প্রমাণ কর যে, [গণিত]
উত্তর: দেওয়া আছে, [গণিত] এবং [গণিত] (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 যে,
উদ্দীপক
A=\left[\[\begin{array}{rrr}2 & -1 & 1 \\ 3 & 1 & -4 \\ 5 & 2 & -3\end{array}\]\right], B=\left[\[\begin{array}{lll}p^{2} & q r & 2 p \q^{2} & r p & 2 q \r^{2} & p q & 2 r\end{array}\]\right].
Dinajpur · 2023
উত্তর
দেওয়া আছে, \left[\[\begin{array}{cc}x-5 & 8 \\ -1 & y+3\end{array}\]\right]=\left[\[\begin{array}{cc}y-1 & 8 \\ -1 & 7\end{array}\]\right]
ম্যাট্রিক্সের সমতা অনুসারে পাই,
বা,
এবং
বা,
বা,
নির্ণেয় মান : .
উত্তর
দেওয়া আছে, \mathrm{A}=\left[\[\begin{array}{rrr}2 & -1 & 1 \\ 3 & 1 & -4 \\ 5 & 2 & -3\end{array}\]\right]
\therefore \quad|\mathrm{A}| =\left|\[\begin{array}{rrr}2 & -1 & 1 \\3 & 1 & -4 \\5 & 2 & -3\end{array}\]\right|
বিদ্যামান।
A_{11}=\left|\[\begin{array}{ll} 1 & -4 \\ 2 & -3 \end{array}\]\right|
A_{12}=-\left|\[\begin{array}{rr} 3 & -4 \\ 5 & -3 \end{array}\]\right|
A_{13}=\left|\[\begin{array}{lr} 3 & 1 \\ 5 & 2 \end{array}\]\right|
A_{21}=-\left|\[\begin{array}{rr} -1 & 1 \\ 2 & -3 \end{array}\]\right|
A_{22}=\left|\[\begin{array}{rr} 2 & 1 \\ 5 & -3 \end{array}\]\right|
A_{23}=-\left|\[\begin{array}{rr} 2 & -1 \\ 5 & 2 \end{array}\]\right|
A_{31}=\left|\[\begin{array}{rr} -1 & 1 \\ 1 & -4 \end{array}\]\right|
A_{32}=-\left|\[\begin{array}{rr} 2 & 1 \\ 3 & -4 \end{array}\]\right|
A_{33}=\left|\[\begin{array}{rr} 2 & -1 \\ 3 & 1 \end{array}\]\right|
\operatorname{Adj} A=\left[\[\begin{array}{lll} A_{11} & A_{12} & A_{13} \A_{21} & A_{22} & A_{23} \A_{31} & A_{32} & A_{33} \end{array}\]\right]^{T}
=\left[\[\begin{array}{rrr} 5 & -11 & 1 \\ -1 & -11 & -9 \\ 3 & 11 & 5 \end{array}\]\right]^{\mathrm{T}}=\left[\[\begin{array}{rrr} 5 & -1 & 3 \\ -11 & -11 & 11 \\ 1 & -9 & 5 \end{array}\]\right]
A^{-1}=\frac{1}{|A|} \operatorname{Adj} A=\frac{1}{22}\left[\[\begin{array}{rrr} 5 & -1 & 3 \\ -11 & -11 & 11 \\ 1 & -9 & 5 \end{array}\]\right]
নির্ণেয় \mathrm{A}^{-1}=\frac{1}{22}\left[\[\begin{array}{rrr} 5 & -1 & 3 \\ -11 & -11 & 11 \\ 1 & -9 & 5 \end{array}\]\right]
উত্তর
দেওয়া আছে, B=\left[\[\begin{array}{lll} p^{2} & \mathrm{qr} & 2 \mathrm{p} \\ \mathrm{q}^{2} & \mathrm{rp} & 2 \mathrm{q} \\ \mathrm{r}^{2} & \mathrm{pq} & 2 \mathrm{r}\end{array}\]\right]
\therefore|B|=\left|\[\begin{array}{lll} p^{2} & \mathrm{qr} & 2 \mathrm{p} \\ \mathrm{q}^{2} & \mathrm{rp} & 2 \mathrm{q} \\ \mathrm{r}^{2} & \mathrm{pq} & 2 \mathrm{r}\end{array}\]\right|
=2\left|\[\begin{array}{lll} \mathrm{p}^{2} & \mathrm{qr} & \mathrm{p} \mathrm{q}^{2} & \mathrm{rp} & \mathrm{q} \\ \mathrm{r}^{2} & \mathrm{pq} & \mathrm{r}\end{array}\]\right|
=2\left|\[\begin{array}{ccc} p^{2}-q^{2} & q r-r p & p-q \q^{2}-r^{2} & r p-p q & q-r \r^{2} & p q & r \end{array}\]\right|
এবং
=2\left|\[\begin{array}{ccc} (p+q)(p-q) & -r(p-q) & p-q \\ (q+r)(q-r) & -p(q-r) & q-r \r^{2} & p q & r \end{array}\]\right|
=2(p-q)(q-r)\left|\[\begin{array}{ccc} p+q & -r & 1 \q+r & -p & 1 \r^{2} & p q & r \end{array}\]\right|
=2(p-q)(q-r)\left|\[\begin{array}{ccc} p-r & p-r & 0 \q+r & -p & 1 \r^{2} & p q & r \end{array}\]\right|\left[r_{1}^{\prime}=r_{1}-r_{2}\right]
=2(p-q)(q-r)(p-r)\left|\[\begin{array}{ccc} 1 & 1 & 0 \q+r & -p & 1 \r^{2} & p q & r \end{array}\]\right|
=2(p-q)(q-r)(p-r)\left|\[\begin{array}{ccc} 1 & 0 & 0 \q+r & -p-q-r & 1 \r^{2} & p q-r^{2} & r \end{array}\]\right|\left[c_{2}{ }^{\prime}=c_{2}-c_{1}\right]
=2(p-q)(q-r)(p-r)\left|\[\begin{array}{cc} -p-q-r & 1 \p q-r^{2} & r \end{array}\]\right|
(প্রমাণিত)