[গণিত] হলে নিয়ামকের সাহায্যে সমাধান কর, যখন [গণিত] .
উত্তর: দেওয়া আছে, [গণিত] এখন, [গণিত] বা, [গণিত] বা, [গণিত] এবং [গণিত] নির্ণেয় সমাধন: [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 নিয়ামকের সাহায্যে সমাধান কর, যখন .
উদ্দীপক
A=\left[\[\begin{array}{ccc}1+\mathrm{m} & 2 & 3 \\ 2 & 3+\mathrm{m} & 1 \\ 3 & 1 & 2+\mathrm{m}\end{array}\]\right], \mathrm{B}=\left[\[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\]\right], \mathrm{C}=\left[\[\begin{array}{l}6 \\ 0 \\ 8\end{array}\]\right]
Rajshahi · 2021
উত্তর
\begin{aligned} & \text{ বামপক্ষ }=\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & \mathrm{p} & \mathrm{p}^2 \\ 1 & \mathrm{p}^2 & \mathrm{p}^4\end{aligned}\right| \\ & =\left|\[\begin{array}{ccc}0 & 0 & 1 \\ 1-\mathrm{p} & \mathrm{p}-\mathrm{p}^2 & \mathrm{p}^2 \\ 1-\mathrm{p}^2 & \mathrm{p}^2-\mathrm{p}^4 & \mathrm{p}^4\end{array}\]\right|\left[\[\begin{array}{c}\mathrm{c}_1^{\prime}=\mathrm{c}_1-\mathrm{c}_2 \text{ এবং } \\ \mathrm{c}_2^{\prime}=\mathrm{c}_2-\mathrm{c}_3 \text{ প্রয়োগ করে }\end{array}\]\right] \\ & =\left|\[\begin{array}{ll}1-\mathrm{p} & \mathrm{p}(1-\mathrm{p}) \\ 1-\mathrm{p}^2 & \mathrm{p}^2\left(1-\mathrm{p}^2\right)\end{array}\]\right| \\ & =\left(1-\mathrm{p}^2\right)\left|\[\begin{array}{cc}1-\mathrm{p} & \mathrm{p}(1-\mathrm{p}) \\ 1 & \mathrm{p}^2\end{array}\]\right| \\ & =\mathrm{p}\left(1-\mathrm{p}^2\right)(1-\mathrm{p})\left|\[\begin{array}{ll}1 & 1 \\ 1 & \mathrm{p}\end{array}\]\right| \\ & =\mathrm{p}\left(1-\mathrm{p}^2\right)(1-\mathrm{p})(\mathrm{p}-1)=\mathrm{p}\left(\mathrm{p}^2-1\right)(\mathrm{p}-1)(\mathrm{p}-1) \text{. } \\ & =\dot{p}\left(\mathrm{p}^2-1\right)(\mathrm{p}-1)^2=\text{ ডানপক } \\ & \therefore\left|\[\begin{array}{lll}1 & 1 & 1 \\ 1 & \mathrm{p} & \mathrm{p}^2 \\ 1 & \mathrm{p}^2 & \mathrm{p}^4\end{array}\]\right|=\mathrm{p}(\mathrm{p}-1)^2\left(\mathrm{p}^2-1\right) \text{ (প্রমাণিত) } \\ & \end{aligned}
\end{array}\end{array}
উত্তর
দেওয়া আছে, A = \left[\[\begin{array}{ccc}1+\mathrm{m} & 2 & 3 \\2 & 3+\mathrm{m} & 1 \\3 & 1 & 2+\mathrm{m}\end{array}\]\right]
\mathrm{m}=0 \text{ হলে, } \mathrm{A}=\left[\[\begin{array}{lll}1 & 2 & 3 \\2 & 3 & 1 \\3 & 1 & 2\end{array}\]\right]
বিদ্যমান।
\therefore \text{ Adj } \mathrm{A}=\left[\[\begin{array}{lll}\mathrm{A}_{11} & \mathrm{~A}_{12} & \mathrm{~A}_{13} \\ \mathrm{~A}_{21} & \mathrm{~A}_{22} & \mathrm{~A}_{23} \mathrm{~A}_{31} & \mathrm{~A}_{32} & \mathrm{~A}_{32}\end{array}\]\right]^{\mathrm{T}}
