Maths

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

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

$x?\left(x\right)={\displaystyle \underset{}{\overset{}{\int }}\left(3t^2-2?\left(t\right)\right)dt,x>-2,By}$  Leibniz theorem

$?\left(x\right)+x?\text{'}\left(x\right)=3x^2-2?\text{'}\left(x\right)\Rightarrow ?\text{'}\left(x\right)+\frac{?\left(x\right)}{x+2}=\frac{3x^2}{x+2}..............\left(i\right)$

$\therefore I.F.=e^{{\displaystyle \int \frac{dx}{x+2}}}=x+2$

Multiplying (i) by I.F. we get,

$\therefore ?\left(x\right)=\frac{x^3+8}{x+2}$

$\therefore ?\left(2\right)=4$

P (E₁)= 0.9 $P\left({\overline{E}}_t\right)=0.1andP\left(E_2\right)=0.8,P\left({\overline{E}}_2\right)=0.2$

$\therefore RequiredprobabilityP=\frac{0.8*0.1}{0.8*0.1+0.2*0.1+0.9*0.2}=\frac{0.8}{2.8}=\frac{2}{7}$

$\therefore 98P=28$

$t_r=\left(3r+4\right)\left(2r+6\right)=6r^2+26r+24$

$\therefore {\displaystyle \sum _{r=1}^{10}{t}_r}=\frac{6*10*11*21}{6}+\frac{26*10*11}{2}+24*10=10*398$

$Mean=\frac{10*398}{10}=398$

Let zc be the centre of the circle

$\frac{z_c-2}{z_c+2}=i\Rightarrow z_c=2i$

Radius of the circle =  $\sqrt[]{4+4}=2\sqrt[]{2}$

Square off minimum distance AB =  ${\left(7\sqrt[]{2}\right)}^2=98$

${\left(\frac{x}{4}-\frac{12}{x^2}\right)}^{12}$

According to question, 12 – 3r = 0 r = 4

$\Rightarrow \frac{3^6}{4^4}*55=\frac{3^6}{4^4}*k\Rightarrow k=55$

$?\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}=r\left(say\right)$

Let P (1 + 2r, 2 + 3r, 1 + 6r) lies on the plane 2x – y + z = 6

$\therefore r=1\Rightarrow P\left(3,5,5\right)$

Distance between P and Q is PQ =  $\sqrt[]{16+36+9}=\sqrt[]{61}$

$\therefore P{Q}^2=61$