প্রমাণ কর যে, [গণিত]
উত্তর: দেওয়া আছে, [গণিত] L.H.S [গণিত] [প্রথমে [গণিত] তারপর নতুন [গণিত] কে [গণিত] নিয়ে [গণিত] কে [গণিত] তে রূপান্তর করি] [গণিত] R.H.S [গণিত] (proved)
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,
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
রাজউক উত্তরা মডেল কলেজ, ঢাকা · 2024
উত্তর
দেওয়া আছে,
M =\left(\[\begin{array}{c}2 \\ 9 \\-3\end{array}\]\right) এবং N=\left(\[\begin{array}{lll} -3 & 5 & 6\end{array}\]\right)
\therefore M N =\left(\[\begin{array}{c} 2 \\ 9 \\ -3 \end{array}\]\right)\left(\[\begin{array}{lll} -3 & 5 & 6 \end{array}\]\right)
=\left(\[\begin{array}{ccc} -6 & 10 & 12 \\ -27 & 45 & 54 \\ -9 & -15 & -18 \end{array}\]\right)
\therefore(\mathrm{MN})^{\mathrm{T}} =\left(\[\begin{array}{ccc} -6 & 10 & 12 \\ -27 & 45 & 54 \\ -9 & -15 & -18 \end{array}\]\right)
=\left(\[\begin{array}{ccc} -6 & -27 & -9 \\ 10 & 45 & -15 \\ 12 & 54 & -18 \end{array}\]\right)
উত্তর
দেওয়া আছে,
\begin{aligned} & \therefore f(\mathrm{P})-5 \mathrm{P}^2-11 \mathrm{P} \text{এবং} \mathrm{P}=\left[\begin{array}{rr}-3 & 2 \\ 3 & -1\end{aligned}\right] \\ & \[\begin{aligned} \therefore \mathrm{P}^2=\mathrm{P} \times \mathrm{P} & =\left[\begin{array}{rr}-3 & 2 \\ 3 & -1\end{aligned}\]\right] \times\left[\[\begin{array}{rr}-3 & 2 \\ 3 & -1\end{array}\]\right] \\ & =\left[\[\begin{array}{rr}9+6 & -6-2 \\ -9-3 & 6+1\end{array}\]\right]=\left[\[\begin{array}{rr}15 & -8 \\ -12 & 7\end{array}\]\right]\end{aligned}
\end{array}
\end{array}
এখন \begin{aligned} & f(\mathrm{P})=6 \mathrm{I}_2=5 \mathrm{P}^2-11 \mathrm{P}+6\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{aligned}\right] \\ &= 5\left[\[\begin{array}{rr}15 & -8 \\ -12 & 7\end{array}\]\right]-11\left[\[\begin{array}{rr}-3 & 2 \\ 3 & -1\end{array}\]\right]+\left[\[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\]\right] \\ &=6\left[\[\begin{array}{rr}75 & -40 \\ -60 & 35\end{array}\]\right]-\left[\[\begin{array}{rr}-33 & 22 \\ 33 & -11\end{array}\]\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[\[\begin{array}{rr}-114 & -62 \\ -93 & 52\end{array}\]\right]\end{aligned}
\end{array}\end{array}
উত্তর
দেওয়া আছে,
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)
L.H.S
=\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|\[\begin{array}{ccc} a^{2} & b^{2} & c^{2} \a^{3} & b^{3} & c^{3} \a^{4} & b^{4} & c^{4} \end{array}\]\right|-\left|\[\begin{array}{ccc} a^{2} & b^{2} & c^{2} \a^{3} & b^{3} & c^{3} \\ 2 a & 2 b & 2 c \end{array}\]\right|
=a^{2} b^{2} c^{2}\left|\[\begin{array}{ccc}1 & 1 & 1 \a & a & a \a^{2} & b^{2} & c^{2}\end{array}\]\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|\[\begin{array}{lll}1 & 1 & 1 \a & a & a \a^{2} & b^{2} & c^{2}\end{array}\]\right|-2 a b c\left|\[\begin{array}{ccc}1 & 1 & 1 \a & b & c \a^{2} & b^{2} & c^{2}\end{array}\]\right|
[প্রথমে তারপর নতুন কে নিয়ে কে তে রূপান্তর করি]
=\left(a^{2} b^{2} c^{2}-2 a b c\right)\left|\[\begin{array}{ccc}1 & 1 & 1 \a & a & a \a^{2} & b^{2} & c^{2}\end{array}\]\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| \quad \[\begin{aligned} & C_{1}{ }^{\prime}=C_{1}-C_{2} \\ & C_{2}{ }^{\prime}=C_{2}-C_{3}\end{aligned}\]
R.H.S
(proved)