সৃজনশীলঃ ‘M=[(b+c)2a2a2b2(c+a)2b2c2c2(a+b)2]…’ চট্টগ্রাম বোর্ড ২০২৫ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $M=\left[\begin{array}{ccc}(b+c{)}^2& a^2& a^2\\ b^2& (c+a{)}^2& b^2\\ c^2& c^2& (a+b{)}^2\end{array}\right]$

বামপক্ষ $=|M|$

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

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

$\left[\begin{array}{c}{c}^2=c_2-c_1\\ c_3=c_3-c_1\end{array}\right]$

$=\left|\begin{array}{ccc}(b+c{)}^2& (a+b+c)(a-b-c)& (a+b+c)(a-b-c)\\ b^2& (a+b+c)(c+a-b)& 0\\ c^2& 0& (a+b+c)(a+b-c)\end{array}\right|$

$=(a+b+c{)}^2\left|\begin{array}{ccc}{b}^2+2bc+c^2& a-b-c& a-b-c\\ b^2& c+a-b& 0\\ c^2& 0& a+b-c\end{array}\right|$

$=(a+b+c{)}^2\left|\begin{array}{ccc}2bc& -2c& -2b\\ b^2& c+a-b& 0\\ c^2& 0& a+b-c\end{array}\right|$

$[r_1^{\prime }=r_1-r_2-r_3]$

$=2(a+b+c{)}^2\left|\begin{array}{ccc}bc& -c& -b\\ b^2& c+a-b& 0\\ c^2& 0& a+b-c\end{array}\right|$

$=2(a+b+c{)}^2\left|\begin{array}{ccc}bc& 0& 0\\ b^2& c+a& \frac{b^2}{c}\\ c^2& \frac{c^2}{b}& a+b\end{array}\right|$

$\left[\begin{array}{c}{c}_2^{\prime }=c_2+\frac{1}{b}{c}_1\\ c_3^{\prime }=c_3+\frac{1}{c}{c}_1\end{array}\right]$

$=\frac{2}{bc}(a+b+c{)}^2\left|\begin{array}{ccc}bc& 0& 0\\ b^2& bc+ab& b^2\\ c^2& c^2& ac+bc\end{array}\right|$

$=\frac{2}{bc}(a+b+c{)}^{2}bc\left|\begin{array}{ccc}bc& 0& 0\\ b& c+a& b\\ c& c& a+b\end{array}\right|$

$=2(a+b+c{)}^2\left\{bc(ca+bc+a^2+ab-bc)\right\}$

$=2(a+b+c{)}^2\left\{bc(a^2+ab+ca)\right\}$

$=2(a+b+c{)}^2[bc\left\{a\right(a+b+c\left)\right\}]$

$=2(a+b+c{)}^2\{abc(a+b+c)\}$

$=2abc(a+b+c{)}^3=$ ডানপক্ষ

$\therefore |M|=2abc(a+b+c{)}^3$. (দেখানো হলো)

দেওয়া আছে, $M=\left[\begin{array}{ccc}(b+c{)}^2& a^2& a^2\\ b^2& (c+a{)}^2& b^2\\ c^2& c^2& (a+b{)}^2\end{array}\right]$

$a=b=c=1$ হলে, $M=\left[\begin{array}{ccc}(1+1{)}^2& 1^2& 1^2\\ 1^2& (1+1{)}^2& 1^2\\ 1^2& 1^2& (1+1{)}^2\end{array}\right]$

$\therefore M=\left[\begin{array}{ccc}4& 1& 1\\ 1& 4& 1\\ 1& 1& 4\end{array}\right]$

$\therefore |M|=\left|\begin{array}{ccc}4& 1& 1\\ 1& 4& 1\\ 1& 1& 4\end{array}\right|$

$=4(16-1)-1(4-1)+1(1-4)$

$=60-3-3=54\ne 0$

$\therefore M^{-1}$ বিদ্যমান।

Adj $M={\left[\begin{array}{ccc}{M}_{11}& M_{12}& M_{13}\\ M_{21}& M_{22}& M_{23}\\ M_{31}& M_{32}& M_{33}\end{array}\right]}^T$

$M$ এর সহগুণকগুলি হচ্ছে:

$M_{11}=\left|\begin{array}{cc}4& 1\\ 1& 4\end{array}\right|=16-1=15$

$M_{12}=-\left|\begin{array}{cc}1& 1\\ 1& 4\end{array}\right|=-(4-1)=-3$

$M_{13}=\left|\begin{array}{cc}1& 4\\ 1& 1\end{array}\right|=1-4=-3$

$M_{21}=-\left|\begin{array}{cc}1& 1\\ 1& 4\end{array}\right|=-(4-1)=-3$

$M_{22}=\left|\begin{array}{cc}4& 1\\ 1& 4\end{array}\right|=16-1=15$

$M_{23}=-\left|\begin{array}{cc}4& 1\\ 1& 1\end{array}\right|=-(4-1)=-3$

$M_{31}=\left|\begin{array}{cc}1& 1\\ 4& 1\end{array}\right|=1-4=-3$

$M_{32}=-\left|\begin{array}{cc}4& 1\\ 1& 1\end{array}\right|=-(4-1)=-3$ 

$M_{33}=\left|\begin{array}{cc}4& 1\\ 1& 4\end{array}\right|=16-1=15$

$\therefore \mathrm{Adj}\left(M\right)={\left[\begin{array}{ccc}15& -3& -3\\ -3& 15& -3\\ -3& -3& 15\end{array}\right]}^{\top }$

$=\left[\begin{array}{ccc}15& -3& -3\\ -3& 15& -3\\ -3& -3& 15\end{array}\right]$

$\therefore M^{-1}=\frac{1}{\left|M\right|}\mathrm{Adj}\left(M\right)$$=\frac{1}{54}\left[\begin{array}{ccc}15& -3& -3\\ -3& 15& -3\\ -3& -3& 15\end{array}\right]$

$=\frac{1}{54}\times 3\left[\begin{array}{ccc}5& -1& -1\\ -1& 5& -1\\ -1& -1& 5\end{array}\right]$

$=\frac{1}{18}\left[\begin{array}{ccc}5& -1& -1\\ -1& 5& -1\\ -1& -1& 5\end{array}\right]$

নির্ণেয় $M^{-1}=\frac{1}{18}\left[\begin{array}{ccc}5& -1& -1\\ -1& 5& -1\\ -1& -1& 5\end{array}\right]$.