[গণিত] হলে দেখাও যে, [গণিত] একটি সমঘাতি ম্যাট্রিক্স।
উত্তর: দেওয়া আছে, [গণিত] সমঘাতী ম্যাট্রিক্স হলে, [গণিত] হবে। এখন, [গণিত] যেহেতু [গণিত] একটি সমঘাতী ম্যাট্রিক্স। (দেখানো হলো)
HSC Higher Math 1st Paper Chapter 1: ম্যাট্রিকস ও নির্ণায়ক CQ — Prosthuti প্রশ্নব্যাংক
HSC · Higher Math 1st Paper · Chapter 1: ম্যাট্রিকস ও নির্ণায়ক · সৃজনশীল
হলে দেখাও যে, একটি সমঘাতি ম্যাট্রিক্স।
উদ্দীপক
দৃশ্যকল্প-১ :
দৃশ্যকল্প-২ : A=\left[\[\begin{array}{rrr}2 & 1 & 1 \\ -2 & 0 & 3 \\ 2 & 3 & 1\end{array}\]\right]
Dhaka · 2025
উত্তর
দেওয়া আছে, C=\left[\[\begin{array}{rr}1 & -1 \\ 0 & 0\end{array}\]\right]
\therefore \quad I-C=\left[\[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\]\right]-\left[\[\begin{array}{rr}1 & -1 \\ 0 & 0\end{array}\]\right]
=\left[\[\begin{array}{ll}1-1 & 0+1 \\ 0-0 & 1-0\end{array}\]\right]=\left[\[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\]\right]
সমঘাতী ম্যাট্রিক্স হলে, হবে।
এখন,
=\left[\[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\]\right]\left[\[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\]\right]
=\left[\[\begin{array}{ll}0+0 & 0+1 \\ 0+0 & 0+1\end{array}\]\right]=\left[\[\begin{array}{ll}0 & 1 \\ 0 & 1\end{array}\]\right]=I-C
যেহেতু
একটি সমঘাতী ম্যাট্রিক্স। (দেখানো হলো)
উত্তর
দেওয়া আছে, \mathrm{A}=\left[\[\begin{array}{rrr}2 & 1 & 1 \\ -2 & 0 & 3 \\ 2 & 3 & 1\end{array}\]\right]
এবং
;
যেখানে, \mathrm{I}_3=\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right]
এখন,
=\left[\[\begin{array}{rrr}2 & 1 & 1 \\ -2 & 0 & 3 \\ 2 & 3 & 1\end{array}\]\right]\left[\[\begin{array}{rrr}2 & 1 & 1 \\ -2 & 0 & 3 \\ 2 & 3 & 1\end{array}\]\right]
=\left[\[\begin{array}{rrr}4-2+2 & 2+0+3 & 2+3+1 \\ -4+0+6 & -2+0+9 & -2+0+3 \\ 4-6+2 & 2+0+3 & 2+9+1\end{array}\]\right]
3A =3\left[\[\begin{array}{rrr}2 & 1 & 1 \\ -2 & 0 & 3 \\ 2 & 3 & 1\end{array}\]\right]=\left[\[\begin{array}{rrr}6 & 3 & 3 \\ -6 & 0 & 9 \\ 6 & 9 & 3\end{array}\]\right]
2 I_3 =2\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right]=\left[\[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\]\right]
=\left[\[\begin{array}{rrr}4 & 5 & 6 \\ 2 & 7 & 1 \\ 0 & 5 & 12\end{array}\]\right]+\left[\[\begin{array}{rrr}6 & 3 & 3 \\ -6 & 0 & 9 \\ 6 & 9 & 3\end{array}\]\right]-\left[\[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\]\right]
=\left[\[\begin{array}{ccc}4+6-2 & 5+3-0 & 6+3-0 \\ 2-6-0 & 7+0-2 & 1+9-0 \\ 0+6-0 & 5+9-0 & 12+3-2\end{array}\]\right]
