সৃজনশীলঃ ‘A=(a2b2c2a3b3c3a4−2ab4−2bc4−2c) P=(−323−1) এবং…’ ২০২৪ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=5x^2-11x$

এবং$\begin{array}{cc}& \therefore f\left(P\right)-5P^2-11P\text{এবং}P=\left[\begin{array}{cc}-3& 2\\ 3& -1\end{array}\right]\\ & \begin{array}{cc}\therefore P^2=P\times P& =\left[\begin{array}{cc}-3& 2\\ 3& -1\end{array}\right]\times \left[\begin{array}{cc}-3& 2\\ 3& -1\end{array}\right]\\ & =\left[\begin{array}{cc}9+6& -6-2\\ -9-3& 6+1\end{array}\right]=\left[\begin{array}{cc}15& -8\\ -12& 7\end{array}\right]\end{array}\end{array}$
এখন $\begin{array}{cc}& f\left(P\right)=6I_2=5P^2-11P+6\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ & =5\left[\begin{array}{cc}15& -8\\ -12& 7\end{array}\right]-11\left[\begin{array}{cc}-3& 2\\ 3& -1\end{array}\right]+\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]\\ & =6\left[\begin{array}{cc}75& -40\\ -60& 35\end{array}\right]-\left[\begin{array}{cc}-33& 22\\ 33& -11\end{array}\right]+\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]\\ & =\left[\begin{array}{cc}75+33+6& -40-22+0\\ -60-33+0& 35+11+6\end{array}\right]=\left[\begin{array}{cc}-114& -62\\ -93& 52\end{array}\right]\end{array}$

দেওয়া আছে,

$A=\left(\begin{array}{ccc}{a}^2& b^2& c^2\\ a^3& b^3& c^3\\ a^4-2a& b^4-2b& c^4-2c\end{array}\right)$

L.H.S

$=\det \left(A\right)$

$=\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a^3& b^3& c^3\\ a^4-2a& b^4-2b& c^4-2c\end{array}\right|$

$=\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a^3& b^3& c^3\\ a^4& b^4& c^4\end{array}\right|-\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ a^3& b^3& c^3\\ 2a& 2b& 2c\end{array}\right|$

$=a^2{b}^2{c}^2\left|\begin{array}{ccc}1& 1& 1\\ a& a& a\\ a^2& b^2& c^2\end{array}\right|-2abc\left|\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ 1& 1& 1\end{array}\right|$

$=a^2{b}^2{c}^2\left|\begin{array}{ccc}1& 1& 1\\ a& a& a\\ a^2& b^2& c^2\end{array}\right|-2abc\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$

[প্রথমে $R_3\to R_2$ তারপর নতুন $R_2$ কে $R_1$নিয়ে $R_1$ কে $R_2$ তে রূপান্তর করি]

$=(a^2{b}^2{c}^2-2abc)\left|\begin{array}{ccc}1& 1& 1\\ a& a& a\\ a^2& b^2& c^2\end{array}\right|$

$=abc(abc-2)\left|\begin{array}{ccc}0& 0& 1\\ a-b& b-c& c\\ a^2-b^2& b^2-c^2& c^2\end{array}\right|\begin{array}{cc}& C_1{}^{\prime }=C_1-C_2\\ & C_2{}^{\prime }=C_2-C_3\end{array}$

$=abc(abc-2)\cdot 1\left\{\right(a-b)(b^2-c^2)-(b-c\left)(a^2-b^2)\right\}$

$=abc(abc-2)(a-b)(b-c)(b+c-a-b)$

$=abc(abc-2)(a-b)(b-c)(c-a)$

$=$ R.H.S

$\therefore \det \left(A\right)=abc(abc-2)(a-b)(b-c)(c-a)$ (proved)