A এর সহগুণকগুলি হচ্ছে
A_{11}=\left|\[\begin{array}{ll}3 & 1 \\1 & 2\end{array}\]\right|=6-1=5
A_{12}=-\left|\[\begin{array}{ll}2 & 1 \\3 & 2\end{array}\]\right|=-(4-3)=-1
A_{13}=\left|\[\begin{array}{ll}2 & 3 \\3 & 1\end{array}\]\right|=2-9=-7
A_{21}=-\left|\[\begin{array}{ll}2 & 3 \\1 & 2\end{array}\]\right|=-(4-3)=-1
A_{22}=\left|\[\begin{array}{ll}1 & 3 \\3 & 2\end{array}\]\right|=2-9=-7
A_{23}=-\left|\[\begin{array}{ll}1 & 2 \\3 & 1\end{array}\]\right|=-(1-6)=5
A_{31}=\left|\[\begin{array}{ll}2 & 3 \\3 & 1\end{array}\]\right|=2-9=-7
A_{32}=-\left|\[\begin{array}{ll}1 & 3 \\2 & 1\end{array}\]\right|=-(1-6)=5
A_{33}=\left|\[\begin{array}{ll}1 & 2 \\2 & 3\end{array}\]\right|=3-4=-1
Adj A=\left[\[\begin{array}{rrr}5 & -1 & -7 \\-1 & -7 & 5 \\-7 & 5 & -1\end{array}\]\right]^T=\left[\[\begin{array}{rrr}5 & -1 & -7 \\-1 & -7 & 5 \\-7 & 5 & -1\end{array}\]\right]
=\frac{1}{-18}\left[\[\begin{array}{rrr}5 & -1 & -7 \\ -1 & -7 & 5 \\ -7 & 5 & -1\end{array}\]\right]=\frac{1}{18}\left[\[\begin{array}{rrr}-5 & 1 & 7 \\ 1 & 7 & -5 \\ 7 & -5 & 1\end{array}\]\right]
\therefore \mathrm{A}^{-1}=\frac{1}{18}\left[\[\begin{array}{rrr}-5 & 1 & 7 \\1 & 7 & -5 \\7 & -5 & 1\end{array}\]\right]
উত্তর
দেওয়া আছে, } A=\left[\[\begin{array}{ccc}1+\mathrm{m} & 2 & 3 \\2 & 3+\mathrm{m} & 1 \\3 & 1 & 2+\mathrm{m}\end{array}\]\right]
\mathrm{m}=1 \text{ হলে, } \mathrm{A}=\left[\[\begin{array}{lll}2 & 2 & 3 \\2 & 4 & 1 \\3 & 1 & 3\end{array}\]\right]
\therefore|\mathrm{A}|=\left|\[\begin{array}{lll}2 & 2 & 3 \\2 & 4 & 1 \\3 & 1 & 3\end{array}\]\right|
এখন,
বা, \left[\[\begin{array}{lll}2 & 2 & 3 \\2 & 4 & 1 \\3 & 1 & 3\end{array}\]\right]\left[\[\begin{array}{l}x \y \z\end{array}\]\right] =\left[\[\begin{array}{l}6 \\0 \\8\end{array}\]\right]
বা, \left[\[\begin{array}{c}2 x+2 y+3 z \\2 x+4 y+z \\3 x+y+3 z\end{array}\]\right]=\left[\[\begin{array}{l}6 \\0 \\8\end{array}\]\right]
\therefore \mathrm{D}=\left|\[\begin{array}{lll}2 & 2 & 3 \\2 & 4 & 1 \\3 & 1 & 3\end{array}\]\right|=|\mathrm{A}|=-14
\mathrm{D}_{\mathrm{x}}=\left|\[\begin{array}{lll}6 & 2 & 3 \\0 & 4 & 1 \\8 & 1 & 3\end{array}\]\right|
D_y=\left|\[\begin{array}{lll}2 & 6 & 3 \\2 & 0 & 1 \\3 & 8 & 3\end{array}\]\right|
\mathrm{D}_{\mathrm{z}}=\left|\[\begin{array}{ccc}2 & 2 & 6 \\2 & 4 & 0 \\3 & 1 & 8\end{array}\]\right|
এবং
নির্ণেয় সমাধন: .