac{\alpha }{2}}}}&{\frac{{\cos \left( {\alpha - \frac{\alpha }{2}} \right)}}{{\cos \frac{\alpha }{2}}}} \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} {\frac{{\cos \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{\frac{{ - \sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \\ {\frac{{\sin \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}}&{\frac{{\cos \frac{\alpha }{2}}}{{\cos \frac{\alpha }{2}}}} \end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}} 1&{ - \tan \frac{\alpha }{2}} \\ {\tan \frac{\alpha }{2}}&1 \end{array}} \right]$
$\therefore$ L.H.S. = R.H.S. Proved.

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