সৃজনশীলঃ ‘A=[a−b−c2a2a2bb−c−a2b2c2cc−a−b]…’ ২০২৫ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $A=\left[\begin{array}{ccc}a-b-c& 2c& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right]$

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

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

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

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

$=(a+b+c)\left|\begin{array}{ccc}0& 0& 1\\ a+b+c& -(a+b+c)& 2b\\ 0& a+b+c& c-a-b\end{array}\right|$

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

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

[প্রথম সারির সাপেক্ষে বিস্তার করে।

$=(a+b+c)\left\{\right(a+b+c{)}^2+0\}=(a+b+c{)}^3$

$\therefore \det \left(A\right)=(a+b+c{)}^3$ (প্রমাণিত)

দেওয়া আছে, $A=\left[\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right]$

$a=b=c=2$ হলে, $A=\left[\begin{array}{ccc}-2& 4& 4\\ 4& -2& 4\\ 4& 4& -2\end{array}\right]$

$\therefore |A|=-2(4-16)-4(-8-16)+4(16+8)$

$=(-2)\times (-12)-4\times (-24)+4\times 24$

$=24+96+96=216\ne 0$

$\therefore A^{-1}$ বিদ্যমান

$AdjA=\left[\begin{array}{ccc}{\hat{A}}_{11}& {\hat{A}}_{12}& {\hat{A}}_{13}\\ {\hat{A}}_{21}& {\hat{A}}_{22}& {\hat{A}}_{23}\\ {\hat{A}}_{31}& {\hat{A}}_{32}& {\hat{A}}_{32}\end{array}\right]$

A এর সহগুণকসমূহ:

$A_{11}=\left|\begin{array}{cc}-2& 4\\ 4& -2\end{array}\right|$

$=4-16=-12$

$A_{12}=-\left|\begin{array}{cc}4& 4\\ 4& -2\end{array}\right|$

$=-(-8-16)=24$

$A_{13}=\left|\begin{array}{cc}4& -2\\ 4& 4\end{array}\right|$

$=16-(-8)=24$

$A_{21}=-\left|\begin{array}{cc}4& 4\\ 4& -2\end{array}\right|$

$=-(-8-16)=24$

$A_{22}=\left|\begin{array}{cc}-2& 4\\ 4& -2\end{array}\right|$

$=4-16=-12$

$A_{23}=-\left|\begin{array}{cc}-2& 4\\ 4& 4\end{array}\right|$

$=-(-8-16)=24$

$A_{31}=\left|\begin{array}{cc}4& 4\\ -2& 4\end{array}\right|$

$=16-(-8)=24.$

$A_{32}=-\left|\begin{array}{cc}-2& 4\\ 4& 4\end{array}\right|$

$=-(-8-16)=24$

$A_{33}=\left|\begin{array}{cc}-2& 4\\ 4& -2\end{array}\right|$

$=4-16=-12$

$AdjA={\left[\begin{array}{ccc}-12& 24& 24\\ 24& -12& 24\\ 24& 24& -12\end{array}\right]}^T$

$=\left[\begin{array}{ccc}-12& 24& 24\\ 24& -12& 24\\ 24& 24& -12\end{array}\right]$

$\therefore A^{-1}=\frac{1}{\left|A\right|}\text{ Adj }A=\frac{1}{216}\left[\begin{array}{ccc}-12& 24& 24\\ 24& -12& 24\\ 24& 24& -12\end{array}\right]$

নির্ণেয় $A^{-1}=\frac{1}{216}\left[\begin{array}{ccc}-12& 24& 24\\ 24& -12& 24\\ 24& 24& -12\end{array}\right]$