প্রমাণ কর : [গণিত]
উত্তর: বামপক্ষ [গণিত] ডানপক্ষ (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 : [গণিত]
উদ্দীপক
দৃশ্যকল্প:
\left.\[\begin{matrix} x+y+z=3 \x+a y+a^{2} z=1 \x+a^{2} y+a^{4} z=m \end{matrix}\]\right\}
C=\left[\[\begin{array}{ll}1 & 2 \\3 & 0\end{array}\]\right]
Mymensingh · 2021
উত্তর
বামপক্ষ
=\left[\[\begin{array}{ccc}x+y & 3(y+z) & x+z \\ 1 & 3 & 1 \z & 3 x & y\end{array}\]\right]
=\left[\[\begin{array}{ccc}x+y+z & 3(x+y+z) & x+y+z \\ 1 & 3 & 1 \z & 3 x & y\end{array}\]\right]
=(x+y+z)\left[\[\begin{array}{ccc}1 & 3 & 1 \\ 1 & 3 & 1 \z & 3 x & y\end{array}\]\right]
ডানপক্ষ
(প্রমাণিত)
উত্তর
প্রদত্ত সমীকরণগুলোঃ
বা, \left(\[\begin{array}{ccc}1 & 1 & 1 \\ 1 & a & a^{2} \\ 1 & a^{2} & a^{4}\end{array}\]\right)\left(\[\begin{array}{l}x \y \z\end{array}\]\right)=\left(\[\begin{array}{c}3 \l \m\end{array}\]\right) যা আকারে
যেখানে,
A=\left(\[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & a & a^{2} \\ 1 & a^{2} & a^{4} \end{array}\]\right), X=\left(\[\begin{array}{l} x \y \z \end{array}\]\right) \text{ এবং } B=\left(\[\begin{array}{c} 3 \l \m \end{array}\]\right)
\therefore \operatorname{det}(A) =\left|\[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & a & a^{2} \\ 1 & a^{2} & a^{4} \end{array}\]\right|
=\left|\[\begin{array}{ccc} 0 & 0 & 1 \\ 1-a & a-a^{2} & a^{2} \\ 1-a^{2} & a^{2}-a^{4} & a^{4} \end{array}\]\right|
=\left|\[\begin{array}{cc} 1-a & a(1-a) \\ 1-a^{2} & a^{2}\left(1-a^{2}\right) \end{array}\]\right|
=(1-a)\left(1-a^{2}\right)\left|\[\begin{array}{cc} 1 & a \\ 1 & a^{2} \end{array}\]\right|
(দেখানো হল)
উত্তর
দেওয়া আছে,
C =\left[\[\begin{array}{ll} 1 & 2 \\ 3 & 0 \end{array}\]\right]
C^{2} =\left[\[\begin{array}{ll} 1 & 2 \\ 3 & 0 \end{array}\]\right]
\cdot\left[\[\begin{array}{ll} 1 & 2 \\ 3 & 0 \end{array}\]\right]
=\left[\[\begin{array}{ll} 1.1+2.3 & 1.2+2.0 \\ 3.1+0.3 & 3.2+0.0 \end{array}\]\right]
=\left[\[\begin{array}{ll} 1+6 & 2+0 \\ 3+0 & 6+0 \end{array}\]\right]
=\left[\[\begin{array}{ll} 7 & 2 \\ 3 & 6 \end{array}\]\right]
এখানে,
=\left[\[\begin{array}{ll} 7 & 2 \\ 3 & 6 \end{array}\]\right]+3\left[\[\begin{array}{ll} 1 & 2 \\ 3 & 0 \end{array}\]\right]-7\left[\[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\]\right]
=\left[\[\begin{array}{ll} 7 & 2 \\ 3 & 6 \end{array}\]\right]+\left[\[\begin{array}{ll} 3 & 6 \\ 9 & 0 \end{array}\]\right]-\left[\[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\]\right]
=\left[\[\begin{array}{ll} 10 & 8 \\ 12 & 6 \end{array}\]\right]-\left[\[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\]\right]
=\left[\[\begin{array}{cc} 3 & 8 \\ 12 & -1 \end{array}\]\right]
নির্ণেয় f(C)=\left[\[\begin{array}{cc} 3 & 8 \\ 12 & -1 \end{array}\]\right]