=\left[\[\begin{array}{rcc}8 & 8 & 9 \\ -4 & 5 & 10 \\ 6 & 14 & 13\end{array}\]\right]
উত্তর
দেওয়া আছে, \mathrm{A}=\left[\[\begin{array}{rrr}2 & 1 & 1 \\ -2 & 0 & 3 \\ 2 & 3 & 1\end{array}\]\right]
এবং \mathrm{B}=\mathrm{A}^{\mathrm{T}}=\left[\[\begin{array}{rrr}2 & 1 & 1 \\ -2 & 0 & 3 \\ 2 & 3 & 1\end{array}\]\right]^{\mathrm{T}}
\therefore \mathrm{B}=\left[\[\begin{array}{rrr}2 & -2 & 2 \\ 1 & 0 & 3 \\ 1 & 3 & 1\end{array}\]\right]
\therefore |\mathrm{B}|=\left|\[\begin{array}{rrr}2 & -2 & 2 \\ 1 & 0 & 3 \\ 1 & 3 & 1\end{array}\]\right|
বিদ্যমান।
Adj B=\left[\[\begin{array}{lll}B_{11} & B_{12} & B_{13} \B_{21} & B_{22} & B_{23} \B_{31} & B_{32} & B_{33}\end{array}\]\right]^T
B এর সহগুণকসমূহ :
\mathrm{B}_{11}=\left|\[\begin{array}{ll}0 & 3 \\ 3 & 1\end{array}\]\right|=0-9=-9
\mathrm{~B}_{12}=-\left|\[\begin{array}{ll}1 & 3 \\ 1 & 1\end{array}\]\right|=-(1-3)=2
\mathrm{~B}_{13}=\left|\[\begin{array}{ll}1 & 0 \\ 1 & 3\end{array}\]\right|=3-0=3
\mathrm{~B}_{21}=-\left|\[\begin{array}{rr}-2 & 2 \\ 3 & 1\end{array}\]\right|=-(-2-6)=8
\mathrm{~B}_{22}=\left|\[\begin{array}{ll}2 & 2 \\ 1 & 1\end{array}\]\right|=2-2=0
\mathrm{~B}_{23}=-\left|\[\begin{array}{rr}2 & -2 \\ 1 & 3\end{array}\]\right|=-(6+2)=-8
B_{31}=\left|\[\begin{array}{rr}-2 & 2 \\ 0 & 3\end{array}\]\right|=-6-0=-6
B_{32}=-\left|\[\begin{array}{ll}2 & 2 \\ 1 & 3\end{array}\]\right|=-(6-2)=-4
B_{33}=\left|\[\begin{array}{rr}2 & -2 \\ 1 & 0\end{array}\]\right|=0+2=2
\therefore \quad \text{ Adj } B=\left[\[\begin{array}{rrr}-9 & 2 & 3 \\ 8 & 0 & -8 \\ -6 & -4 & 2\end{array}\]\right]^T
=\left[\[\begin{array}{rrr}-9 & 8 & -6 \\ 2 & 0 & -4 \\ 3 & -8 & 2\end{array}\]\right]
Adj Adj
=\frac{1}{-16}\left[\[\begin{array}{rrr}-9 & 8 & -6 \\ 2 & 0 & -4 \\ 3 & -8 & 2\end{array}\]\right]
=\left[\[\begin{array}{ccc}\frac{-9}{-16} & \frac{8}{-16} & \frac{-6}{-16} \\ \frac{2}{-16} & \frac{0}{-16} & \frac{-4}{-16} \\ \frac{3}{-16} & \frac{-8}{-16} & \frac{2}{-16}\end{array}\]\right]
=\left[\[\begin{array}{ccc}\frac{9}{16} & -\frac{1}{2} & \frac{3}{8} \\ -\frac{1}{8} & 0 & \frac{1}{4} \\ -\frac{3}{16} & \frac{1}{2} & -\frac{1}{8}\end{array}\]\right]
নির্ণেয় B^{-1}=\left[\[\begin{array}{ccc}\frac{9}{16} & -\frac{1}{2} & \frac{3}{8} \\ -\frac{1}{8} & 0 & \frac{1}{4} \\ -\frac{3}{16} & \frac{1}{2} & -\frac{1}{8}\end{array}\]\right]