সৃজনশীলঃ ‘(i) |a+b+2cabcb+c+2abcac+a+2b|−P=0.(ii) M=[2−42−211−1−43]…’ ২০২৬ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$\begin{array}{cc}\therefore & \left|\begin{array}{ccc}a+b+2c& a& b\\ c& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|-P=0\\ \therefore & \left|\begin{array}{ccc}a+b+2c& a& b\\ c& b+c+2a& b\\ c& a& c+a+2b\end{array}\right|\\ & =\left|\begin{array}{ccc}a+b+c& -a-b-c& 0\\ 0& a+b+c& -a-b-c\\ c& a& c+a+2b\end{array}\right|[\because r_1^{\prime }=r_1-r_2,r_2^{\prime }=r_2-r_3]\end{array}$

$\begin{array}{cc}\therefore & (a+b+c{)}^2\left|\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ c& a& c+a+2b\end{array}\right|\\ =& (a+b+c{)}^2\{1(c+a+2b+a)+1(0+c)+0\}\\ =& (a+b+c{)}^2(2a+2b+2c)\\ =& 2(a+b+c{)}^2(a+b+c)\\ =& 2(a+b+c{)}^3\end{array}$

বামপক্ষ $=0.5P=0.5\times 2(a+b+c{)}^3$

$=(a+b+c{)}^3=5$ ডানপক্ষ

$\therefore 0.5P=(a+b+c{)}^3$. (দেখানো হলো)

দেওয়া আছে, $M=\left[\begin{array}{ccc}2& -4& 2\\ -2& 1& 1\\ -1& -4& 3\end{array}\right]$

এখন, $\mathrm{IM}=M=\left[\begin{array}{ccc}2& -4& 2\\ -2& 1& 1\\ -1& -4& 3\end{array}\right]$

$\begin{array}{cc}\therefore |\mathrm{IM}|& =\left|\begin{array}{ccc}2& -4& 2\\ -2& 1& 1\\ -1& -4& 3\end{array}\right|\\ & =2(3+4)+4(-6+1)+2(8+1)\\ & =14-20+18=12\ne 0\end{array}$

∴ (IM) ${}^{-1}$ নির্ণয়যোগ্য ।

(IM) এর সহগুণকসমূহ :

$\begin{array}{cc}& (\mathrm{IM}{)}_{11}=\left|\begin{array}{cc}1& 1\\ -4& 3\end{array}\right|=3+4=7\\ & (\mathrm{IM}{)}_{12}=-\left|\begin{array}{cc}-2& 1\\ -1& 3\end{array}\right|=-(-6+1)=5\\ & (\mathrm{IM}{)}_{13}=\left|\begin{array}{cc}-2& 1\\ -1& -4\end{array}\right|=8+1=9\\ & (\mathrm{IM}{)}_{21}=-\left|\begin{array}{cc}-4& 2\\ -4& 3\end{array}\right|=-(-12+8)=4\\ & (\mathrm{IM}{)}_{22}=\left|\begin{array}{cc}2& 2\\ -1& 3\end{array}\right|=6+2=8\end{array}$

$\begin{array}{cc}& (\mathrm{IM}{)}_{23}=-\left|\begin{array}{cc}2& -4\\ -1& -4\end{array}\right|=-(-8-4)=12\\ & (\mathrm{IM}{)}_{31}=\left|\begin{array}{cc}-4& 2\\ 1& 1\end{array}\right|=-4-2=-6\\ & (\mathrm{IM}{)}_{32}=-\left[\begin{array}{cc}2& 2\\ -2& 1\end{array}\right]=-(2+4)=-6\\ & (\mathrm{IM}{)}_{33}=\left|\begin{array}{cc}2& -4\\ -2& 1\end{array}\right|=2-8=-6\end{array}$

$\begin{array}{cc}& \therefore \text{ Adj (IM) }={\left[\begin{array}{ccc}7& 5& 9\\ 4& 8& 12\\ -6& -6& -6\end{array}\right]}^T=\left[\begin{array}{ccc}7& 4& -6\\ 5& 8& -6\\ 9& 12& -6\end{array}\right]\\ & \therefore (\mathrm{IM}{)}^{-1}=\frac{1}{\left|\mathrm{IM}\right|}\text{ Adj (IM) }=\frac{1}{12}\left[\begin{array}{ccc}7& 4& -6\\ 5& 8& -6\\ 9& 12& -6\end{array}\right]\end{array}$

নির্ণেয় $(IM{)}^{-1}=\frac{1}{12}\left[\begin{array}{ccc}7& 4& -6\\ 5& 8& -6\\ 9& 12& -6\end{array}\right]$.