সৃজনশীলঃ ‘দৃশ্যকল্প-০১ : A=[aij]3×3;…’ ২০২৬ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $A={\left[A_{ij}\right]}_{3\times 3}$

$\therefore A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$ ;

$a_{ij}=\{\begin{array}{c}(i+j{)}^2;i+j<5\\ i+j;i+j\ge 5\end{array}$

$a_{11}=(1+1{)}^2=4,$

$a_{12}=(1+2{)}^2=9,$

$a_{13}=(1+3{)}^2=16$

$a_{21}=(2+1{)}^2=9,$

$a_{22}=(2+2{)}^2=16,$

$a_{23}=2+3=5$

$a_{31}=(3+1{)}^2=16,$

$a_{32}=3+2=5,$

$a_{33}=3+3=6$

$\therefore A=\left[\begin{array}{ccc}4& 9& 16\\ 9& 16& 5\\ 16& 5& 6\end{array}\right]$

এখন $|A|=\left[\begin{array}{ccc}4& 9& 16\\ 9& 16& 5\\ 16& 5& 6\end{array}\right]$

$=4(96-25)+9(80-54)+16(45-256)$

$=284+234-3376=-2858\ne 0$

$\therefore A^{-1}$ বিদ্যমান।

A এর সহগুণকসমূহ :

$A_{11}=\left|\begin{array}{cc}16& 5\\ 5& 6\end{array}\right|$

$=96-25=71$

$A_{12}=-\left|\begin{array}{cc}9& 5\\ 16& 6\end{array}\right|$

$=(54-80)=26$

$A_{13}=\left|\begin{array}{cc}9& 16\\ 16& 5\end{array}\right|$

$=45-256=-211$

$A_{21}=-\left|\begin{array}{cc}9& 16\\ 5& 6\end{array}\right|$

$=-(54-80)=26$

$A_{22}=\left|\begin{array}{cc}4& 16\\ 16& 6\end{array}\right|$

$=24-256=-232$ 

$A_{23}=-\left|\begin{array}{cc}4& 9\\ 16& 5\end{array}\right|$

$=-(20-144)=124$

$A_{31}=\left|\begin{array}{cc}9& 16\\ 16& 5\end{array}\right|$

$=45-256=-211$

$A_{32}=-\left|\begin{array}{cc}4& 16\\ 9& 5\end{array}\right|$

$=-(20-144)=124$

$A_{33}=\left|\begin{array}{cc}4& 9\\ 9& 16\end{array}\right|$

$=64-81=-17$

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

$={\left[\begin{array}{ccc}71& 26& -211\\ 26& -232& 124\\ -211& 124& -17\end{array}\right]}^T=\left[\begin{array}{ccc}71& 26& -211\\ 26& -232& 124\\ -211& 124& -17\end{array}\right]$

 $\therefore A^{-1}=\frac{1}{\left|A\right|}\text{ Adj }A=\frac{1}{-2858}\left[\begin{array}{ccc}71& 26& -211\\ 26& -232& 124\\ -211& 124& -17\end{array}\right]$

$=\frac{1}{2858}\left[\begin{array}{ccc}-71& -26& 211\\ -26& 232& -124\\ 211& -124& 17\end{array}\right]$

এখন, $\mathrm{AB}=\mathrm{BA}=I_3$ হলে,

$B=A^{-1}=\left[\begin{array}{ccc}-71& -26& 211\\ -26& 232& -124\\ 211& -124& 17\end{array}\right]$

নির্ণেয় $B=\frac{1}{2858}\left[\begin{array}{ccc}-71& -26& 211\\ -26& 232& -124\\ 211& -124& 17\end{array}\right]$.

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)