Question

Ans it (ans-null matrix)​

Answer 1

$\huge\tt{\red{Given}}$

★ A matrix P=$\left[\begin{array}{ c c }\sec\alpha&\tan\alpha\\-\cot\alpha&\cos\alpha \end{array}\right]$

★A matrix Q=$\left[\begin{array}{c c}-\cos\alpha&\tan\alpha\\-\cot\alpha&-\sec\alpha \end{array}\right]$

$\huge\tt{\red{To\:\:Find}}$

★$2P^{-1}+Q$

$\huge\tt{\red{Concept\:\:Used}}$

★We will find the inverse of matrix P and then add it with Q.

$\huge\tt{\red{Answer}}$

We have,

P=$\left[\begin{array}{ c c}\sec\alpha&\tan\alpha\\-\cot\alpha&\cos\alpha \end{array}\right]$

So,

$P^{-1}$=$\left[\begin{array}{c c}\cos\alpha&-\tan\alpha\\\cot\alpha&\sec\alpha \end{array}\right]\frac{1}{cos\alpha\times sec\alpha+tan\alpha.cot\alpha}$

$\large\orange{\boxed{\red{A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]} ,\purple{A^{-1}=\dfrac{1}{ad-bc}\left[\begin{array}{cc}d&-b\\-c&a\end{array}\right]}}}$

=$\left[\begin{array}{ c c}\cos\alpha&-\tan\alpha\\\cot\alpha&\sec\alpha \end{array}\right]\frac{1}{cos\alpha\times \frac{1}{\cos\alpha}+tan\alpha.\times\frac{1}{\tan\alpha}}$

=$\left[\begin{array}{ c c}\cos\alpha&-\tan\alpha\\\cot\alpha&\sec\alpha \end{array}\right]\frac{1}{\cancel{cos\alpha}\times \frac{1}{\cancel{\cos\alpha}}+\cancel{tan\alpha}.\times\frac{1}{\cancel{\tan\alpha}}}$

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

________________________________

Therefore,

$2P^{-1}$

=$2\times\left[\begin{array}{c c}\cos\alpha&-\tan\alpha\\\cot\alpha&\sec\alpha \end{array}\right]\frac{1}{2}$

=$\cancel{2}\times\frac{1}{\cancel{2}}\left[\begin{array}{c c}\cos\alpha&-\tan\alpha\\\cot\alpha&\sec\alpha \end{array}\right]$

=$\left[\begin{array}{ c c}\cos\alpha&-\tan\alpha\\\cot\alpha&\sec\alpha \end{array}\right]$

________________________________

Now,

$2P^{-1}+Q$

=$\left[\begin{array}{c c}\cos\alpha&-\tan\alpha\\\cot\alpha&\sec\alpha \end{array}\right] + \left[\begin{array}{c c c}-\cos\alpha&\tan\alpha\\-\cot\alpha&-\sec\alpha \end{array}\right]$

=$\left[\begin{array}{c}(cos\alpha-cos\alpha) &(tan\alpha-tan\alpha) \\(cot\alpha-cot\alpha) &(sec\alpha-sec\alpha) \end{array}\right]$

=$\left[\begin{array}{cc} 0&0\\0&0\end{array}\right]$

${\underline{\boxed{\red{.\degree.2P^{-1}+Q=\left[\begin{array}{cc} 0&0\\0&0\end{array}\right]}}}}$

Answer 2

Given

* A matrix P=sec a

tan a

| - cot a

cos a

*A matrix Q=- cos a

- cota

tan a

tan a

- sec a

To Find

*2P-1 +Q

Concept Used

*We will find the inverse of matrix P and then add it with Q.

P=

sec a

- cot a

So,

P-1 =

cosa ( costa

cosa | cota

tana cosa

- tan a sec a

- tan a seca

cosa -tan a cot a sec

cosa x seca + tana.cota

-1

cosa X

cosa X

ad

- bc

+ tana. X

tan

+ tana. XT

COS a - tan a

cot a sec a

2P-1

=2 x

= 2 X

1

cot a

tan a

- sec

2P-1+Q

cos a

cot a

- tan a

sec a

(cosa - cosa) (cota - cota)

.2P+Q =

$\\ \\ \\ \\ thank \: bye \: friend$