[গণিত] হলে [গণিত] নির্ণয় কর।
উত্তর: দেওয়া আছে, [গণিত] হলে, [গণিত] ∵ [গণিত] বিদ্যমান। এখন, [গণিত] নির্ণেয় [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
হলে নির্ণয় কর।
উদ্দীপক
A=\left[\[\begin{array}{ccc}3+x & 4 & 2 \\ 4 & 2+x & 3 \\ 2 & 3 & 4+x\end{array}\]\right], B=\left[\[\begin{array}{l}x \y \z\end{array}\]\right], C=\left[\[\begin{array}{r}-1 \\ 0 \\ 2\end{array}\]\right]
মাইলস্টোন কলেজ, ঢাকা · 2025
উত্তর
বামপক্ষ =\left|\[\begin{array}{ccc}x+y & 3(y+z) & z+x \\ 1 & 3 & 1 \z & 3 x & y\end{array}\]\right|
=\left|\[\begin{array}{ccc}x+y+z & 3(x+y+z) & x+y+z \\ 1 & 3 & 1 \z & 3 x & y\end{array}\]\right|
প্রয়োগ করে]
=(x+y+z)\left|\[\begin{array}{rrr}1 & 3 & 1 \\ 1 & 3 & 1 \z & 3 x & y\end{array}\]\right|
ডানপক্ষ
\therefore\left|\[\begin{array}{ccc}x+y & 3(x+z) & z+x \\ 1 & 3 & 1 \z & 3 x & z\end{array}\]\right|=0 (প্রমাণিত)
উত্তর
দেওয়া আছে, A=\left[\[\begin{array}{ccc}3+x & 4 & 2 \\ 4 & 2+x & 3 \\ 2 & 3 & 4+x\end{array}\]\right]
হলে, A=\left[\[\begin{array}{lll}2 & 4 & 2 \\ 4 & 1 & 3 \\ 2 & 3 & 3\end{array}\]\right]
∵
বিদ্যমান।
এখন, A_{11}=\left|\[\begin{array}{ll}1 & 3 \\ 3 & 3\end{array}\]\right|=3-9=-6
\begin{aligned} & A_{12}=-\left|\begin{array}{ll}4 & 3 \\ 2 & 3\end{aligned}\right|=-(12-6)=-6 \\ & A_{13}=\left|\[\begin{array}{ll}4 & 1 \\ 2 & 3\end{array}\]\right|=12-2=10 \\ & A_{21}=-\left|\[\begin{array}{ll}4 & 2 \\ 3 & 3\end{array}\]\right|=-(12-6)=-6 \\ & A_{22}=\left|\[\begin{array}{ll}2 & 2 \\ 2 & 3\end{array}\]\right|=6-4=2 \\ & A_{23}=-\left|\[\begin{array}{ll}2 & 4 \\ 2 & 3\end{array}\]\right|=-(6-8)=2 \\ & A_{31}=\left|\[\begin{array}{ll}4 & 2 \\ 1 & 3\end{array}\]\right|=12-2=10\end{aligned}
\end{array}
\begin{aligned} & A_{32}=-\left|\begin{array}{ll}2 & 2 \\ 4 & 3\end{aligned}\right|=-(6-8)=2 \\ & A_{33}=\left|\[\begin{array}{ll}2 & 4 \\ 4 & 1\end{array}\]\right|=2-16=-14\end{aligned}
\end{array}
\operatorname{Adj} A=\left[\[\begin{array}{lll}A_{11} & A_{12} & A_{13} \A_{21} & A_{22} & A_{23} \A_{31} & A_{32} & A_{33}\end{array}\]\right]^{\mathrm{T}}
=\left[\[\begin{array}{ccc}-6 & -6 & 10 \\ -6 & 2 & 2 \\ 10 & 2 & -14\end{array}\]\right]^T=\left[\[\begin{array}{ccc}-6 & -6 & 10 \\ -6 & 2 & 2 \\ 10 & 2 & -14\end{array}\]\right]
A^{-1}=\frac{1}{|A|} A d j A=\frac{1}{-16}\left[\[\begin{array}{ccc}-6 & -6 & 10 \\ -6 & 2 & 2 \\ 10 & 2 & -14\end{array}\]\right]=\left[\[\begin{array}{ccc}\frac{3}{8} & \frac{3}{8} & \frac{-5}{8} \\ \frac{3}{8} & \frac{-1}{8} & \frac{-1}{8} \\ \frac{-5}{8} & \frac{-1}{8} & \frac{7}{8}\end{array}\]\right]
নির্ণেয় A^{-1}=\left[\[\begin{array}{ccc}\frac{3}{8} & \frac{3}{8} & \frac{-5}{8} \\ \frac{3}{8} & \frac{-1}{8} & \frac{-1}{8} \\ \frac{-5}{8} & \frac{-1}{8} & \frac{7}{8}\end{array}\]\right]\end{array}
উত্তর
দেওয়া আছে, P=\left[\[\begin{array}{rrr}2 & 1 & 1 \\ -1 & 2 & 0 \\ 3 & 0 & 1\end{array}\]\right], B=\left[\[\begin{array}{l}x \y \z\end{array}\]\right] এবং C=\left[\[\begin{array}{r}-1 \\ 0 \\ 2\end{array}\]\right]
এখানে,
বা, \left[\[\begin{array}{ccc}2 & 1 & 1 \\ -1 & 2 & 0 \\ 3 & 0 & 1\end{array}\]\right]^t \left[\[\begin{array}{l}x \y \z\end{array}\]\right]=\left[\[\begin{array}{r}-1 \\ 0 \\ 2\end{array}\]\right] বা, \left[\[\begin{array}{rrr}2 & -1 & 3 \\ 1 & 2 & 0 \\ 1 & 0 & 1\end{array}\]\right]\left[\[\begin{array}{l}x \y \z\end{array}\]\right]=\left[\[\begin{array}{r}-1 \\ 0 \\ 2\end{array}\]\right]
\therefore\left[\[\begin{array}{c}2 x-y+3 z \x+2 y \x+z\end{array}\]\right]=\left[\[\begin{array}{r}-1 \\ 0 \\ 2\end{array}\]\right]
ম্যাট্রিক্সের সমতা অনুসারে,
ও এর সহগগুচ্ছ নিয়ে গঠিত নির্ণায়ক,
D=\left|\[\begin{array}{rrr}2 & -1 & 3 \\ 1 & 2 & 0 \\ 1 & 0 & 1\end{array}\]\right|
D_x=\left|\[\begin{array}{rrr}-1 & -1 & 3 \\ 0 & 2 & 0 \\ 2 & 0 & 1\end{array}\]\right|
D_y=\left|\[\begin{array}{rrr}2 & -1 & 3 \\ 1 & 0 & 0 \\ 1 & 2 & 1\end{array}\]\right|
D_z=\left|\[\begin{array}{rrr}2 & -1 & -1 \\ 1 & 2 & 0 \\ 1 & 0 & 2\end{array}\]\right|
এবং
নির্ণেয় সমাধান: