If A=left[begin{array}{cc}0 & -tan frac{α}{2} tan frac{α}{2} & 0end{array}right] and I is

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3 and 4 .Determinants and Matrices

hard

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
$I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

Option A

Option B

Option C

Option D

Solution

$L.H.S.$

$I+A$

$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{array}\right]$

$\left[\begin{array}{cc}1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1\end{array}\right]$ ………. $(1)$

$R.H.S$ :

$(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

$=\left(\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{array}\right]\right)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

$=\left[\begin{array}{cc}1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1\end{array}\right]\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

$=\left[\begin{array}{cc}\cos \alpha+\sin \alpha \tan \frac{\alpha}{2} & -\sin \alpha+\cos \alpha \tan \frac{\alpha}{2} \\ -\cos \alpha \tan \frac{\alpha}{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha}{2}+\cos \alpha\end{array}\right]$ ………. $(2)$

$1-2 \sin ^2 \frac{\alpha}{2}+2 \sin \frac{\alpha}{2}-\cos \frac{\alpha}{2} \tan \frac{\alpha}{2} -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\left(2 \cos ^2 \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2}$

$-\left(2 \cos ^2 \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2}+1-2 \sin ^2 \frac{\alpha}{2}$

$1-2 \sin ^2 \frac{\alpha}{2}+2 \sin ^2 \frac{\alpha}{2} 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}-\tan \frac{\alpha}{2}$

$-2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} 2 \sin ^2 \frac{\alpha}{2}+1-2 \sin ^2 \frac{\alpha}{2}$

$=\left[\begin{array}{cc}1 -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} 1\end{array}\right]$

Thus, from $( 1 )$ and $( 2 )$, we get $L.H.S. = R.H.S.$

Standard 12

Mathematics

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easy

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medium

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hard

(JEE MAIN-2023)

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 $I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

Solution

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Let $A=\left[\begin{array}{cc}i & -i \\ -i & i\end{array}\right], i=\sqrt{-1}$.Then, the system of linear equations $A^{8}\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}8 \\ 64\end{array}\right]$ has :

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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
$I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$

$A = \left[ {\begin{array}{{20}{c}}4&6&{ – 1}\\3&0&2\\1&{ – 2}&5\end{array}} \right]$,$B = \left[ {\begin{array}{{20}{c}}2&4\\0&1\\{ – 1}&2\end{array}} \right],\,\,C = \left[ {\begin{array}{*{20}{c}}3\\1\\2\end{array}} \right]$, then the expression which is not defined is