সৃজনশীলঃ ‘M=[2xyx2y2x2y22xyy22xyx2]…’ ২০২৪ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

$x=0$ ও $y=1$হলে,

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

$\begin{array}{cccc}\left|m\right|& =-1& & \\ a_{11}& =0\left|\begin{array}{cc}1& 0\\ 0& 0\end{array}\right|& a_{12}& =0\left|\begin{array}{cc}0& 0\\ 1& 0\end{array}\right|\\ & =0& & =0\end{array}$

$\begin{array}{cccc}{a}_{13}& =1\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|& a_{21}& =0\left|\begin{array}{cc}0& 1\\ 0& 0\end{array}\right|\\ & =0& & =0\end{array}$

$\begin{array}{cccc}{a}_{22}& =1\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|& a_{23}& =0\left|\begin{array}{cc}0& 0\\ 1& 0\end{array}\right|\\ & =-1& & =0\end{array}$

$\begin{array}{cccc}{a}_{31}& =1\left|\begin{array}{cc}0& 1\\ 1& 0\end{array}\right|& a_{32}& =0\left|\begin{array}{cc}0& 1\\ 0& 0\end{array}\right|\\ & =0& & =0\end{array}$

$a_{33}=0\left|\begin{array}{cc}0& 0\\ 0& 1\end{array}\right|=0$

$\therefore \mathrm{Adj}\left(M\right)={\left[\begin{array}{ccc}0& 0& -1\\ 0& -1& 0\\ -1& 0& 0\end{array}\right]}^T=\left[\begin{array}{ccc}0& 0& -1\\ 0& -1& 0\\ -1& 0& 0\end{array}\right]$

$\therefore M^{-1}=\frac{Ad\dot{j}\left(M\right)}{\left(M\right)}=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right]=M$

$\therefore M^{-1}=M$ (Proved)

$L.H.S=\left|\begin{array}{ccc}2xy& x^2& y^2\\ x^2& y^2& 2xy\\ y^2& 2xy& x^2\end{array}\right|$

$=\left|\begin{array}{ccc}2xy+x^2+y^2& x^2& y^2\\ x^2+y^2+2xy& y^2& 2xy\\ y^2+x^2+2xy& 2xy& x^2\end{array}\right|\left(\begin{array}{c}{R}_1^{\prime }=R_1+R_2+R_3\\ R_2^{\prime }=R_1+R_2+R_3\\ R_3^{\prime }=R_1+R_2+R_3\end{array}\right)$

$=\left|\begin{array}{ccc}(x+y{)}^2& x^2& y^2\\ (x+y{)}^2& y^2& 2xy\\ (x+y{)}^2& 2xy& x^2\end{array}\right|$

$=(x+y{)}^2\left|\begin{array}{ccc}1& x^2& y^2\\ 1& y^2& 2xy\\ 1& 2xy& x^2\end{array}\right|$

$=(x+y{)}^2\left|\begin{array}{ccc}0& x^2-y^2& y^2-2xy\\ 0& y^2-2xy& 2xy-x^2\\ 1& 2xy& x^2\end{array}\right|\left[\begin{array}{c}{R}_1^{\prime }=R_1-R_2\\ R_2^{\prime }=R_2-R_3\end{array}\right]$

$=(x+y{)}^2\{(x^2-y^2)(2xy-x^2)-{(y^2-2xy)}^2\}$

$=(x+y{)}^2\{2x^{3}y-x^4-2x{y}^3+y^2{x}^2-y^4+4x{y}^3-4x{y}^2\}$

$(x+y{)}^2(-1)\{x^4-2x^{3}y+x^2{y}^2+2x^2{y}^2-2x{y}^3+y^4\}$

$=-(x+y{)}^2{(x^2-xy)}^2+2(x^2-xy)y^2+y^4)$

$=-(x+y{)}^2{(x^2-xy+y^2)}^2$

$=-{(x^3+y^3)}^2$

R.H.S

$\therefore \text{ L.H.S }=\text{ R.H.S }$