সৃজনশীলঃ ‘→M=[1213−3−1210]…’ দিনাজপুর বোর্ড ২০১৭ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

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

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

$=\left[\begin{array}{ccc}1+6+2& 2-6+1& 1-2+0\\ 3-9-2& 6+9-1& 3+3-0\\ 2+3+0& 4-3+0& 2-1+0\end{array}\right]=\left[\begin{array}{ccc}9& -3& -1\\ -8& 14& 6\\ 5& 1& 1\end{array}\right]$

 প্রদত্ত রাশি $=M^2-3M+\mathrm{MI}$

$=\left[\begin{array}{ccc}9& -3& -1\\ -8& 14& 6\\ 5& 1& 1\end{array}\right]-3\left[\begin{array}{ccc}1& 2& 1\\ 3& -3& -1\\ 2& 1& 0\end{array}\right]+\left[\begin{array}{ccc}1& 2& 1\\ 3& -3& -1\\ 2& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}9& -3& -1\\ -8& 14& 6\\ 5& 1& 1\end{array}\right]-\left[\begin{array}{ccc}3& 6& 3\\ 9& -9& -3\\ 6& 3& 0\end{array}\right]+\left[\begin{array}{ccc}1& 2& 1\\ 3& -3& -1\\ 2& 1& 0\end{array}\right]$

$=\left[\begin{array}{ccc}9-3+1& -3-6+2& -1-3+1\\ -8-9+3& 14+9-3& 6+3-1\\ 5-6+2& 1-3+1& 1-0+0\end{array}\right]$

$=\left[\begin{array}{ccc}7& -7& -3\\ -14& 20& 8\\ 1& -1& 1\end{array}\right]\text{ (Ans.) }$

$|M|=\left|\begin{array}{ccc}1& 2& 1\\ 3& -3& -1\\ 2& 1& 0\end{array}\right|$

$=1(-0+1)-2(0+2)+1(3+6)$

$=1-4+9=6\ne 0$

$\therefore M^{-1}$ নির্ণয়যোগ্য।

$\mathrm{adj}M={\left[\begin{array}{ccc}\left|\begin{array}{cc}-3& -1\\ 1& 0\end{array}\right|& -\left|\begin{array}{cc}3& -1\\ 2& 0\end{array}\right|& \left|\begin{array}{cc}3& -3\\ 2& 1\end{array}\right|\\ -\left|\begin{array}{cc}2& 1\\ 1& 0\end{array}\right|& \left|\begin{array}{cc}1& 1\\ 2& 0\end{array}\right|& -\left|\begin{array}{cc}1& 2\\ 2& 1\end{array}\right|\\ \left|\begin{array}{cc}2& 1\\ -3& -1\end{array}\right|& -\left|\begin{array}{cc}1& 1\\ 3& -1\end{array}\right|& \left|\begin{array}{cc}1& 2\\ 3& -3\end{array}\right|\end{array}\right]}^t$

$={\left[\begin{array}{ccc}1& -2& 9\\ 1& -2& 3\\ 1& 4& -9\end{array}\right]}^t=\left[\begin{array}{ccc}1& 1& 1\\ -2& -2& 4\\ 9& 3& -9\end{array}\right]$

$\therefore M^{-1}=\frac{1}{\left|M\right|}$ adj $M=\frac{1}{6}\left[\begin{array}{ccc}1& 1& 1\\ -2& -2& 4\\ 9& 3& -9\end{array}\right]$ (Ans.)