k এর মান কত হলে [গণিত] ম্যাট্রিক্সটি প্রতিসম হবে?
উত্তর: [গণিত] ম্যাট্রিক্স প্রতিসম [গণিত] এখানে, [গণিত] Ans.
HSC Higher Math 1st Paper Chapter 1: ম্যাট্রিকস ও নির্ণায়ক CQ — Prosthuti প্রশ্নব্যাংক
HSC · Higher Math 1st Paper · Chapter 1: ম্যাট্রিকস ও নির্ণায়ক · সৃজনশীল
k এর মান কত হলে ম্যাট্রিক্সটি প্রতিসম হবে?
উদ্দীপক
M=\left[\[\begin{array}{ccc}2 x y & x^2 & y^2 \x^2 & y^2 & 2 x y \y^2 & 2 x y & x^2\end{array}\]\right]
আদমজী ক্যান্টনমেন্ট কলেজ, ঢাকা · 2024
উত্তর
ম্যাট্রিক্স প্রতিসম
A=\left[\[\begin{array}{cc} 3 & k-1 \\ 2 k+3 & k-2 \end{array}\]\right]
A^{\top}=\left[\[\begin{array}{cc} 3 & 2 k+3 \k-1 & k-2 \end{array}\]\right]
\Rightarrow\left[\[\begin{array}{cc} 3 & k-1 \\ 2 k+3 & k-2 \end{array}\]\right]=\left[\[\begin{array}{cc} 3 & 2 k+3 \k-1 & k-2 \end{array}\]\right]
এখানে,
Ans.
উত্তর
ও হলে,
M=\left[\[\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\]\right]
\begin{aligned}|m| & =-1 & & \a_{11} & =0\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{aligned}\right| & a_{12} & =0\left|\[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\]\right| \\ & =0 & & =0\end{aligned}
\end{array}
\begin{aligned} a_{13} & =1\left|\begin{array}{ll}0 & 1 \\ 1 & 0\end{aligned}\right| & a_{21} & =0\left|\[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\]\right| \\ & =0 & & =0\end{aligned}
\end{array}
\begin{aligned} a_{22} & =1\left|\begin{array}{ll}0 & 1 \\ 1 & 0\end{aligned}\right| & a_{23} & =0\left|\[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\]\right| \\ & =-1 & & =0\end{aligned}
\end{array}
\begin{aligned} a_{31} & =1\left|\begin{array}{ll}0 & 1 \\ 1 & 0\end{aligned}\right| & a_{32} & =0\left|\[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\]\right| \\ & =0 & & =0\end{aligned}
\end{array}
a_{33}=0\left|\[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\]\right|=0
\therefore \operatorname{Adj}(M)=\left[\[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\]\right]^{T}=\left[\[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{array}\]\right]
\therefore M^{-1}=\frac{A d \dot{j}(M)}{(M)}=\left[\[\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\]\right]=M
(Proved)\end{array}
উত্তর
L.H.S =\left|\[\begin{array}{ccc}2 x y & x^{2} & y^{2} \x^{2} & y^{2} & 2 x y \y^{2} & 2 x y & x^{2} \end{array}\]\right|
=\left|\[\begin{array}{ccc} 2 x y+x^{2}+y^{2} & x^{2} & y^{2} \x^{2}+y^{2}+2 x y & y^{2} & 2 x y \y^{2}+x^{2}+2 x y & 2 x y & x^{2} \end{array}\]\right|\left(\[\begin{array}{l} R_{1}^{\prime}=R_{1}+R_{2}+R_{3} \R_{2}^{\prime}=R_{1}+R_{2}+R_{3} \R_{3}^{\prime}=R_{1}+R_{2}+R_{3} \end{array}\]\right)
=\left|\[\begin{array}{ccc} (x+y)^{2} & x^{2} & y^{2} \\ (x+y)^{2} & y^{2} & 2 x y \\ (x+y)^{2} & 2 x y & x^{2} \end{array}\]\right|
=(x+y)^{2}\left|\[\begin{array}{ccc} 1 & x^{2} & y^{2} \\ 1 & y^{2} & 2 x y \\ 1 & 2 x y & x^{2} \end{array}\]\right|
=(x+y)^{2}\left|\[\begin{array}{ccc} 0 & x^{2}-y^{2} & y^{2}-2 x y \\ 0 & y^{2}-2 x y & 2 x y-x^{2} \\ 1 & 2 x y & x^{2} \end{array}\]\right|\left[\[\begin{array}{l} R_{1}^{\prime}=R_{1}-R_{2} \R_{2}^{\prime}=R_{2}-R_{3} \end{array}\]\right]
R.H.S