Matrices

$f\left(x\right)=a{x}^2+bx+c$  

g (x) = px + q

  $f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

  $9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

Kindly consider the following figure

B = (I – adjA)⁵

Kindly consider the following figure

B = (I – adjA)⁵

System of equation is

$\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill \left|\lambda \right|\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left[\begin{array}{l}-2\\ 4\\ 4\lambda -4\end{array}\right]$

R₁ – 2 R₂, R₃ – R₂

$\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill -2\hfill & \hfill \left|\lambda \right|-1\hfill \end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{l}-10\\ 4\\ 4\lambda -8\end{array}\right)$

System of equation will have no solution for = -7.

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

$\alpha =-5$

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

from Cayley – Hamilton Theorem

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

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

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

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

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

A = [1]
[0 1]
Now A² = [1] [1] = [1 2]
[0 1] [0 1] [0 1]
A³ = A².A = [1 2] [1] = [1 3]
[0 1] [0 1] [0 1]
Similarly A²? ¹¹ = [1 2011]
[0 1]