সৃজনশীলঃ ‘A=[3−1−11−311−13]…’ যশোর বোর্ড ২০২৫ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

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

এখন, $A^2=A.A$

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

$=\left[\begin{array}{ccc}9-1-1& -3+3+1& -3-1-3\\ 3-3+1& -1+9-1& -1-3+3\\ 3-1+3& -1+3-3& -1-1+9\end{array}\right]$

$=\left[\begin{array}{ccc}7& 1& -7\\ 1& 7& -1\\ 5& -1& 7\end{array}\right]$

প্রদত্ত রাশি, $=A^2-2A+3I$

$=\left[\begin{array}{ccc}7& 1& -7\\ 1& 7& -1\\ 5& -1& 7\end{array}\right]-2\left[\begin{array}{ccc}3& -1& -1\\ 1& -3& 1\\ 1& -1& 3\end{array}\right]+3\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}7& 1& -7\\ 1& 7& -1\\ 5& -1& 7\end{array}\right]-\left[\begin{array}{ccc}6& -2& -2\\ 2& -6& 2\\ 2& -2& 6\end{array}\right]+\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]$

$=\left[\begin{array}{ccc}7& -6+3& 1+2+0\\ 1& -7+2+0& 7+6+3\\ 5-2+0& -1+2+0& 7-6+3\end{array}\right]$

$=\left[\begin{array}{ccc}4& 3& -5\\ -1& 16& -3\\ 3& 1& 4\end{array}\right]$

$\therefore A^2-2A+3I=\left[\begin{array}{ccc}4& 3& -5\\ -1& 16& -3\\ 3& 1& 4\end{array}\right]$

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

এখন, $AB=I$

বা, $A^{-1}\cdot AB=A^{-1}\cdot I$

বা, I. $B=A^{-1}[\because A^{-1}\cdot A=I$ 

এবং $A^{-1}\cdot I=A^{-1}]$

$\therefore B=A^{-1}$ ........ (1)

এখন, $|A|=\left|\begin{array}{ccc}3& -1& -1\\ 1& -3& 1\\ 1& -1& 3\end{array}\right|$

$=3(-9+1)+1(3-1)-1(-1+3)$

$=-24+2-2=-24\ne 0$

$\therefore A^{-1}$ বিদ্যমান।

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

A এর সহগুণকসমূহ

$A_{11}=\left|\begin{array}{cc}-3& 1\\ -1& 3\end{array}\right|=-9+1=-8$

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

$A_{13}=\left|\begin{array}{cc}1& -3\\ 1& -1\end{array}\right|=-1+3=2$

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

$A_{22}=\left|\begin{array}{cc}3& -1\\ 1& 3\end{array}\right|=9+1=10$

$A_{23}=-\left|\begin{array}{cc}3& -1\\ 1& -1\end{array}\right|=-(-3+1)=2$

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

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

$A_{33}=\left|\begin{array}{cc}3& -1\\ 1& -3\end{array}\right|=-9+1=-8$

$\therefore AdjA={\left[\begin{array}{ccc}-8& -2& 2\\ 4& 10& 2\\ -4& -4& -8\end{array}\right]}^T$

$=\left[\begin{array}{ccc}-8& 4& -4\\ -2& 10& -4\\ 2& 2& -8\end{array}\right]$

$\therefore A^{-1}=\frac{1}{\left|A\right|}\text{ Adj }A=\frac{1}{-24}\left[\begin{array}{ccc}-8& 4& -4\\ -2& 10& -4\\ 2& 2& -8\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{-8}{-24}& \frac{4}{-24}& \frac{-4}{-24}\\ \frac{-2}{-24}& \frac{10}{-24}& \frac{-4}{-24}\\ \frac{2}{-24}& \frac{2}{-24}& \frac{-8}{-24}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{3}& -\frac{1}{6}& \frac{1}{6}\\ \frac{1}{12}& -\frac{5}{12}& \frac{1}{6}\\ -\frac{1}{12}& -\frac{1}{12}& \frac{1}{3}\end{array}\right]$

(1)নং হতে পাই, $B=A^{-1}=\left[\begin{array}{ccc}\frac{1}{3}& -\frac{1}{6}& \frac{1}{6}\\ \frac{1}{12}& -\frac{5}{12}& \frac{1}{6}\\ -\frac{1}{12}& -\frac{1}{12}& \frac{1}{3}\end{array}\right]$

নির্ণেয় $B=\left[\begin{array}{ccc}\frac{1}{3}& -\frac{1}{6}& \frac{1}{6}\\ \frac{1}{12}& -\frac{5}{12}& \frac{1}{6}\\ -\frac{1}{12}& -\frac{1}{12}& \frac{1}{3}\end{array}\right]$