প্রমাণ কর যে, det(A) = abc(abc - 2) (a - b)(b - c) (c - a).
উত্তর: [গণিত] (প্রমাণিত)
HSC Higher Math 1st Paper Chapter 1: ম্যাট্রিকস ও নির্ণায়ক CQ — Prosthuti প্রশ্নব্যাংক
HSC · Higher Math 1st Paper · Chapter 1: ম্যাট্রিকস ও নির্ণায়ক · সৃজনশীল
প্রমাণ কর যে, det(A) = abc(abc - 2) (a - b)(b - c) (c - a).
উদ্দীপক
A=\left[\[\begin{array}{ccc}a^2 & b^2 & c^2 \a^3 & b^3 & c^3 \a^4-2 a & b^4-2 b & c^4-2 c\end{array}\]\right], P=\left[\[\begin{array}{cc}-3 & 2 \\3 & -1\end{array}\]\right] এবং
Chittagong · 2022
উত্তর
মনেকরি (3, –11) বিন্দু থেকে 5 একক দূরবর্তী যেকোনো একটি বিন্দু (x,y)
শর্তমতে,
উত্তর
দেওয়া আছে,
P=\left[\[\begin{array}{cc}-3 & 2 \\3 & -1\end{array}\]\right]
\therefore P^2=p \cdot p=\left[\[\begin{array}{cc}-3 & 2 \\3 & -1\end{array}\]\right]\left[\[\begin{array}{cc}-3 & 2 \\3 & -1\end{array}\]\right]
\quad=\left[\[\begin{array}{cc}9+6 & -6-2 \\-9-3 & 6+1\end{array}\]\right]
\quad=\left[\[\begin{array}{cc}15 & -8 \\-12 & 7\end{array}\]\right]
এখন,
=5\left[\[\begin{array}{cc}15 & -8 \\-12 & 7\end{array}\]\right]-11\left[\[\begin{array}{cc}-3 & 2 \\3 & -1\end{array}\]\right]+6\left[\[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\]\right]
=\left[\[\begin{array}{cc}75 & -40 \\-60 & 35\end{array}\]\right]-\left[\[\begin{array}{cc}-33 & 22 \\33 & -11\end{array}\]\right]+\left[\[\begin{array}{ll}6 & 0 \\0 & 6\end{array}\]\right]
=\left[\[\begin{array}{cc}75+33+6 & -40-22+0 \\-60-33+0 & 35+11+6\end{array}\]\right]
=\left[\[\begin{array}{cc}114 & -62 \\-93 & 52\end{array}\]\right] Ans
উত্তর
A=\left[\[\begin{array}{ccc}a^2 & b^2 & c^2 \a^3 & b^3 & c^3 \a^4-2 a & b^4-2 b & c^4-2 c\end{array}\]\right]
\therefore \operatorname{det}(A)= |A|=\left|\[\begin{array}{ccc}a^2 & b^2 & c^2 \a^3 & b^3 & c^3 \a^4-2 a & b^4-2 b & c^4-2 c\end{array}\]\right|
=\left|\[\begin{array}{lll}a^2 & b^2 & c^2 \a^3 & b^3 & c^3 \a^4 & b^4 & c^4\end{array}\]\right|-\left|\[\begin{array}{lll}a^2 & b^2 & c^2 \a^3 & b^3 & c^3 \\ 2 a & 2 b & 2 c\end{array}\]\right|
=a^2 b^2 c^2\left|\[\begin{array}{ccc}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2\end{array}\]\right|-a b c\left|\[\begin{array}{ccc}a & b & c \a^2 & b^2 & c^2 \\ 2 & 2 & 2\end{array}\]\right|
=a^2 b^2 c^2\left|\[\begin{array}{lll}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2\end{array}\]\right|-2 a b c\left|\[\begin{array}{ccc}a & b & c \a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\]\right|
=a^2 b^2 c^2\left|\[\begin{array}{ccc}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2\end{array}\]\right|-2 a b c\left|\[\begin{array}{lll}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2\end{array}\]\right|
=\left(a^2 b^2 c^2-2 a b c\right)\left|\[\begin{array}{lll}1 & 1 & 1 \a & b & c \a^2 & b^2 & c^2\end{array}\]\right|
=a b c(a b c-2)\left|\[\begin{array}{ccc}0 & 0 & 1 \a-b & b-c & c \a^2-b^2 & b^2-c^2 & c^2\end{array}\]\right|
= a b c(a b c-2)(a-b)(b-c)\left|\[\begin{array}{cc}1 & 1 \a+b & b+c\end{array}\]\right|
(প্রমাণিত)