C–1 নির্ণেয় কর।
উত্তর: দেওয়া আছে, [গণিত] বিদ্যমান [গণিত]
HSC Higher Math 1st Paper Chapter 1: ম্যাট্রিকস ও নির্ণায়ক CQ — Prosthuti প্রশ্নব্যাংক
HSC · Higher Math 1st Paper · Chapter 1: ম্যাট্রিকস ও নির্ণায়ক · সৃজনশীল
C–1 নির্ণেয় কর।
উদ্দীপক
\mathrm{A}=\left[\[\begin{array}{ll}4 & 3 \\ 0 & 1\end{array}\]\right], \mathrm{B}=\left[\[\begin{array}{cc}\frac{1}{4} & n \\ 0 & p\end{array}\]\right] এবং \mathrm{C}=\left[\[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 0 \\ 2 & 0 & 4\end{array}\]\right]
Comilla · 2021
উত্তর
প্রশ্নমতে,
P Q=\left[\[\begin{array}{c}1 \\ -2 \\ 3\end{array}\]\right]=\left[\[\begin{array}{lll}-2 & -1 & 0\end{array}\]\right]
=\left[\[\begin{array}{ccc}-2 & -1 & 0 \\ 4 & 2 & 0 \\ -6 & -3 & 0\end{array}\]\right]
=\left[\[\begin{array}{ccc}1 \cdot(-2) & 1 \cdot(-1) & 1 \cdot 0 \\ (-2) \cdot(-2) & -2 \cdot(-1) & -2 \cdot 0 \\ 3 \cdot(-2) & 3 \cdot(-1) & 3 \cdot 0\end{array}\]\right]
\therefore[\mathrm{PQ}]^{\mathrm{T}}=\left[\[\begin{array}{ccc}-2 & 4 & 6 \\ -1 & 2 & -3 \\ 0 & 0 & 0\end{array}\]\right]
উত্তর
দেওয়া আছে,
\mathrm{A}=\left[\[\begin{array}{ll}4 & 3 \\ 0 & 1\end{array}\]\right], \mathrm{B}=\left[\[\begin{array}{ll}\frac{1}{4} & n \\ 0 & p\end{array}\]\right]
\mathrm{AB}=\left[\[\begin{array}{ll}4 & 3 \\ 0 &1\end{array}\]\right]\left[\[\begin{array}{ll}\frac{1}{4} & n \\ 0 & p\end{array}\]\right]
=\left[\[\begin{array}{cc}1 & 4 n+3 p \\ 0 & p\end{array}\]\right]
প্রশ্নমতে,
\left[\[\begin{array}{cc}1 & 4 n+3 p \\ 0 & p\end{array}\]\right]=\left[\[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\]\right]
উভয়পক্ষকে তুলনা করে পাই,
এবং
P এর মান (i) নং এ বসিয়ে,
উত্তর
দেওয়া আছে, C = [ \left.\[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 0 \\ 2 & 0 & 4\end{array}\]\right]
\therefore|C|=\left[\[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 0 \\ 2 & 0 & 4\end{array}\]\right]
বিদ্যমান
C_{11}=\left[\[\begin{array}{ll}2 & 0 \\ 0 & 4\end{array}\]\right] \quad C_{12}=\left[\[\begin{array}{ll}1 & 0 \\ 2 & 4\end{array}\]\right] \quad C_{13}=\left[\[\begin{array}{ll}1 & 2 \\ 2 & 0\end{array}\]\right]
=\left(\[\begin{array}{cc}8-0) & =-(4-0)\end{array}\]=(0-4)\right.
C_{21}=\left[\[\begin{array}{ll}1 & 2 \\ 0 & 4\end{array}\]\right] \quad C_{22}=\left[\[\begin{array}{ll}0 & 2 \\ 2 & 4\end{array}\]\right] \quad C_{23}=\left[\[\begin{array}{ll}0 & 1 \\ 2 & 0\end{array}\]\right]
C_{31}=\left[\[\begin{array}{ll}1 & 2 \\ 2 & 0\end{array}\]\right] \quad C_{12}=\left[\[\begin{array}{ll}1 & 0 \\ 2 & 4\end{array}\]\right] \quad C_{13}=\left[\[\begin{array}{ll}1 & 2 \\ 2 & 0\end{array}\]\right]
=\operatorname{Adj} C=\left[\[\begin{array}{ccc}8 & -4 & -4 \\ -4 & -4 & 2 \\ -4 & 2 & -1\end{array}\]\right]^{t}
=\left[\[\begin{array}{ccc}8 & -4 & -4 \\ -4 & -4 & 2 \\ -4 & 2 & -1\end{array}\]\right]
\therefore C^{-1}=\frac{1}{|C|} \operatorname{adj} C=\frac{1}{-12}\left[\[\begin{array}{ccc}8 & -4 & -4 \\ -4 & -4 & 2 \\ -4 & 2 & -1\end{array}\]\right]