সৃজনশীলঃ ‘A=[121−120343],B=[(b+c)2a2a2b2(c+a)2b2c2c2(a+b)2]…’ ২০২৬ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

[বিদ্র.: C, l, m, n এর পরিবর্তে যথাক্রমে B, a, b, c হবে] 

দেওয়া আছ, $C=\left(\begin{array}{ccc}(m+n{)}^2& l^2& l^2\\ m^2& (n+l{)}^2& m^2\\ n^2& n^2& (l+m{)}^2\end{array}\right)$ 

$\begin{array}{cc}\therefore |C|& =\left|\begin{array}{ccc}(m+n{)}^2& l^2& l^2\\ m^2& (n+l{)}^2& m^2\\ n^2& n^2& (l+m{)}^2\end{array}\right|\\ & =\left|\begin{array}{ccc}m+n{)}^2& l^2-(m+n{)}^2& l^2-(m+n{)}^2\\ m^2& (n+l{)}^2-m^2& m^2-m^2\\ n^2& n^2-n^2& (l+m{)}^2-n^2\end{array}\right|\end{array}$ 

$[c_2^{\prime }=c_2-c_1$ এবং$c_3^{\prime }=c_3-c_1$ প্রয়োগ করে$]$ 

$\begin{array}{cc}& =\left|\begin{array}{ccc}(m+n{)}^2& (l+m+n)(l-m-n)& (l+m+n)(l-m-n)\\ m^2& (l+m+n)(l-m+n)& 0\\ n^2& 0& (l+m+n)(l+m-n)\end{array}\right|\\ & =(l+m+n{)}^2\left|\begin{array}{ccc}(m+n{)}^2& l-m-n& l-m-n\\ m^2& l-m+n& 0\\ n^2& 0& l+m-n\end{array}\right|\\ & =(l+m+n{)}^2\left|\begin{array}{ccc}2mn& -2n& -2m\\ m^2& l-m+n& 0\\ n^2& 0& l+m-n\end{array}\right|\end{array}$ 

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

$\begin{array}{cc}& =2(l+m+n{)}^2\left|\begin{array}{ccc}mn& -n& -m\\ m^2& l-m+n& 0\\ n^2& 0& l+m-n\end{array}\right|\\ & =2(l+m+n{)}^2\left|\begin{array}{ccc}mn& 0& 0\\ m^2& n+l& \frac{m^2}{n}\\ n^2& 0& \frac{n^2}{m}\end{array}\right|\end{array}$ 

$[c_2^{\prime }=c_2+\frac{1}{y}{c}_1$ এবং $c_3^{\prime }=c_3+\frac{1}{z}{c}_1$ প্রয়োগ করে$]$ 

$\begin{array}{cc}& =2(l+m+n{)}^{2}mn\left|\begin{array}{cc}n+l& \frac{m^2}{n}\\ \frac{n^2}{m}& l+m\end{array}\right|\\ & =2(l+m+n{)}^{2}mn(nl+l^2+mn+lm-mn)\\ & =2mn(l+m+n{)}^2\cdot l(l+m+n)=2lmn(l+m+n{)}^3\end{array}$ 

দেওয়া আছে, $A=\left[\begin{array}{ccc}1& 2& 1\\ -1& 2& 0\\ 3& 4& 3\end{array}\right]$ 

$\begin{array}{cc}\therefore A^T& =\left[\begin{array}{ccc}1& -1& 3\\ 2& 2& 4\\ 1& 0& 3\end{array}\right]\\ \therefore \left|A^T\right|& =\left|\begin{array}{ccc}1& -1& 3\\ 2& 2& 4\\ 1& 0& 3\end{array}\right|\\ & =1(6-0)+1(6-4)+3(0-2)\\ & =6+2-6=2\ne 0\end{array}$ 

$\therefore {\left(A^T\right)}^{-1}$ বিদ্যমান। 

A ${}^T$ ম্যাট্রিক্সের সহগুণকসমূহ : 

$\begin{array}{cc}& A_{11}=\left|\begin{array}{cc}2& 4\\ 0& 3\end{array}\right|=6-0=6\\ & A_{12}=-\left|\begin{array}{cc}2& 4\\ 1& 3\end{array}\right|=-(6-4)=-2\\ & A_{13}=\left|\begin{array}{cc}2& 2\\ 1& 0\end{array}\right|=0-2=-2\end{array}$ 

$\begin{array}{cc}& A_{21}=-\left|\begin{array}{cc}-1& 3\\ 0& 3\end{array}\right|=-(-3-0)=3\\ & A_{22}=\left|\begin{array}{cc}1& 3\\ 1& 3\end{array}\right|=3-3=0\\ & A_{23}=-\left|\begin{array}{cc}1& -1\\ 1& 0\end{array}\right|=-(0+1)=-1\\ & A_{31}=\left|\begin{array}{cc}-1& 3\\ 2& 4\end{array}\right|=-4-6=-10\\ & A_{32}=-\left|\begin{array}{cc}1& 3\\ 2& 4\end{array}\right|=-(4-6)=2\\ & A_{33}=\left|\begin{array}{cc}1& -1\\ 2& 2\end{array}\right|=2+2=4\\ \therefore & {\mathrm{Adj}}^T=\left[\begin{array}{ccc}6& -2& -2\\ 3& 0& -1\\ -10& 2& 4\end{array}\right]=\left[\begin{array}{ccc}6& 3& -10\\ -2& 0& 2\\ -2& -1& 4\end{array}\right]\\ \therefore & {\left(A^T\right)}^{-1}=\frac{1}{\left|A^T\right|}\mathrm{Adj}{A}^T\end{array}$ 

নির্ণেয় ${\left(A^T\right)}^{-1}=\left[\begin{array}{ccc}3& \frac{3}{2}& -5\\ -1& 0& 1\\ -1& -\frac{1}{2}& 2\end{array}\right]$.