দেখাও যে, [গণিত] .
উত্তর: দেওয়া আছে, [গণিত] (দেখানো হলো) [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
দেখাও যে, .
উদ্দীপক
A=\left(\[\begin{array}{lll}1 & -2 & 3\end{array}\]\right),
X=\left(\[\begin{array}{lll}x & y & z\end{array}\]\right), B=\left(\[\begin{array}{rrr}1 & -2 & 3 \\ 1 & 5 & 0 \\ 4 & -2 & 1\end{array}\]\right)
\mathrm{C}=\left(\[\begin{array}{ccc}(\mathrm{m}+\mathrm{n})^{2} & l^{2} & l^{2} \\ \mathrm{~m}^{2} & (\mathrm{n}+l)^{2} & \mathrm{~m}^{2} \\ \mathrm{n}^{2} & \mathrm{n}^{2} & (l+\mathrm{m})^{2}\end{array}\]\right)
Jessore · 2023
উত্তর
দেওয়া আছে, 3\left(\[\begin{array}{cc}1 & -1 \\ 2 & 4\end{array}\]\right)+\mathrm{E}=\mathrm{I}_{2}
বা, \left(\[\begin{array}{ll} 3 & -3 \\ 6 & 12 \end{array}\]\right)+E=\left(\[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\]\right)
বা, E=\left(\[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\]\right)-\left(\[\begin{array}{ll} 3 & -3 \\ 6 & 12 \end{array}\]\right)
বা, E=\left(\[\begin{array}{ll} 1-3 & 0+3 \\ 0-6 & 1-12 \end{array}\]\right)
\therefore E=\left(\[\begin{array}{rr} -2 & 3 \\ -6 & -11 \end{array}\]\right)
নির্ণেয় E=\left(\[\begin{array}{rr}-2 & 3 \\ -6 & -11\end{array}\]\right)
উত্তর
দেওয়া আছে, \mathrm{A}=\left(\[\begin{array}{lll}1 & -2 & 3\end{array}\]\right)
A^{T}=\left(\[\begin{array}{lll} 1 & -2 & 3 \end{array}\]\right)^{T}=\left(\[\begin{array}{r} 1 \\ -2 \\ 3 \end{array}\]\right)
B=\left(\[\begin{array}{lll} 1 & -2 & 3 \\ 1 & 5 & 0 \\ 4 & -2 & 1 \end{array}\]\right)
এবং X=\left(\[\begin{array}{lll}x & y & z\end{array}\]\right)
\therefore X^{T}=\left(\[\begin{array}{lll}x & y & z\end{array}\]\right)^{T}=\left(\[\begin{array}{l}x \y \z\end{array}\]\right)
এখন,
বা, \left(\[\begin{array}{lrr}1 & -2 & 3 \\ 1 & 5 & 0 \\ 4 & -2 & 1\end{array}\]\right)\left(\[\begin{array}{l}x \y \z\end{array}\]\right)=\left(\[\begin{array}{r}1 \\ -2 \\ 3\end{array}\]\right)
y=\left(\[\begin{array}{l}x-2 y+3 z \x+5 y+0 z \\ 4 x-2 y+z\end{array}\]\right)=\left(\[\begin{array}{r}1 \\ -2 \\ 3\end{array}\]\right)
ম্যট্রিক্সের সমতা অনুসারে পাই,
ও এর সহগগুচ্ছ নিয়ে গঠিত নির্ণায়ক,
D =\left|\[\begin{array}{rrr} 1 & -2 & 3 \\ 1 & 5 & 0 \\ 4 & -2 & 1 \end{array}\]\right|
এখন, D_{x}=\left|\[\begin{array}{rrr} 1 & -2 & 3 \\ -2 & 5 & 0 \\ 3 & -2 & 1\end{array}\]\right|=1(5+0)+2(-2-0)+3(4-15)
D_{y}=\left|\[\begin{array}{rrr} 1 & 1 & 3 \\ 1 & -2 & 0 \\ 4 & 3 & 1 \end{array}\]\right|
D_{z} =\left|\[\begin{array}{rrr} 1 & -2 & 1 \\ 1 & 5 & -2 \\ 4 & -2 & 3 \end{array}\]\right|
নির্ণেয় সমাধান : .
