Matrix

$\left[\begin{array}{cc}\hfill 0\hfill & \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 0\hfill \end{array}\right],I_2+A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 1\hfill \end{array}\right],I_2-A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill -tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 1\hfill \end{array}\right]$           

  $\left(I_2+A\right){\left(I_2-A\right)}^{-1}=\left[\begin{array}{cc}\hfill a\hfill & \hfill -b\hfill \\ \hfill b\hfill & \hfill a\hfill \end{array}\right]$          

  $a^2+b^2=\left|\left(I_2+A\right){\left(I_2-A\right)}^{-1}\right|=se{c}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.*co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1$          

$\therefore 13\left(a^2+b^2\right)=13*1=13$

  $A={\left[a_{ij}\right]}_{3*3}=\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$             

$a_{i1}+a_{i2}+a_{i3}=1;i=1,2,3$            

$LetX=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]them$  

given $\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$ $\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$  

->AX = X .(i)

replace x by A x we have

A (AX) = AX

->A²X = AX = X .(ii)

Again replace X by AX

A³X = AX = X.

As  $X=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right],$ Sum of all entries in A³ = sum of entries in X = 1 +1 + 1 = 3

x + y + z = 4

3x + 2y + 5z = 3

$9x+4y+\left(28+\left[\lambda \right]\right)z=\left[\lambda \right]$           

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 9\hfill & \hfill 4\hfill & \hfill 28+\left[\lambda \right]\hfill \end{array}\right|=56+2\left[\lambda \right]-20-\left(84+3\left[\lambda \right]-45\right)+\left(-6\right)$

$=-\left[\lambda \right]-9$              

$If\left[\lambda \right]+9\ne 0$ then unique solution

& if  $\left[\lambda \right]+9=0then{\Delta }_1={\Delta }_2={\Delta }_3=0$ so infinite solution will exist

Hence $\lambda \in R$  

$\left|A\right|=\left|\begin{array}{ccc}\hfill \left[x+1\right]\hfill & \hfill \left[x+2\right]\hfill & \hfill \left[x+3\right]\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x+3\right]\hfill & \hfill \left[x+3\right]\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x+2\right]\hfill & \hfill \left[x+4\right]\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \left[x\right]+1\hfill & \hfill \left[x\right]+2\hfill & \hfill \left[x\right]+3\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x\right]+3\hfill & \hfill \left[x\right]+3\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x\right]+2\hfill & \hfill \left[x\right]+4\hfill \end{array}\right|$

$R_1\to R_1-R_3\&R_2\to R_2-R_3$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill \left[x\right]\hfill & \hfill \left[x\right]+2\hfill & \hfill \left[x\right]+4\hfill \end{array}\right|=192$

$\Rightarrow \left[x\right]=62$

$\Rightarrow x\in [62,63)$              

$A^2=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^3=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$

$A^{2025}-A^{2020}$

$=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2024\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)-\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2019\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)=A^6-A$

$\Delta =\left|\begin{array}{ccc}\hfill -k\hfill & \hfill 3\hfill & \hfill -14\hfill \\ \hfill -15\hfill & \hfill 4\hfill & \hfill -k\hfill \\ \hfill -4\hfill & \hfill 1\hfill & \hfill 3\hfill \end{array}\right|$

$=\left(-k\right)\left(12-k\right)+3\left(4k+45\right)-14\left(-15+16\right)$

$\Delta =0\Rightarrow k=\pm 11$             

 For k = -11,

->11x + 3y – 14z = 25

-4x + y + 3z = 4

$\left\{\begin{array}{l}-11x+3y-14z=25\\ -15x+4y-11z=3\\ -4x+y+3z=4\end{array}\right\}\to$ No solution for k = $\pm$ 11

Let A = $\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \\ \hfill 9\hfill & \hfill h\hfill & \hfill i\hfill \end{array}\right]$  

Now A^TA

trace will be    $a^2+b^2+c^2+d^2+e^2+f^2+9^2+h^2=6$

total ways = 

Kindly go through the solution

Use characteristic equation = 0

$\left|adj\left(adj\left(A\right)\right)\right|={\left|A\right|}^{2^2}={\left|A\right|}^4$

$\therefore {\left|A\right|}^4=\left|\begin{array}{ccc}\hfill 14\hfill & \hfill 28\hfill & \hfill -14\hfill \\ \hfill -14\hfill & \hfill 14\hfill & \hfill 28\hfill \\ \hfill 25\hfill & \hfill -14\hfill & \hfill 14\hfill \end{array}\right|$

$={\left(14\right)}^3\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|$

$={\left(14\right)}^3\left(3-2\left(-5\right)-1\left(-1\right)\right)$

${\left|A\right|}^4={\left(14\right)}^4\Rightarrow \left|A\right|=14$

x + 2y + z = 2

$\alpha x+3y-z=\alpha$

$-\alpha x+y+2z=-\alpha$            

$\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \alpha \hfill & \hfill 3\hfill & \hfill -1\hfill \\ \hfill -\alpha \hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=1\left(6+1\right)-2\left(2\alpha -\alpha \right)+1\left(\alpha +3\alpha \right)=7+2a$            

$\alpha =-\frac{7}{2}$