[গণিত] নির্ণয় কর; যেখানে [গণিত] একটি একক ম্যাট্রিক্স।
উত্তর: দেওয়া আছে, [গণিত] এখন, [গণিত] প্রদত্ত রাশি, [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
নির্ণয় কর; যেখানে একটি একক ম্যাট্রিক্স।
উদ্দীপক
A=\left[\[\begin{array}{rrr}3 & -1 & -1 \\ 1 & -3 & 1 \\ 1 & -1 & 3\end{array}\]\right]
Jessore · 2025
উত্তর
ধরি, P=\left[\[\begin{array}{cc}a & a+1 \\ -a+1 & -a\end{array}\]\right]
হলে, \mathrm{P}=\left[\[\begin{array}{cc} 3 & 3+1 \\ -3+1 & -3\end{array}\]\right]=\left[\[\begin{array}{rr} 3 & 4 \\-2 & -3\end{array}\]\right] .
ম্যাট্রিক্সটি অভেদঘাতী ম্যাট্রিক্স হবে যদি হয়।
এখন,
=\left[\[\begin{array}{rr}3 & 4 \\ -2 & -3\end{array}\]\right]\left[\[\begin{array}{rr}3 & 4 \\ -2 & -3\end{array}\]\right]
=\left[\[\begin{array}{rr}9-8 & 12-12 \\ -6+6 & -8+9\end{array}\]\right]=\left[\[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\]\right]=I
যেহেতু
সুতরাং P=\left[\[\begin{array}{cc}a & a+1 \\ -a+1 & -a\end{array}\]\right] একটি অভেদঘাতী ম্যাট্রিক্স;
যখন . ('দেখানো হলো)
উত্তর
দেওয়া আছে, A=\left[\[\begin{array}{rrr}3 & -1 & -1 \\ 1 & -3 & 1 \\ 1 & -1 & 3\end{array}\]\right]
এখন,
=\left[\[\begin{array}{rrr}3 & -1 & -1 \\ 1 & -3 & 1 \\ 1 & -1 & 3\end{array}\]\right]\left[\[\begin{array}{rrr}3 & -1 & -1 \\ 1 & -3 & 1 \\ 1 & -1 & 3\end{array}\]\right]
=\left[\[\begin{array}{llll}9-1-1 & -3+3+1 & -3-1-3 \\ 3-3+1 & -1+9-1 & -1-3+3 \\ 3-1+3 & -1+3-3 & -1-1+9\end{array}\]\right]
=\left[\[\begin{array}{rrr}7 & 1 & -7 \\ 1 & 7 & -1 \\ 5 & -1 & 7\end{array}\]\right]
প্রদত্ত রাশি,
=\left[\[\begin{array}{rrr}7 & 1 & -7 \\ 1 & 7 & -1 \\ 5 & -1 & 7\end{array}\]\right]-2\left[\[\begin{array}{rrr}3 & -1 & -1 \\ 1 & -3 & 1 \\ 1 & -1 & 3\end{array}\]\right]+3\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right]
=\left[\[\begin{array}{rrr}7 & 1 & -7 \\ 1 & 7 & -1 \\ 5 & -1 & 7\end{array}\]\right]-\left[\[\begin{array}{rrr}6 & -2 & -2 \\ 2 & -6 & 2 \\ 2 & -2 & 6\end{array}\]\right]+\left[\[\begin{array}{lll}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\]\right]
=\left[\[\begin{array}{rrr}7 & -6+3 & 1+2+0 \\ 1 & -7+2+0 & 7+6+3 \\ 5-2+0 & -1+2+0 & 7-6+3\end{array}\]\right]
=\left[\[\begin{array}{rrr}4 & 3 & -5 \\ -1 & 16 & -3 \\ 3 & 1 & 4\end{array}\]\right]
\therefore A^2-2 A+3 I=\left[\[\begin{array}{rcr}4 & 3 & -5 \\ -1 & 16 & -3 \\ 3 & 1 & 4\end{array}\]\right]
উত্তর
দেওয়া আছে, A=\left[\[\begin{array}{rrr}3 & -1 & -1 \\ 1 & -3 & 1 \\ 1 & -1 & 3\end{array}\]\right]
এখন,
বা,
বা, I.
