[গণিত] নির্ণয় কর।
উত্তর: দেওয়া আছে, [গণিত] প্রদত্ত রাশি [গণিত] (Ans.)
HSC Higher Math 1st Paper Chapter 1: ম্যাট্রিকস ও নির্ণায়ক CQ — Prosthuti প্রশ্নব্যাংক
HSC · Higher Math 1st Paper · Chapter 1: ম্যাট্রিকস ও নির্ণায়ক · সৃজনশীল
[গণিত] নির্ণয় কর।
উদ্দীপক
\partial . \Delta \mathrm{A}=\left[\[\begin{array}{rrr}1 & 2 & -3 \\ -2 & 1 & 0\end{array}\]\right], \mathrm{B}=\left[\[\begin{array}{rr}-1 & 2 \\ -3 & 7 \\ 5 & 0\end{array}\]\right], \mathrm{C}=\left[\[\begin{array}{rr}5 & 1 \\ 1 & -3\end{array}\]\right]
\mathrm{D}=\left[\[\begin{array}{rrr} 1 & 1 & -1 \\ 0 & -2 & 2 \\ 1 & 2 & 3 \end{array}\]\right]
Comilla · 2017
উত্তর
\left[\[\begin{array}{cc}\mathrm{x}^2 & 2 \mathrm{x} \\ 5 & 3\end{array}\]\right] ম্যাট্রিক্সটি ব্যতিক্রমী হলে, \left|\[\begin{array}{cc}\mathrm{x}^2 & 2 \mathrm{x} \\ 5 & 3\end{array}\]\right|=0
বা, বা,
(Ans.)
উত্তর
দেওয়া আছে, \mathrm{A}=\left[\[\begin{array}{rrr}1 & 2 & -3 \\ -2 & 1 & 0\end{array}\]\right], \mathrm{B}=\left[\[\begin{array}{rr}-1 & 2 \\ -3 & 7 \\ 5 & 0\end{array}\]\right]
A B=\left[\[\begin{array}{ccc} 1 & 2 & -3 \\ -2 & 1 & 0 \end{array}\]\right]\left[\[\begin{array}{cc} -1 & 2 \\ -3 & 7 \\ 5 & 0 \end{array}\]\right]
=\left[\[\begin{array}{cc} -1-6-15 & 2+14-0 \\ 2-3+0 & -4+7+0 \end{array}\]\right]=\left[\[\begin{array}{cc} -22 & 16 \\ -1 & 3 \end{array}\]\right]
C^2=C \cdot C=\left[\[\begin{array}{cc} 5 & 1 \\ 1 & -3 \end{array}\]\right]\left[\[\begin{array}{cc} 5 & 1 \\ 1 & -3 \end{array}\]\right]=\left[\[\begin{array}{cc} 25+1 & 5-3 \\ 5-3 & 1+9 \end{array}\]\right]=\left[\[\begin{array}{cc} 26 & 2 \\ 2 & 10 \end{array}\]\right]
প্রদত্ত রাশি
=\left[\[\begin{array}{cc} -22 & 16 \\ -1 & 3 \end{array}\]\right]-\left[\[\begin{array}{cc} 26 & 2 \\ 2 & 10 \end{array}\]\right]+\left[\[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\]\right]
=\left[\[\begin{array}{cc} -22-26+2 & 16-2+0 \\ -1-2+0 & 3-10+2 \end{array}\]\right]=\left[\[\begin{array}{cc} -46 & 14 \\ -3 & -5 \end{array}\]\right]
(Ans.)
উত্তর
|\mathrm{D}|=\left|\[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -2 & 2 \\ 1 & 2 & 3 \end{array}\]\right| =1(-6-4)-1(0-2)+(-1)(0+2)
নির্ণয়যোগ্য।
\therefore \operatorname{adj} D=\left[\[\begin{array}{lll}\left|\begin{array}{cc}-2 & 2 \\ 2 & 3\end{array}\right| & -\left|\begin{array}{ll}0 & 2 \\ 1 & 3\end{array}\right| & \left|\begin{array}{cc}0 & -2 \\ 1 & 2\end{array}\right| \\ -\left|\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right| & \left|\begin{array}{cc}1 & -1 \\ 1 & 3\end{array}\right| & -\left|\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right| \\ \left|\begin{array}{cc}1 & -1 \\ -2 & 2\end{array}\right| & -\left|\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right| & \left|\begin{array}{cc}1 & 1 \\ 0 & -2\end{array}\right|\end{array}\]\right]^{t}
=\left[\[\begin{array}{ccc}-10 & 2 & 2 \\ -5 & 4 & -1 \\ 0 & -2 & -2\end{array}\]\right]^{\mathrm{t}}=\left[\[\begin{array}{ccc}-10 & -5 & 0 \\ 2 & 4 & -2 \\ 2 & -1 & -2\end{array}\]\right]
adj \mathrm{D}=\frac{1}{-10}\left[\[\begin{array}{ccc}-10 & -5 & 0 \\ 2 & 4 & -2 \\ 2 & -1 & -2\end{array}\]\right]
=\frac{1}{10}\left[\[\begin{array}{rrr}10 & 5 & 0 \\ -2 & -4 & 2 \\ -2 & 1 & 2\end{array}\]\right] (Ans.)