Maths Matrices

Kindly consider the following figure

B = (I – adjA)⁵

Kindly consider the following figure

B = (I – adjA)⁵

$\underset{x\to \frac{1}{\sqrt[]{2}}}{lim}\frac{sin\left(co{s}^{-1}x\right)-x}{1-tan\left(co{s}^{-1}x\right)}$

Let $co{s}^{-1}x=\frac{\pi }{4}+\theta$

= $\underset{\theta \to \infty }{lim}\frac{\sqrt[]{2}sin\theta }{-2tan\theta }\left(1-tan\theta \right)=-\frac{1}{\sqrt[]{2}}$

Kindly go through the solution

B = (I – adjA)⁵

$N=M^2+M^4+.....+M^{98}$

$=\left(-{\alpha }^{2}I\right)+{\left(-{\alpha }^{2}I\right)}^2+....+{\left(-{\alpha }^{2}I\right)}^{49}$

$=I\left(-{\alpha }^2+{\alpha }^4-{\alpha }^6+....-{\alpha }^{98}\right)$

N =  $-I\left({\alpha }^2-{\alpha }^4+{\alpha }^6.......+{\alpha }^{98}\right)$

$=-I\frac{{\alpha }^2\left(1-{\left(-{\alpha }^2\right)}^{49}\right)}{1-\left(-{\alpha }^2\right)}$

N =  $\frac{-I{\alpha }^2\left(1+{\alpha }^{98}\right)}{1+{\alpha }^2}$  

Now  $\left(I-m^2\right)N=-2I$

$\left(I+{\alpha }^{2}I\right)\frac{\left(-I{\alpha }^2\cdot (1+{\alpha }^{98}\right)}{1+{\alpha }^2}=-2I$  

-> a¹⁰⁰ + a² = 2

->a = $\pm 1$

$\left|\begin{array}{ccc}\hfill 1-\alpha \hfill & \hfill 6\hfill & \hfill 6\hfill \\ \hfill 4\hfill & \hfill -1-\alpha \hfill & \hfill 4\hfill \\ \hfill 2\alpha \hfill & \hfill 2\alpha \hfill & \hfill \alpha -5\hfill \end{array}\right|=0$

$\alpha =-5$

$\left|A-xI\right|=\left|\begin{array}{cc}\hfill 1/2-x\hfill & \hfill 1/2\hfill \\ \hfill a\hfill & \hfill b-x\hfill \end{array}\right|=x^2-\left(b+\frac{1}{2}\right)x+\frac{b-a}{2}$

from Cayley – Hamilton Theorem

$A^2-\left(b+\frac{1}{2}\right)A+\left(\frac{b-a}{2}\right)I=0$

$\Rightarrow A^3=\left({\left(b+\frac{1}{2}\right)}^2-\left(\frac{b-a}{2}\right)\right)A-\left(b+\frac{1}{2}\right)\left(\frac{b-a}{2}\right)I$

$\therefore {\left(b+\frac{1}{2}\right)}^2-\left(\frac{b-a}{2}\right)=1\&\left(b+\frac{1}{2}\right)\left(\frac{b-a}{2}\right)=0$

$\therefore \left(a,b\right)=\left(\frac{1}{2},\frac{1}{2}\right),\left(-\frac{3}{2},-\frac{3}{2}\right),\left(\frac{3}{2},-\frac{1}{2}\right)$

$A^2{\left|A\right|}^2-{\left|B\right|}^{3.2}B$

$=25A^2-\left|B\right|.B$

$\begin{array}{l}=25A^2+\left|B\right|A^2\\ =0\end{array}$

A² - 4A + 4I = 0
(A - 2I)² = 0 ⇒ A-2I ≠ 0
B = 297A - 594I
= 297 (A - 2I)