দেখাও যে, [গণিত] .
উত্তর: দেওয়া আছে, [গণিত] (দেখানো হলো)
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[\[\begin{array}{lll}p & p^{2} & p^{3}+1 \q & q^{2} & \mathbf{q}^{3}+1 \r & r^{2} & r^{3}+1\end{array}\]\right], X=\left[\[\begin{array}{l}x \y \z\end{array}\]\right] এবং B=\left[\[\begin{array}{r}0 \\ -2 \\ -2\end{array}\]\right].
রংপুর ক্যাডেট কলেজ · 2025
উত্তর
দেওয়া আছে, \mathrm{C}=\left[\[\begin{array}{rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\]\right]
আমরা জানি, দুই ক্রমের অব্যতিক্রমী ম্যাট্রিক্স \left[\[\begin{array}{ll}a & b \c & d\end{array}\]\right] এর বিপরীত ম্যাট্রিক্স =\frac{1}{a d-b c}\left[\[\begin{array}{rr}d & -b \\ -c & a\end{array}\]\right]
\therefore C^{-1} =\frac{1}{\cos \theta \cdot \cos \theta-\sin \theta(-\sin \theta)}\left[\[\begin{array}{c}\cos \theta \\ -\sin \theta \\ -\cos \theta\end{array}\]\right]
=\frac{1}{\cos^2 \theta+\sin^2 \theta}\left[\[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\]\right]
=\left[\[\begin{array}{rr}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\]\right]\left[\because \sin^2 \theta+\cos^2 \theta=1\right] \\ \therefore C^{-1}
=\left[\[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\]\right] \cdot\left[\[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\]\right]
=\left[\[\begin{array}{cc}\cos^2 \theta+\sin^2 \theta & \sin \theta \cos \theta-\sin \theta \cos \theta \\ \sin \theta \cos^2-\sin \theta \cos \theta & \sin^2 \theta+\cos^2 \theta\end{array}\]\right]
=\left[\[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\]\right]\left[\because \sin^2 \theta+\cos^2 \theta=1\right]
নির্ণেয় \mathrm{CC}^{-1}:\left[\[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\]\right].
উত্তর
দেওয়া আছে, \mathrm{A}=\left[\[\begin{array}{lll}\mathrm{p} & \mathrm{p}^{2} & \mathrm{p}^{3}+1 \\ \mathrm{q} & \mathrm{q}^{2} & \mathrm{q}^{3}+1 \\ \mathrm{r} & \mathrm{r}^{2} & \mathrm{r}^{3}+1\end{array}\]\right], \mathrm{X}=\left[\[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\]\right] এবং \mathrm{B}=\left[\[\begin{array}{r}0 \\ -2 \\ -2\end{array}\]\right]
এখন, এবং হলে,
\mathrm{A}=\left[\[\begin{array}{rrr}1 & 1 & 2 \\ -1 & 1 & 0 \\ 2 & 4 & 9\end{array}\]\right]
এখন,
বা, \left[\[\begin{array}{rrr}1 & 1 & 2 \\ -1 & 1 & 0 \\ 2 & 4 & 9\end{array}\]\right]\left[\[\begin{array}{l}x \y \z\end{array}\]\right]=\left[\[\begin{array}{r}0 \\ -2 \\ -2\end{array}\]\right]
বা, \left[\[\begin{array}{c}x+y+2 z \\ -x+y+0 . z \\ 2 x+4 y+9 z\end{array}\]\right]=\left[\[\begin{array}{r}0 \\ -2 \\ -2\end{array}\]\right]
ম্যাট্রিক্সের সমতা অনুসারের পাই,
ও z এর সহগগুচ্ছ নিয়ে গঠিত নির্ণায়ক,
D=\left|\[\begin{array}{rrr}1 & 1 & 2 \\ -1 & 1 & 0 \\ 2 & 4 & 9\end{array}\]\right|
\therefore \quad \mathrm{D}_{\mathrm{x}}=\left|\[\begin{array}{rrr}0 & 1 & 2 \\ -2 & 1 & 0 \\ -2 & 4 & 9\end{array}\]\right|
D_y=\left|\[\begin{array}{rrr}1 & 0 & 2 \\ -1 & -2 & 0 \\ 2 & -2 & 9\end{array}\]\right|
\mathrm{D}_{\mathrm{z}}=\left|\[\begin{array}{rrr}1 & 1 & 0 \\ -1 & 1 & -2 \\ 2 & 4 & -2\end{array}\]\right|
এবং
নির্ণেয় সমাধান :
উত্তর
দেওয়া আছে, \mathrm{A}=\left[\[\begin{array}{lll}\mathrm{p} & \mathrm{p}^{2} & \mathrm{p}^{3}+1 \\ \mathrm{q} & \mathrm{q}^{2} & \mathrm{q}^{3}+1 \r & \mathrm{r}^{2} & \mathrm{r}^{3}+1\end{array}\]\right]
\therefore |A| =\left|\[\begin{array}{lll}p & p^2 & p^3+1 \q & q^2 & q^3+1 \r & r^2 & r^3+1\end{array}\]\right|=\left|\[\begin{array}{lll}p & p^2 & p^3 \q & q^2 & q^3 \r & r^2 & r^3\end{array}\]\right|+\left|\[\begin{array}{lll}p & p^2 & 1 \q & q^2 & 1 \r & r^2 & 1\end{array}\]\right|
=p q r\left|\[\begin{array}{lll}1 & p & p^2 \\ 1 & q & q^2 \\ 1 & r & r^2\end{array}\]\right|+\left|\[\begin{array}{lll}p & p^2 & 1 \q & q^2 & 1 \r & r^2 & 1\end{array}\]\right|
=\operatorname{pqr}\left|\[\begin{array}{lll}1 & p & p^2 \\ 1 & q & q^2 \\ 1 & r & r^2\end{array}\]\right|+\left|\[\begin{array}{ccc}1 & p & p^2 \\ 1 & q & q^2 \\ 1 & r & r^2\end{array}\]\right|
=(p q r+1)\left|\[\begin{array}{lll}1 & p & p^2 \\ 1 & q & q^2 \\ 1 & r & r^2\end{array}\]\right|
=(p q r+1)\left|\[\begin{array}{ccc}0 \mathbf{p}-q & p^2-q^2 \\ 0 & q-r & q^2-r^2 \\ 1 & r & r^2\end{array}\]\right|\left[\[\begin{array}{l}r_1^{\prime}=r_1-r_2, \r_2^{\prime}=r_2-r_3\end{array}\]\right]
=(p q r+1)\left|\[\begin{array}{ll}p-q & p^2-q^2 \q-r & q^2-r^2\end{array}\]\right|
=(p q r+1)(p-q)(q-r)\left|\[\begin{array}{ll}1 & p+q \\ 1 & q+r\end{array}\]\right|
(দেখানো হলো)