[গণিত] ফাংশনটির রেঞ্জ নির্ণয় কর।
উত্তর: ধরি, [গণিত] বা, [গণিত] বা, [গণিত] হবে যদি ও কেবল যদি [গণিত] হয়। [গণিত] ফাংশনটির রেঞ্জ = [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
ফাংশনটির রেঞ্জ নির্ণয় কর।
উদ্দীপক
\mathrm{A}=\left[\[\begin{array}{lll} 2 & 3 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 0 \end{array}\]\right], \mathrm{B}=\mathrm{A}^{\mathrm{t}}, f(\mathrm{x})=\mathrm{x}^2-4 \mathrm{x}
Chittagong · 2017
উত্তর
ধরি,
বা,
বা,
হবে যদি ও কেবল যদি হয়।
ফাংশনটির রেঞ্জ = (Ans.)
উত্তর
দেওয়া আছে,
\therefore \mathrm{B}=\mathrm{A}^{\mathrm{t}}=\left[\[\begin{array}{lll} 2 & 3 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 0 \end{array}\]\right]=\left[\[\begin{array}{lll} 2 & 1 & 1 \\ 3 & 2 & 2 \\ 1 & 2 & 0 \end{array}\]\right]
\therefore \mathrm{B}^2=\left[\[\begin{array}{lll} 2 & 1 & 1 \\ 3 & 2 & 2 \\ 1 & 2 & 0 \end{array}\]\right]\left[\[\begin{array}{lll} 2 & 1 & 1 \\ 3 & 2 & 2 \\ 1 & 2 & 0 \end{array}\]\right]
=\left[\[\begin{array}{lll} 4+3+1 & 2+2+2 & 2+2+0 \\ 6+6+2 & 3+4+4 & 3+4+0 \\ 2+6+0 & 1+4+0 & 1+4+0 \end{array}\]\right]=\left[\[\begin{array}{ccc} 8 & 6 & 4 \\ 14 & 11 & 7 \\ 8 & 5 & 5 \end{array}\]\right]
\therefore f(\mathrm{~B})=\mathrm{B}^2-4 \mathrm{~B}=\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.) }
উত্তর
|\mathrm{B}| =\left|\[\begin{array}{lll} 2 & 1 & 1 \\ 3 & 2 & 2 \\ 1 & 2 & 0 \end{array}\]\right| ['খ' হতে প্রাপ্ত]
নির্ণয়যোগ্য।
\begin{aligned} & \mathrm{B}_{11}=(-1)^{1+1}\left|\begin{array}{ll}2 & 2 \\ 2 & 0\end{aligned}\right|=0-4=-4 \\ & \mathrm{~B}_{12}=(-1)^{1+2}\left|\[\begin{array}{ll}3 & 2 \\ 1 & 0\end{array}\]\right|=-(0-2)=2 \\ & \mathrm{~B}_{13}=(-1)^{1+3}\left|\[\begin{array}{ll}3 & 2 \\ 1 & 2\end{array}\]\right|=6-2=4 \\ & \mathrm{~B}_{21}=(-1)^{2+1}\left|\[\begin{array}{ll}1 & 1 \\ 2 & 0\end{array}\]\right|=-(0-2)=2 \\ & \mathrm{~B}_{22}=(-1)^{2+2}\left|\[\begin{array}{ll}2 & 1 \\ 1 & 0\end{array}\]\right|=0-1=-1 \\ & \mathrm{~B}_{23}=(-1)^{2+3}\left|\[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\]\right|=-(4-1)=-3 \\ & \mathrm{~B}_{31}=(-1)^{3+1}\left|\[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\]\right|=2-2=0 \\ & \mathrm{~B}_{32}=(-1)^{3+2}\left|\[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\]\right|=-(4-3)=-1 \\ & \mathrm{~B}_{33}=(-1)^{3+3}\left|\[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\]\right|=4-3=1\end{aligned}
\end{array}
\begin{aligned} & \therefore \mathrm{B}^{-1}=\frac{1}{|\mathrm{~B}|} \operatorname{adjB}=\frac{1}{-2}\left[\begin{array}{ccc}-4 & 2 & 4 \\ 2 & -1 & -3 \\ 0 & -1 & 1\end{aligned}\right]^{\mathrm{t}} \\ &=\frac{-1}{2}\left[\[\begin{array}{ccc}-4 & 2 & 0 \\ 2 & -1 & -1 \\ 4 & -3 & 1\end{array}\]\right] \text{ (Ans.) }\end{aligned}
\end{array}\end{array}