প্রমাণ কর যে, [গণিত]
উত্তর: দেওয়া আছে, [গণিত] এবং [গণিত] . [গণিত] L.H.S [গণিত] L.H.S [গণিত] R.H.S
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
\mathrm{P}=\left[\[\begin{array}{ccc}4 & -1 & 3 \\ 0 & 7 & 5 \\ 6 & -2 & 2\end{array}\]\right] এবং, \mathrm{Q}=\left[\[\begin{array}{ccc}0 & 4 & 3 \\ -3 & -4 & -5 \\ -2 & 1 & 2\end{array}\]\right]
Dhaka · 2023
উত্তর
দেওয়া আছে,
P=\left[\[\begin{array}{ccc}4 & -1 & 3 \\ 0 & 7 & 5 \\ 6 & -2 & 2\end{array}\]\right] এবং, Q=\left[\[\begin{array}{ccc}0 & 4 & 3 \\ -3 & -4 & -5 \\ -2 & 1 & 2\end{array}\]\right]
\therefore P+Q=\left[\[\begin{array}{ccc}4 & -1 & 3 \\ 0 & 7 & 5 \\ 6 & -2 & 2\end{array}\]\right]+\left[\[\begin{array}{ccc}0 & 4 & 3 \\ -3 & -4 & -5 \\ -2 & 1 & 2\end{array}\]\right]
=\left[\[\begin{array}{ccc}4 & 3 & 6 \\ -3 & 3 & 0 \\ 4 & -1 & 4\end{array}\]\right]
ম্যাট্রিক্সের ট্রেস
উত্তর
দেওয়া আছে,
P=\left[\[\begin{array}{rrr}4 & -1 & 3 \\ 0 & 7 & 5 \\ 6 & -2 & 2\end{array}\]\right] এবং Q=\left[\[\begin{array}{rrr}0 & 4 & 3 \\ -3 & -4 & -5 \\ -2 & 1 & 2\end{array}\]\right].
\therefore p^{t}=\left[\[\begin{array}{ccc}4 & 0 & 6 \\ -1 & 7 & -2 \\ 3 & 5 & 2\end{array}\]\right] \therefore Q^{t}=\left[\[\begin{array}{ccc}0 & -3 & -2 \\ 4 & -4 & 1 \\ 3 & -5 & 2\end{array}\]\right]
\therefore P Q=\left[\[\begin{array}{ccc}4 & -1 & 3 \\ 0 & 7 & 5 \\ 6 & -2 & 2\end{array}\]\right] \times\left[\[\begin{array}{ccc}0 & 4 & 3 \\ -3 & -4 & -5 \\ -2 & 1 & 2\end{array}\]\right]
=\left[\[\begin{array}{ccc}0+3-6 & 16+4+3 & 12+5+6 \\ 0-2-10 & 0-28+5 & 0-35+10 \\ 0+6-4 & 24+8+2 & 18+10+4\end{array}\]\right]
=\left[\[\begin{array}{ccc}-3 & 23 & 23 \\ -31 & -23 & -25 \\ 2 & 34 & 32\end{array}\]\right]
L.H.S
=\left[\[\begin{array}{ccc}-3 & 23 & 23 \\ -31 & -23 & -25 \\ 2 & 34 & 32\end{array}\]\right]
=\left[\[\begin{array}{ccc}-3 & -31 & 2 \\ 23 & -23 & 34 \\ 23 & -25 & 32\end{array}\]\right]
=\left[\[\begin{array}{ccc}0 & -3 & -2 \\ 4 & -4 & 1 \\ 3 & -5 & 2\end{array}\]\right] \times\left[\[\begin{array}{ccc}4 & 0 & 6 \\ -1 & 7 & -2 \\ 3 & 5 & 2\end{array}\]\right]
=\left[\[\begin{array}{ccc}0+3-6 & 0-21-10 & 0+6-4 \\ 16+4+3 & 0-28+5 & 24+8+2 \\ 12+5+6 & 0-35+10 & 18+10+4\end{array}\]\right]
=\left[\[\begin{array}{ccc}-3 & -31 & 2 \\ 23 & -23 & 34 \\ 23 & -25 & 32\end{array}\]\right]
L.H.S R.H.S
উত্তর
দেওয়া আছে,
এবং P=\left[\[\begin{array}{ccc}4 & -1 & 3 \\ 0 & 7 & 5 \\ 6 & -2 & 2\end{array}\]\right]
\therefore|p|=\left|\[\begin{array}{ccc}4 & -1 & 3 \\ 0 & 7 & 5 \\ 6 & -2 & 2\end{array}\]\right|
, যা একটি অব্যতিক্রমী ম্যাট্রিক্স
এর সহগুণক :
\therefore \operatorname{adj} P=\left[\[\begin{array}{ccc}24 & 35 & -42 \\ -4 & -10 & 2 \\ -26 & -20 & 28\end{array}\]\right]^{T}
=\left[\[\begin{array}{ccc}24 & -4 & -26 \\ 35 & -10 & -20 \\ -42 & 2 & 28\end{array}\]\right]
=\frac{1}{-60}\left[\[\begin{array}{ccc}24 & -4 & -26 \\ 35 & -10 & -20 \\ -42 & 2 & 28\end{array}\]\right]
=\frac{2 \times 2 \times 5}{-60}\left[\[\begin{array}{ccc}12 & -2 & -13 \\ 7 & -2 & -4 \\ -21 & 1 & 14\end{array}\]\right]
=\frac{-1}{3}\left[\[\begin{array}{ccc}12 & -2 & -13 \\ 7 & -2 & -4 \\ -21 & 1 & 14\end{array}\]\right]
\therefore R=I \times P^{-1}=\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right] \times \frac{-1}{3}\left[\[\begin{array}{ccc}12 & -2 & -13 \\ 7 & -2 & -4 \\ -21 & 1 & 14\end{array}\]\right]
=-\frac{1}{3}\left[\[\begin{array}{ccc}12 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 14\end{array}\]\right]
=\left[\[\begin{array}{ccc}-4 & 0 & 0 \\ 0 & \frac{2}{3} & 0 \\ 0 & 0 & -\frac{14}{3}\end{array}\]\right]