সৃজনশীলঃ ‘P=[1213−3−1210],Q=[ap3−m3aq3−m3ar3−m3p2q2r2pqr]…’ ২০২৪ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

$|P|=\left|\begin{array}{ccc}1& 2& 1\\ 3& -3& -1\\ 2& 1& 0\end{array}\right|=1(-0+1)-2(0+2)+1(3+6)$

$=1-4+9=6\ne 0$

$\therefore P^{-1}$নির্ণয়যোগ্য।

$\therefore \mathrm{adj}P=\left[\begin{array}{ccc}\left|\begin{array}{cc}-3& -1\\ 1& 0\end{array}\right|& -\left|\begin{array}{cc}3& -1\\ 2& 0\end{array}\right|& \left|\begin{array}{cc}3& -3\\ 2& 1\end{array}\right|\\ -\left|\begin{array}{cc}2& 1\\ 1& 0\end{array}\right|& \left|\begin{array}{cc}1& 1\\ 2& 0\end{array}\right|& -\left|\begin{array}{cc}1& 2\\ 2& 1\end{array}\right|\\ \left|\begin{array}{cc}2& 1\\ -3& -1\end{array}\right|& -\left|\begin{array}{cc}1& 1\\ 3& -1\end{array}\right|& \left|\begin{array}{cc}1& 2\\ 3& -3\end{array}\right|\end{array}\right]$

$={\left[\begin{array}{ccc}1& -2& 9\\ 1& -2& 3\\ 1& 4& -9\end{array}\right]}^t=\left[\begin{array}{ccc}1& 1& 1\\ -2& -2& 4\\ 9& 3& -9\end{array}\right]$

$\therefore P^{-1}=\frac{1}{\left|P\right|}$ adj $P=\frac{1}{6}\left[\begin{array}{ccc}1& 1& 1\\ -2& -2& 4\\ 9& 3& -9\end{array}\right]$ (Ans.)

বামপক্ষ $=|Q|$

$=\left|\begin{array}{ccc}a{p}^3-m^3& a{q}^3-m^3& a{r}^3-m^3\\ p^2& q^2& r^2\\ p& q& r\end{array}\right|$

$=\left|\begin{array}{ccc}a{p}^3& a{q}^3& a{r}^3\\ p^2& q^2& r^2\\ p& q& r\end{array}\right|-\left|\begin{array}{ccc}{m}^3& m^3& m^3\\ p^2& q^2& r^2\\ p& q& r\end{array}\right|$

$=\mathrm{apqr}\left|\begin{array}{ccc}{p}^2& q^2& r^2\\ p& q& r\\ 1& 1& 1\end{array}\right|-m^3\left|\begin{array}{ccc}1& 1& 1\\ p^2& q^2& r^2\\ p& q& r\end{array}\right|$

$=-\mathrm{apqr}\left|\begin{array}{ccc}{p}^2& q^2& r\\ p^2& q^2& r^2\\ 1& 1& 1\end{array}\right|+m^3\left|\begin{array}{ccc}1& 1& 1\\ p& q& r\\ p^2& q^2& r^2\end{array}\right|$

$=\mathrm{apqr}\left|\begin{array}{ccc}p& q& r\\ 1& 1& 1\\ p^2& q^2& r^2\end{array}\right|+m^3\left|\begin{array}{ccc}1& 1& 1\\ p& q& r\\ p^2& q^2& r^2\end{array}\right|$

$=-\mathrm{apqr}\left|\begin{array}{ccc}1& 1& 1\\ p& q& r\\ p^2& q^2& r^2\end{array}\right|+m^3\left|\begin{array}{ccc}1& 1& 1\\ p& q& r\\ p^2& q^2& r^2\end{array}\right|$

$=(-\mathrm{apqr}+m^3)\left|\begin{array}{ccc}1& 1& 1\\ p& q& r\\ p^2& q^2& r^2\end{array}\right|$

$=(-apqr+m^3)\left|\begin{array}{ccc}0& 0& 1\\ p-q& q-r& r\\ p^2-q^2& q^2-r^2& r^2\end{array}\right|\left[\begin{array}{c}{c}_1^{\prime }=c_1-c_2\\ c_2^{\prime }=c_2-c_3\end{array}\right]$

$=(-apqr+m^3)\left|\begin{array}{cc}p-q& q-r\\ (p+q)(p-q)& (q+r)(q-r)\end{array}\right|$

$=(-apqr+m^3)(p-q)(q-r)\left|\begin{array}{cc}1& 1\\ p+q& q+r\end{array}\right|$

$=(-$ apqr $+m^3)(p-q)(q-r)(q+r-p-q)$

$=(-apqr+m^3)(p-q)(q-r)(r-p)$

$=-(apqr-m^3)(p-q)(q-r)(r-p)$

$=(apqr-m^3)(p-q)(q-r)(p-r)$

$=$ডানপক্ষ

$\therefore |Q|=(apqr-m^3)(p-q)(q-r)(p-r)$ (প্রমাণিত)