সৃজনশীলঃ ‘A=[3−42−210−1−11]…’ দিনাজপুর বোর্ড ২০২১ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$AB=BA=I_3$

অর্থাৎ, $B=A^{-1}$

দেওয়া আছে,

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

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

$=3(1-0)+4(-2-0)+2(2+1)$

$=3-8+6=1\ne 0$

$\therefore A^{-1}$ বিদ্যমান।

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

$\left|A\right|$ এর সহগুণকসমূহঃ

$A_{11}=\left|\begin{array}{cc}1& 0\\ -1& 1\end{array}\right|=(1-0)=1$

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

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

$A_{21}=\left|\begin{array}{cc}-4& 2\\ -1& 1\end{array}\right|=-(-4+2)=2$

$A_{22}=\left|\begin{array}{cc}3& 2\\ -1& 1\end{array}\right|=(3+2)=5$

$A_{23}=\left|\begin{array}{cc}3& -4\\ -1& -1\end{array}\right|=-(-3-4)=7$

$A_{31}=\left|\begin{array}{cc}-4& 2\\ 1& 0\end{array}\right|=(0-2)=-2$

$A_{32}=\left|\begin{array}{cc}3& 2\\ -2& 0\end{array}\right|=-(0+4)=-4$

$A_{33}=\left|\begin{array}{cc}3& -4\\ -2& 1\end{array}\right|=(3-8)=-5$

দেওয়া আছে,

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

এখন, $C=A^t$ হলে,

$C={\left[\begin{array}{ccc}3& -4& 2\\ -2& 1& 0\\ -1& -1& 1\end{array}\right]}^t$

$\therefore C=\left[\begin{array}{ccc}3& -2& -1\\ -4& 1& -1\\ 2& 0& 1\end{array}\right]$

$\therefore C^2=C.C$

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

$C^2=\left[\begin{array}{ccc}15& -8& -2\\ -18& 9& 2\\ 8& -4& -1\end{array}\right]$

প্রদত্ত রাশি $=C^2-5C+6I$

$=\left[\begin{array}{ccc}15& -8& -2\\ -18& 9& 2\\ 8& -4& -1\end{array}\right]-5\left[\begin{array}{ccc}3& -2& -1\\ -4& 1& -1\\ 2& 0& 1\end{array}\right]+6\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}15& -8& -2\\ -18& 9& 2\\ 8& -4& -1\end{array}\right]-\left[\begin{array}{ccc}15& -10& -5\\ -20& 5& -5\\ 10& 0& 5\end{array}\right]+\left[\begin{array}{ccc}6& 0& 0\\ 0& 6& 0\\ 0& 0& 6\end{array}\right]$

$=\left[\begin{array}{ccc}6& 2& 3\\ 2& 10& 7\\ -2& -4& 0\end{array}\right]$

$\therefore C^2-5C+6I=\left[\begin{array}{ccc}6& 2& 3\\ 2& 10& 7\\ -2& -4& 0\end{array}\right]$