[গণিত] নির্ণয় কর, যেখানে [গণিত] অভেদ মাট্রিক্স।
উত্তর: [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
নির্ণয় কর, যেখানে অভেদ মাট্রিক্স।
উদ্দীপক
A=\left[\[\begin{array}{ccc}12 & 3 & 0 \\1 & 2 & 1 \\6 & 1 & 0\end{array}\]\right], B=\left[\[\begin{array}{lll}1 & 0 & 1 \\0 & 2 & 0 \\3 & 0 & 1\end{array}\]\right], C=\left[\[\begin{array}{l}x \y \z\end{array}\]\right], D=\left[\[\begin{array}{l}1 \\2 \\ 1\end{array}\]\right]
Rajshahi · 2019
উত্তর
|A|=\left|\[\begin{array}{ccc}12 & 3 & 0 \\ 1 & 2 & 1 \\ 6 & 1 & 0\end{array}\]\right|
এর তম ভুক্তির সহগুণক =\left|\[\begin{array}{cc}12 & 0 \\ 1 & 1\end{array}\]\right|
নির্ণেয় তম ভুক্তির সহগুণক -12.
উত্তর
\begin{aligned} & A^2=A \cdot A \\ & =\left[\begin{array}{ccc}12 & 3 & 0 \\ 1 & 2 & 1 \\ 6 & 1 & 0\end{aligned}\right]\left[\[\begin{array}{ccc}12 & 3 & 0 \\ 1 & 2 & 1 \\ 6 & 1 & 0\end{array}\]\right] \\ & =\left[\[\begin{array}{ccc}12 \times 12+3 \times 1+0 \times 6 & 12 \times 3+3 \times 2+0 \times 1 & 12 \times 0+3 \times 1+0 \times 0 \\ 1 \times 12+2 \times 1+1 \times 6 & 1 \times 3+2 \times 2+1 \times 1 & 1 \times 0+2 \times 1+1 \times 0 \\ 6 \times 12+1 \times 1+0 \times 6 & 6 \times 3+1 \times 2+0 \times 1 & 6 \times 0+1 \times 1+0 \times 0\end{array}\]\right] \\ & =\left[\[\begin{array}{ccc}147 & 42 & 3 \\ 20 & 8 & 2 \\ 73 & 20 & 1\end{array}\]\right] \\ & \text{ এখন, } 5 A^2-3 I=5\left[\[\begin{array}{ccc}147 & 42 & 3 \\ 20 & 8 & 2 \\ 73 & 20 & 1\end{array}\]\right]-3\left[\[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right] \\ & =\left[\[\begin{array}{ccc}735 & 210 & 15 \\ 100 & 40 & 10 \\ 365 & 100 & 5\end{array}\]\right]-\left[\[\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\]\right]=\left[\[\begin{array}{ccc}732 & 210 & 15 \\ 100 & 37 & 10 \\ 365 & 100 & 2\end{array}\]\right]\end{aligned}
\end{array}\end{array}
উত্তর
\begin{aligned} & \text{ (গ) } B C=D \text{ হলে, } \\ & {\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 0 \\ 3 & 0 & 01\end{aligned}\right]\left[\[\begin{array}{l}x \y \z\end{array}\]\right]=\left[\[\begin{array}{l}1 \\ 2 \\ 1\end{array}\]\right] \text{ বा, }\left[\[\begin{array}{c}x+0 . y+z \\ 0 . x+2 y+0 . z \\ 3 x+0 . y+z\end{array}\]\right]=\left[\[\begin{array}{l}1 \\ 2 \\ 1\end{array}\]\right]} \\ & \therefore x+0 . y+z=1 \\ & 0 . x+2 y+0 . z=2 \\ & 3 x+0 . y+z=1 \\ & \left.\therefore \Delta=\mid \[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 0 \\ 3 & 0 & 1\end{array}\]\right]=1(2-0)+0+1(0-6)=2-6=-4 \\ & \Delta_x=\left[\[\begin{array}{lll}1 & 0 & 1 \\ 2 & 2 & 0 \\ 1 & 0 & 1\end{array}\]\right]=1(2-0)+0+1(0-2)=2-2=0 \\ & \Delta_y=\left[\[\begin{array}{lll}1 & 1 & 1 \\ 0 & 2 & 0 \\ 3 & 1 & 1\end{array}\]\right]=1(2-0)-1(0-0)+1(0-6)=2-6=-4\end{aligned}
\end{array}
এবং \Delta_z=\left[\[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 2 \\ 3 & 0 & 1\end{array}\]\right]=1(2-0)-0+1(0-6)=2-6=-4
এবং
নির্ণেয় সমাধান, \end{array}