[গণিত] হলে [গণিত] নির্ণেয় কর।
উত্তর: দৃশ্যকল্প-১ : অনুসারে, [গণিত] দেওয়া আছে, [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
হলে নির্ণেয় কর।
উদ্দীপক
দৃশ্যকল্প-১ :
দৃশ্যকল্প-২ : \left[\[\begin{array}{ccc}1 & \mathrm{~m} & \mathrm{n} \l^{2} & \mathrm{~m}^{2} & \mathrm{n}^{2} \l^{3}-1 & \mathrm{~m}^{3}-1 & \mathrm{n}^{3}-1\end{array}\]\right]
Combined · 2018
উত্তর
\left|\[\begin{array}{ll}x & 2 \x & 2\end{array}\]\right|=2 x-2 x=0
এর সকল বাস্তব মানের জন্য \left(\[\begin{array}{ll}\mathrm{x} & 2 \\ \mathrm{x} & 2\end{array}\]\right) ম্যাট্রিক্সটি ব্যতিক্রমী হবে।
উত্তর
দৃশ্যকল্প-১ : অনুসারে,
দেওয়া আছে, =\left(\[\begin{array}{rrr}2 & 1 & 5 \\ -1 & 4 & 3 \\ 4 & -7 & 5\end{array}\]\right)
\therefore A^{2}=\left(\[\begin{array}{rrr}2 & 1 & 5 \\ -1 & 4 & 3 \\ 4 & -7 & 5\end{array}\]\right)\left(\[\begin{array}{rrr}2 & 1 & 5 \\ -1 & 4 & 3 \\ 4 & -7 & 5\end{array}\]\right)
=\left(\[\begin{array}{ccc}4-1+20 & 2+4-35 & 10+3+25 \\ -2-4+12 & -1+16-21 & -5+12+15 \\ 8+7+20 & 4-28-35 & 20-21+25\end{array}\]\right)
=\left(\[\begin{array}{lll}23 & -29 & 38 \\ 6 & -6 & 22 \\ 35 & -59 & 24\end{array}\]\right)
\therefore f(\mathrm{~A})=3\left(\[\begin{array}{lll}23 & -29 & 38 \\ 6 & -6 & 22 \\ 35 & -59 & 24\end{array}\]\right)+5\left(\[\begin{array}{rrr}2 & 1 & 5 \\ -1 & 4 & 3 \\ 4 & -7 & 5\end{array}\]\right)
=\left(\[\begin{array}{ccc}69 & -87 & 114 \\ 18 & -18 & 66 \\ 105 & -177 & 72\end{array}\]\right)+\left(\[\begin{array}{ccc}10 & 5 & 25 \\ -5 & 20 & 15 \\ 20 & -35 & 25\end{array}\]\right)
=\left(\[\begin{array}{ccc}79 & -82 & 139 \\ 13 & 2 & 81 \\ 125 & -212 & 97\end{array}\]\right)
উত্তর
দেওয়া আছে, \left[\[\begin{array}{ccc}l & \mathrm{~m} & \mathrm{n} \l^2 & \mathrm{~m}^2 & \mathrm{n}^2 \l^3-1 & \mathrm{~m}^3-1 & \mathrm{n}^3-1\end{array}\]\right]
\therefore|\mathrm{B}|=\left[\[\begin{array}{ccc}1 & \mathrm{~m} & \mathrm{n} \l^{2} & \mathrm{~m}^{2} & \mathrm{n}^{2} \l^{3}-1 & \mathrm{~m}^{3}-1 & \mathrm{n}^{3}-1\end{array}\]\right]
=\left[\[\begin{array}{ccc}l & \mathrm{~m} & \mathrm{n} \l^{2} & \mathrm{~m}^{2} & \mathrm{n}^{2} \l^{3} & \mathrm{~m}^{3} & \mathrm{n}^{3}\end{array}\]\right]-\left|\[\begin{array}{ccc}l & \mathrm{~m} & \mathrm{n} \l^{2} & \mathrm{~m}^{2} & \mathrm{n}^{2} \\ 1 & 1 & 1\end{array}\]\right|
=\operatorname{lmn}\left|\[\begin{array}{ccc}1 & 1 & 1 \l & \mathrm{~m} & \mathrm{n} \l^{2} & \mathrm{~m}^{2} & \mathrm{n}^{2}\end{array}\]\right|+\left|\[\begin{array}{ccc}l & \mathrm{~m} & \mathrm{n} \\ 1 & 1 & 1 \l^{2} & \mathrm{~m}^{2} & \mathrm{n}^{2}\end{array}\]\right|
=\operatorname{lmn}\left|\[\begin{array}{ccc}1 & 1 & 1 \l & \mathrm{~m} & \mathrm{n} \l^{2} & \mathrm{~m}^{2} & \mathrm{n}^{2}\end{array}\]\right|-\left|\[\begin{array}{ccc}1 & 1 & 1 \l & \mathrm{~m} & \mathrm{n} \l^{2} & \mathrm{~m}^{2} & \mathrm{n}^{2}\end{array}\]\right|
=(\operatorname{lmn}-1)\left|\[\begin{array}{ccc}1 & 1 & 1 \l & \mathrm{~m} & \mathrm{n} \l^{2} & \mathrm{~m}^{2} & \mathrm{n}^{2}\end{array}\]\right|
=(\operatorname{lmn}-1)\left|\[\begin{array}{ccc}0 & 0 & 1 \l-\mathrm{m} & \mathrm{m}-\mathrm{n} & \mathrm{n} \l^{2}-\mathrm{m}^{2} & \mathrm{~m}^{2}-\mathrm{n}^{2} & \mathrm{n}^{2}\end{array}\]\right|
এবং
=(l \mathrm{mn}-1)\left|\[\begin{array}{cc}l-\mathrm{m} & \mathrm{m}-\mathrm{n} \\ \quad (l-\mathrm{m})(l+\mathrm{m}) & (\mathrm{m}+\mathrm{n})(\mathrm{m}-\mathrm{n})\end{array}\]\right|
=(l m n-1)(l-m)(m-n)\left|\[\begin{array}{cc}1 & 1 \l+m & m+n\end{array}\]\right|
. (প্রমাণিত)