সৃজনশীলঃ ‘A=[231122120],B=At,f(x)=x2−4x…’ চট্টগ্রাম বোর্ড ২০১৭ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=x^2-4x$

$\therefore f\left(B\right)=B^2-4B$

$\therefore B=A^t=\left[\begin{array}{ccc}2& 3& 1\\ 1& 2& 2\\ 1& 2& 0\end{array}\right]=\left[\begin{array}{ccc}2& 1& 1\\ 3& 2& 2\\ 1& 2& 0\end{array}\right]$

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

$=\left[\begin{array}{ccc}4+3+1& 2+2+2& 2+2+0\\ 6+6+2& 3+4+4& 3+4+0\\ 2+6+0& 1+4+0& 1+4+0\end{array}\right]=\left[\begin{array}{ccc}8& 6& 4\\ 14& 11& 7\\ 8& 5& 5\end{array}\right]$

$\therefore f\left(B\right)=B^2-4B=\left[\begin{array}{ccc}8& 6& 4\\ 14& 11& 7\\ 8& 5& 5\end{array}\right]-\left[\begin{array}{ccc}8& 4& 4\\ 12& 8& 8\\ 4& 8& 0\end{array}\right]$

$=\left[\begin{array}{ccc}0& 2& 0\\ 2& 3& -1\\ 4& -3& 5\end{array}\right]\text{ (Ans.) }$

$|B|=\left|\begin{array}{ccc}2& 1& 1\\ 3& 2& 2\\ 1& 2& 0\end{array}\right|$ [‘খ' হতে প্রাপ্ত] 

$=2(0-4)-1(0-2)+1(6-2)$

$=-8+2+4=-2\ne 0$

$\therefore B^{-1}$ নির্ণয়যোগ্য।

$\begin{array}{cc}& B_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}2& 2\\ 2& 0\end{array}\right|=0-4=-4\\ & B_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}3& 2\\ 1& 0\end{array}\right|=-(0-2)=2\\ & B_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}3& 2\\ 1& 2\end{array}\right|=6-2=4\\ & B_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}1& 1\\ 2& 0\end{array}\right|=-(0-2)=2\\ & B_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}2& 1\\ 1& 0\end{array}\right|=0-1=-1\\ & B_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}2& 1\\ 1& 2\end{array}\right|=-(4-1)=-3\\ & B_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}1& 1\\ 2& 2\end{array}\right|=2-2=0\\ & B_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}2& 1\\ 3& 2\end{array}\right|=-(4-3)=-1\\ & B_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}2& 1\\ 3& 2\end{array}\right|=4-3=1\end{array}$

$\begin{array}{cc}& \therefore B^{-1}=\frac{1}{\left|B\right|}\mathrm{adjB}=\frac{1}{-2}{\left[\begin{array}{ccc}-4& 2& 4\\ 2& -1& -3\\ 0& -1& 1\end{array}\right]}^t\\ & =\frac{-1}{2}\left[\begin{array}{ccc}-4& 2& 0\\ 2& -1& -1\\ 4& -3& 1\end{array}\right]\text{ (Ans.) }\end{array}$