এবং
এখন, |A|=\left|\[\begin{array}{rrr}3 & -1 & -1 \\ 1 & -3 & 1 \\ 1 & -1 & 3\end{array}\]\right|
বিদ্যমান।
\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]^T
A এর সহগুণকসমূহ
A_{11}=\left|\[\begin{array}{rr}-3 & 1 \\ -1 & 3\end{array}\]\right|=-9+1=-8
A_{12}=-\left|\[\begin{array}{rr}1 & 1 \\ 1 & 3\end{array}\]\right|=-(3-1)=-2
A_{13}=\left|\[\begin{array}{rr}1 & -3 \\ 1 & -1\end{array}\]\right|=-1+3=2
A_{21}=-\left|\[\begin{array}{rr}-1 & -1 \\ -1 & 3\end{array}\]\right|=-(-3-1)=4
A_{22}=\left|\[\begin{array}{rr}3 & -1 \\ 1 & 3\end{array}\]\right|=9+1=10
A_{23}=-\left|\[\begin{array}{rr}3 & -1 \\ 1 & -1\end{array}\]\right|=-(-3+1)=2
A_{31}=\left|\[\begin{array}{rr}-1 & -1 \\ -3 & 1\end{array}\]\right|=-1-3=-4
A_{32}=-\left|\[\begin{array}{lr}3 & -1 \\ 1 & 1\end{array}\]\right|=-(3+1)=-4
A_{33}=\left|\[\begin{array}{ll}3 & -1 \\ 1 & -3\end{array}\]\right|=-9+1=-8
\therefore Adj A=\left[\[\begin{array}{rrr}-8 & -2 & 2 \\ 4 & 10 & 2 \\ -4 & -4 & -8\end{array}\]\right]^{\mathrm{T}}
=\left[\[\begin{array}{rcr}-8 & 4 & -4 \\ -2 & 10 & -4 \\ 2 & 2 & -8\end{array}\]\right]
\therefore A^{-1}=\frac{1}{|A|} \text{ Adj } A=\frac{1}{-24}\left[\[\begin{array}{rcc}-8 & 4 & -4 \\ -2 & 10 & -4 \\ 2 & 2 & -8\end{array}\]\right]
=\left[\[\begin{array}{lll}\frac{-8}{-24} & \frac{4}{-24} & \frac{-4}{-24} \\ \frac{-2}{-24} & \frac{10}{-24} & \frac{-4}{-24} \\ \frac{2}{-24} & \frac{2}{-24} & \frac{-8}{-24}\end{array}\]\right]=\left[\[\begin{array}{rrr}\frac{1}{3} & -\frac{1}{6} & \frac{1}{6} \\ \frac{1}{12} & -\frac{5}{12} & \frac{1}{6} \\ -\frac{1}{12} & -\frac{1}{12} & \frac{1}{3}\end{array}\]\right]
(1)নং হতে পাই, B=A^{-1}=\left[\[\begin{array}{rrr}\frac{1}{3} & -\frac{1}{6} & \frac{1}{6} \\ \frac{1}{12} & -\frac{5}{12} & \frac{1}{6} \\ -\frac{1}{12} & -\frac{1}{12} & \frac{1}{3}\end{array}\]\right]
নির্ণেয় B=\left[\[\begin{array}{rrr}\frac{1}{3} & -\frac{1}{6} & \frac{1}{6} \\ \frac{1}{12} & -\frac{5}{12} & \frac{1}{6} \\ -\frac{1}{12} & -\frac{1}{12} & \frac{1}{3}\end{array}\]\right]