[গণিত] নির্ণয় কর(যদি বিদ্যমান থাকে)।
উত্তর: [গণিত] বিপরীত ম্যাট্রিক্স 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
নির্ণয় কর(যদি বিদ্যমান থাকে)।
উদ্দীপক
A=\left[\[\begin{array}{ccc}1 & 5 & -2 \\ 4 & 3 & 7 \\ -3 & 4 & 5\end{array}\]\right] এবং B=\left[\[\begin{array}{ccc}2 & -3 & 1 \\ 7 & 2 & 5 \\ -1 & 8 & 9\end{array}\]\right] দুটি ম্যাট্রিক্স
Sylhet · 2021
উত্তর
দেওয়া আছে,
A=\left[\[\begin{array}{ccc}1 & 5 & -2 \\ 4 & 3 & 7 \\ -3 & 4 & 5\end{array}\]\right] \quad B=\left[\[\begin{array}{ccc}2 & -3 & 1 \\ 7 & 2 & 5 \\ -1 & 8 & 9\end{array}\]\right]
2 A-B=2\left[\[\begin{array}{ccc}1 & 5 & -2 \\ 4 & 3 & 7 \\ -3 & 4 & 5\end{array}\]\right]-\left[\[\begin{array}{ccc}2 & -3 & 1 \\ 7 & 2 & 5 \\ -1 & 8 & 9\end{array}\]\right]
=\left[\[\begin{array}{ccc}2 & 10 & -4 \\ 8 & 6 & 14 \\ -6 & 8 & 10\end{array}\]\right]-\left[\[\begin{array}{ccc}2 & -3 & 1 \\ 7 & 2 & 5 \\ -1 & 8 & 9\end{array}\]\right]
=\left[\[\begin{array}{ccc}2-2 & 10+3 & -4-1 \\ 8-7 & 6-2 & 14-5 \\ -6+1 & 8-8 & 10-9\end{array}\]\right]
=\left[\[\begin{array}{ccc}0 & 13 & -5 \\ 1 & 4 & 9 \\ -5 & 0 & 1\end{array}\]\right]
(2 A-B)^{T}=\left[\[\begin{array}{ccc}0 & 1 & -5 \\ 13 & 4 & 0 \\ -5 & 9 & 1\end{array}\]\right]
উত্তর
দেওয়া আছে,
A=\left[\[\begin{array}{ccc}1 & 5 & -2 \\ 4 & 3 & 7 \\ -3 & 4 & 5\end{array}\]\right]
A^{2}=\left[\[\begin{array}{ccc}1 & 5 & -2 \\ 4 & 3 & 7 \\ -3 & 4 & 5\end{array}\]\right]\left[\[\begin{array}{ccc}1 & 5 & -2 \\ 4 & 3 & 7 \\ -3 & 4 & 5\end{array}\]\right]
=\left[\[\begin{array}{ccc}1+20+6 & 5+15-8 & -2+35-10 \\ 4+12-21 & 20+9+28 & -8+21+35 \\ -3+16-15 & -15+12+20 & 6+28+25\end{array}\]\right]
\begin{aligned} & =\left[\begin{array}{lll}27 & 12 & 23 \\ -5 & 57 & 48 \\ -2 & 17 & 59\end{aligned}\right] \\ & \therefore f(A)=3 A^2+2 A-5 I \\ & 3=\left[\[\begin{array}{ccc}27 & 12 & 23 \\ -5 & 57 & 48 \\ -2 & 17 & 59\end{array}\]\right]+2\left[\[\begin{array}{ccc}1 & 5 & -2 \\ 4 & 3 & 7 \\ -3 & 4 & 5\end{array}\]\right]-5\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right] \\ & =\left[\[\begin{array}{ccc}81 & 36 & 69 \\ -15 & 171 & 144 \\ -6 & 51 & 177\end{array}\]\right]+\left[\[\begin{array}{ccc}2 & 10 & -4 \\ 8 & 6 & 14 \\ -6 & 8 & 10\end{array}\]\right]-\left[\[\begin{array}{ccc}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\]\right] \\ & =\left[\[\begin{array}{ccc}81+2-5 & 36+10-0 & 69-4-0 \\ -15+8-0 & 171+6+5 & 144+14-0 \\ -6-6-0 & 51+8-0 & 177+10-5\end{array}\]\right] \\ & =\left[\[\begin{array}{ccc}78 & 46 & 65 \\ -7 & 172 & 158 \\ -12 & 59 & 182\end{array}\]\right]\end{aligned}
\end{array}\end{array}
উত্তর
\begin{aligned}& B=\left[\begin{array}{ccc}2 & -3 & 1 \\7 & 2 & 5 \\-1 & 8 & 9\end{aligned}\right] \\& |B|=2(18-40)+3(63+5)+1(56+2) \\& =-44+204+58 \\& =218 \neq 0\end{aligned}
\end{array}
বিপরীত ম্যাট্রিক্স বিদ্যমান।
\begin{aligned} & B_{11}=\left[\begin{array}{ll}2 & 5 \\ 8 & 9\end{aligned}\right] \\ & =(18-40) \\ & =-22\end{aligned}
\end{array}
\begin{aligned} & B_{12}=-\left[\begin{array}{cc}7 & 5 \\ -1 & 9\end{aligned}\right] \\ & =-(63+5) \\ & =-68\end{aligned}
\end{array}
\begin{aligned} & B_{13}=\left[\begin{array}{cc}7 & 2 \\ -1 & 8\end{aligned}\right] \\ & =(56+2) \\ & =58\end{aligned}
\end{array}
\begin{aligned} & B_{21}=-\left[\begin{array}{cc}-3 & 1 \\ 5 & 9\end{aligned}\right] \\ & =-(-27-8) \\ & =35\end{aligned}
\end{array}
\begin{aligned} & B_{22}=\left[\begin{array}{cc}2 & 1 \\ -1 & 9\end{aligned}\right] \\ & =(18+1) \\ & =19\end{aligned}
\end{array}
\begin{aligned} & B_{23}=\left[\begin{array}{cc}2 & -3 \\ -1 & 8\end{aligned}\right] \\ & =-(16-3) \\ & =-13\end{aligned}
\end{array}
\begin{aligned} & B_{31}=\left[\begin{array}{cc}-3 & 1 \\ 2 & 5\end{aligned}\right] \\ & =(-15-2) \\ & =-17\end{aligned}
\end{array}
\begin{aligned} & B_{32}=-\left[\begin{array}{ll}2 & 1 \\ 7 & 5\end{aligned}\right] \\ & =-(10-7) \\ & =-3\end{aligned}
\end{array}
\begin{aligned} & B_{33}=\left[\begin{array}{cc}2 & -3 \\ 7 & 2\end{aligned}\right] \\ & =(4+21) \\ & =25\end{aligned}
\end{array}
\begin{aligned} & \therefore \operatorname{adj} B=\left[\begin{array}{ccc}-22 & -68 & 58 \\ 35 & 19 & -13 \\ -17 & -3 & 25\end{aligned}\right]^T \\ & =\left[\[\begin{array}{ccc}-22 & 35 & -13 \\ -68 & 19 & -3 \\ 58 & -13 & 25\end{array}\]\right] \\ & B^{-1}=\frac{1}{|B|} \operatorname{adj} B=\frac{1}{218}\left[\[\begin{array}{ccc}-22 & 35 & -13 \\ -68 & 19 & -3 \\ 58 & -13 & 25\end{array}\]\right]\end{aligned}
\end{array}\end{array}\end{array}\end{array}