Maths Matrices

$A=\left[\begin{array}{ccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill y\hfill & \hfill z\hfill & \hfill x\hfill \\ \hfill z\hfill & \hfill x\hfill & \hfill y\hfill \end{array}\right],\left|A\right|=3xyz-\left(x^3+y^3+z^3\right)=-\left(x+y+z\right)\left[{\left(x+y+z\right)}^2-3\left(xy+yz+zx\right)\right]$

A² = l

A. A’ = l    (as A = A’)

$\therefore x^2+y^2+z^2=1andxy+yz+zx=0$         

$x^3+y^3+z^3=3\times 2+1\times \left(1-0\right)=7$             

$\left[\begin{array}{cc}\hfill 0\hfill & \hfill -tan\raisebox{1ex}{$\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.\times co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1$          

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

2x + 3y + 2z = 9 . (i)

3x + 2y + 2z = 9     . (ii)

x – y + 4z = 8          . (iii)

(ii) – (i) ⇒ x = y

Then (iii) ⇒ z = 2

(i) ⇒ 5x + 4 = 9 ⇒ (x = 1, y = 1, z = 2) unique solution

x + 2y + z = 2

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

$-\alpha x+y+2z=-\alpha$

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

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

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

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

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

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 +

  $lo{g}_{9^{\frac{1}{2}}}x+lo{g}_{9^{\frac{1}{3}}}x+lo{g}_{9^{\frac{1}{4}}}x+.......+lo{g}_{9^{\frac{1}{22}}}x=504$

=> $2lo{g}_{9}x+3lo{g}_{9}x+4lo{g}_{9}x+........+22lo{g}_{9}x=504$

=> (1 + 2 + 3 + .+ 22) log₉ x – log₉ x = 504 Þ x = 81

$I-A^3-3A+3A^2=I-A^3$

=> 3A² – 3A = 0

=> 3A (A – I) = 0

=>A² = A

$\left[\begin{array}{cc}\hfill a^2\hfill & \hfill ab+bd\hfill \\ \hfill 0\hfill & \hfill d^2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill 0\hfill & \hfill d\hfill \end{array}\right]$    

Total number of ways = 8

$A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

B = 7A20 – 20A7 + 2l

$A^2\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$A^3=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -2\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$A^4=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -4\hfill & \hfill 6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$a_{13}ofA,A^2,A^3,.......$

are 0, 1, 3, 6,.

For 0, 1, 3, 6, .

$S_n=0+1+3+6+....+t_n$

$S_n=0++3+....+t_{n-1}+t_n$

$t_n=1+2+3+...+\left(t_n-t_{n-1}\right)$

$=\frac{\left(n-1\right).n}{2}$

$b_{13}=7\left(190\right)-20\left(21\right)+0$

= 1330 – 420 = 910