বিস্তার না করে প্রমান কর, [গণিত]
উত্তর: [গণিত] = R. H.S
HSC Higher Math 1st Paper Chapter 1: ম্যাট্রিকস ও নির্ণায়ক CQ — Prosthuti প্রশ্নব্যাংক
HSC · Higher Math 1st Paper · Chapter 1: ম্যাট্রিকস ও নির্ণায়ক · সৃজনশীল
বিস্তার না করে প্রমান কর,
উদ্দীপক
S=\left(\[\begin{array}{cc}-1 & 1 \\ 2 & -3\end{array}\]\right), T=\left(\[\begin{array}{cc}3 & -5 \\ -1 & 2\end{array}\]\right)
U =\left|\[\begin{array}{ccc}a & b & c \\ 2 a^3+1 & 2 b^3+1 & 2 c^3+1 \a^2 & b^2 & c^2\end{array}\]\right|
Jessore · 2022
উত্তর
L \cdot H \cdot S=\left|\[\begin{array}{ccc}a & -x & a+x \b & -y & b+y \c & -z & c+z\end{array}\]\right|
=\left|\[\begin{array}{lll}a+x & -x & -a+x \b+x & -y & b+y \c+x & -z & c+z\end{array}\]\right|
= R. H.S
উত্তর
দেওয়া আছে,
S=\left(\[\begin{array}{cc}-1 & 1 \\2 & -3\end{array}\]\right)
T=\left(\[\begin{array}{cc}3 & -5 \\-1 & 2\end{array}\]\right)
S T=\left(\[\begin{array}{cc}-1 & 1 \\2 & -3\end{array}\]\right)\left(\[\begin{array}{cc}3 & -5 \\-1 & 2\end{array}\]\right)
=\left(\[\begin{array}{cc}-3-1 & 5+2 \\6+3 & -10-6\end{array}\]\right)
=\left(\[\begin{array}{cc}-4 & 7 \\9 & -16\end{array}\]\right)
\begin{aligned} & |S T|=64-6 \mathrm{~B} \\ & =1 \\ & (S T)^{-1}=\left(\begin{array}{ll}-16 & -7 \\ -9 & -4\end{aligned}\right) \\ & T=\left(\[\begin{array}{cc}3 & -5 \\ -1 & 2\end{array}\]\right) \\ & |T|=6-5=1 \\ & \therefore T^{-1}=\left(\[\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\]\right) \\ & s=\left(\[\begin{array}{cc}-1 & 1 \\ 2 & -3\end{array}\]\right) \\ & |s|=3-2 \\ & =1 \\ & s^{-1}=\left(\[\begin{array}{ll}-3 & -1 \\ -2 & -1\end{array}\]\right) \\ & T^{-1} S^{-1}=\left(\[\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\]\right)\left(\[\begin{array}{ll}-3 & -1 \\ -2 & -1\end{array}\]\right) \\ & =\left(\[\begin{array}{ll}-6-10 & -2-5 \\ -3-6 & -1-3\end{array}\]\right) \\ & =\left(\[\begin{array}{ll}-16 & -7 \\ -9 & -4\end{array}\]\right) \\ & (S T)^{-1}-T^{-1} S^{-1}=\left(\[\begin{array}{ll}-16 & -7 \\ -9 & -4\end{array}\]\right)-\left(\[\begin{array}{cc}-16 & -7 \\ -9 & -4\end{array}\]\right) \\ & =\left(\[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\]\right) \\ & \therefore(S T)^{-1}-T^{-1} S^{-1} \\ & \end{aligned}
\end{array}
একটি শূন্য ম্যাট্রিক্স।\end{array}
উত্তর
প্রমান কর যে,
U=\left|\[\begin{array}{ccc}a & b & c \\2 a^3+1 & 2 b^3+1 & 2 c^3+1 \a^2 & b^2 & c^2\end{array}\]\right|
=\left|\[\begin{array}{ccc}a & b & c \\2 a^3 & 2 b^3 & 2 c^3 \a^2 & b^2 & c^2\end{array}\]\right|+\left|\[\begin{array}{ccc}a & b & c \\1 & 1 & 1 \a^2 & b^2 & c^2\end{array}\]\right|
=2 a b c\left|\[\begin{array}{ccc}1 & 1 & 1 \a^2 & b^2 & c^2 \a & b & c\end{array}\]\right|-\left|\[\begin{array}{ccc}1 & 1 & 1 \mathrm{a} & \mathrm{b} & c \a^2 & b^2 & c^2\end{array}\]\right|
=2 a b c\left|\[\begin{array}{ccc}1 & 1 & 1 \a^2 & b^2 & c^2 \a & b & c\end{array}\]\right|+\left|\[\begin{array}{ccc}1 & 1 & 1 \a^2 & b^2 & c^2 \a & b & c\end{array}\]\right|
=(2 a b c+1)\left|\[\begin{array}{ccc}1 & 1 & 1 \a^2 & b^2 & c^2 \a & b & c\end{array}\]\right|
=(2 a b c+1)\left|\[\begin{array}{ccc}1-1 & 1-1 & 1 \a^2-b^2 & b^2-c^2 & c^2 \a-b & b-c & c\end{array}\]\right|
=(2 a b c+1)\left|\[\begin{array}{ccc}0 & 0 & 1 \a^2-b^2 & b^2 c^2 & c^2 \a-b & b-c & c\end{array}\]\right|
=(2 a b c+1)\left|\[\begin{array}{ll}a^2-b^2 & b^2-c^2 \a-b & b-c\end{array}\]\right|
=(2 a b c+1)\left|\[\begin{array}{ll}(a+b)(a-b) & (b+c)(b-c) \a-b & b-c\end{array}\]\right|
=(2 a b c+1)(a-b)(b-c)\left|\[\begin{array}{ll}a+b & b+c \\1 & 1\end{array}\]\right|
(Proved)