Question

Solve this question. Question in attachment .​

Answer 1

GiveN :-

• Two matrices P and Q are given to us such that ,
• $\sf P = \left[ \begin{array}{cc} \sf sec\alpha & \sf tan\alpha \\ -\sf cot\alpha & \sf cos\alpha \end{array}\right]$
• $\sf Q = \left[ \begin{array}{cc} -\sf cos\alpha & \sf tan\alpha \\ -\sf cot\alpha & -\sf sec\alpha \end{array}\right]$

To FinD :-

• The value of $\sf 2P^{-1} + Q$

SolutioN :-

Two matrices are given and we need to find , 2P-¹ + Q . Let's Firstly find out Inverse of Matrix P . We know that Inverse of matrix say $\sf A= \left[ \begin{array}{cc}\sf a &\sf b\\ \sf c & \sf d \end{array}\right]$ is given by ,

$\sf:\implies \pink{ A^{-1}= \dfrac{1}{ad-bc}\left[\begin{array}{cc} \sf d & \sf -b \\\sf -c &\sf a \end{array}\right]}$

$\red{\bigstar}$ So that :-

$\sf:\implies P^{-1}= \dfrac{1}{\sf (sec\alpha)(cos\alpha)-(-cot\alpha)(tan\alpha)} \left[\begin{array}{cc} \sf cos\alpha & \sf -tan\alpha \\ \sf cot\alpha &\sf sec\alpha \end{array}\right]$

$\sf:\implies P^{-1}= \dfrac{1}{\sf (\cancel{sec\alpha}) \bigg(\dfrac{1}{\sf \cancel{sec\alpha}}\bigg) + \sf (\cancel{tan\alpha} )\bigg( \dfrac{1}{\cancel{tan\alpha}} \bigg) } \left[ \begin{array}{cc} \sf cos\alpha & \sf -tan\alpha \\ \sf cot\alpha &\sf sec\alpha \end{array}\right]$

$\sf:\implies P^{-1}= \dfrac{1}{1+1}\left[ \begin{array}{cc} \sf cos\alpha & \sf -tan\alpha \\ \sf cot\alpha &\sf sec\alpha \end{array}\right]$

$\sf:\implies\boxed{\pink{ \sf P^{-1}= \dfrac{1}{2}\left[ \begin{array}{cc} \sf cos\alpha & \sf -tan\alpha \\ \sf cot\alpha &\sf sec\alpha \end{array}\right] }}$

$\rule{200}2$

Now we need to find the value of 2P-¹ + Q. So on Multiplying P-¹ by 2 and adding Q to it , we have ,

$\sf:\implies 2P^{-1}+Q = 2\times\dfrac{1}{2} \left[ \begin{array}{cc} \sf cos\alpha & \sf -tan\alpha \\ \sf cot\alpha &\sf sec\alpha \end{array}\right] + \sf \left[ \begin{array}{cc} -\cos\alpha & \tan\alpha \\ -\cot\alpha & -\sec\alpha \end{array}\right] \\\\\sf:\implies 2P^{-1}+Q = \left[ \begin{array}{cc} \sf cos\alpha & \sf -tan\alpha \\ \sf cot\alpha &\sf sec\alpha \end{array}\right] + \sf \left[ \begin{array}{cc} -\cos\alpha & \tan\alpha \\ -\cot\alpha & -\sec\alpha \end{array}\right] \\\\\sf:\implies 2P^{-1}+Q = \left[\begin{array}{cc} \sf (cos \alpha-cos\alpha )& \sf (tan\alpha-tan\alpha) \\\sf (cot\alpha - cot\alpha) & \sf (sec\alpha-sec\alpha)\end{array}\right] \\\\\sf:\implies \underset{\blue{\sf Required\ Answer\lgroup Null \ Matrix \rgroup}}{\underbrace{\boxed{\pink{\frak{ 2P^{-1}+Q =\left[\begin{array}{cc} \sf 0 &\sf 0 \\\sf 0 &\sf 0 \end{array}\right] }}}}}$