উত্তর
দেওয়া আছে, \mathrm{C}=\left(\[\begin{array}{ccc}(\mathrm{m}+\mathrm{n})^{2} & l^{2} & l^{2} \\ \mathrm{~m}^{2} & (\mathrm{n}+l)^{2} & \mathrm{~m}^{2} \\ \mathrm{n}^{2} & \mathrm{n}^{2} & (l+\mathrm{m})^{2}\end{array}\]\right)
\therefore|\mathrm{C}|=\left|\[\begin{array}{ccc} (\mathrm{m}+\mathrm{n})^{2} & l^{2} & l^{2} \\ \mathrm{~m}^{2} & \left(\mathrm{n}+l^{2}\right. & \mathrm{m}^{2} \\ \mathrm{n}^{2} & \mathrm{n}^{2} & (l+\mathrm{m})^{2} \end{array}\]\right|
=\left|\[\begin{array}{ccc}(m+n)^{2} & l^{2}-(m+n)^{2} & l^{2}-(m+n)^{2} \m^{2} & \left(n+n^{2}-m^{2}\right. & m^{2}-m^{2} \n^{2} & n^{2}-n^{2} & (l+m)^{2}-n^{2}\end{array}\]\right|\left[\[\begin{array}{l}c_{2}^{\prime}=c_{2}-c_{1} \text{ এবং } \c_{3}^{\prime}=c_{3}-c_{1} \text{ প্রয়োগ করে }\end{array}\]\right]
=\left|\[\begin{array}{ccc} (\mathrm{m}+\mathrm{n})^{2} & (l+\mathrm{m}+\mathrm{n})(l-\mathrm{m}-\mathrm{n}) & (l+\mathrm{m}+\mathrm{n})(l-\mathrm{m}-\mathrm{n}) \\ \mathrm{m}^{2} & (l+\mathrm{m}+\mathrm{n})(l-\mathrm{m}+\mathrm{n}) & 0 \\ \mathrm{n}^{2} & (l+\mathrm{m}+\mathrm{n})(l+\mathrm{m}-\mathrm{n}) \end{array}\]\right|
\begin{aligned} & =\left|\begin{array}{ccc}(\mathrm{m}+\mathrm{n})^2 & (l+\mathrm{m}+\mathrm{n})(l-\mathrm{m}-\mathrm{n}) & (l+\mathrm{m}+\mathrm{n})(l-\mathrm{m}-\mathrm{n}) \\ \mathrm{m}^2 & (l+\mathrm{m}+\mathrm{n})(l-\mathrm{m}+\mathrm{n}) & 0 \\ \mathrm{n}^2 & 0 & (l+\mathrm{m}+\mathrm{n})(l+\mathrm{m}-\mathrm{n})\end{aligned}\right| \\ & =(l+\mathrm{m}+\mathrm{n})^2\left|\[\begin{array}{ccc}(\mathrm{m}+\mathrm{n})^2 & l-\mathrm{m}-\mathrm{n} & l-\mathrm{m}-\mathrm{n} \\ \mathrm{m}^2 & l-\mathrm{m}+\mathrm{n} & 0 \\ \mathrm{n}^2 & 0 & l+\mathrm{m}-\mathrm{n}\end{array}\]\right| \\ & =(l+\mathrm{m}+\mathrm{n})^2\left|\[\begin{array}{ccc}2 \mathrm{mn} & -2 \mathrm{n} & -2 \mathrm{~m} \\ \mathrm{~m}^2 & l-\mathrm{m}+\mathrm{n} & 0 \\ \mathrm{n}^2 & 0 & l+\mathrm{m}-\mathrm{n}\end{array}\]\right| \\ & {\left[r_1^{\prime}=r_1-r_2-r_3 \text{ প্রয়োগ করে] }\right.} \\ & =2(l+\mathrm{m}+\mathrm{n})^2\left|\[\begin{array}{ccc}\mathrm{mn} & -\mathrm{n} & -\mathrm{m} \\ \mathrm{m}^2 & l-\mathrm{m}+\mathrm{n} & 0 \\ \mathrm{n}^2 & 0 & l+\mathrm{m}-\mathrm{n}\end{array}\]\right| \\ & =2(l+\mathrm{m}+\mathrm{n})^2\left|\[\begin{array}{ccc}\mathrm{mn} & 0 & 0 \\ \mathrm{~m}^2 & \mathrm{n}+l & \frac{\mathrm{m}^2}{\mathrm{n}} \\ \mathrm{n}^2 & \frac{\mathrm{n}^2}{\mathrm{~m}} & l+\mathrm{m}\end{array}\]\right|\left[\[\begin{array}{l}\mathrm{c}_2^{\prime}=\mathrm{c}_2+\frac{1}{\mathrm{y}} \mathrm{c}_1 \text{ এবং } \\ \mathrm{c}_3^{\prime}=\mathrm{c}_3+\frac{1}{\mathrm{z}} \mathrm{c}_1 \text{ প্রয়োগ করে } \end{array}\]\right] \\ & \end{aligned}
\end{array}
=2(l+\mathrm{m}+\mathrm{n})^{2} \mathrm{mn}\left|\[\begin{array}{cc} \mathrm{n}+l & \frac{\mathrm{m}^{2}}{\mathrm{n}} \\ \frac{\mathrm{n}^{2}}{\mathrm{~m}} & l+\mathrm{m} \end{array}\]\right|
(দেখানো হলো)\end{array}