[গণিত] হলে [গণিত] ম্যাট্রিক্স নির্ণয় কর যেখানে [গণিত] .
উত্তর: দেওয়া আছে,, [গণিত] এখন, [গণিত] হলে, [গণিত] আবার, [গণিত] এখন, [গণিত] বিদ্যমান। D এর সহগুণকসমূহ : [গণিত] নির্ণেয় [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
হলে ম্যাট্রিক্স নির্ণয় কর যেখানে .
উদ্দীপক
A =\left(\[\begin{array}{ccc}(q+r)^2 & (r+p)^2 & (p+q)^2 \p^2 & q^2 & r^2 \\ 1 & 1 & 1\end{array}\]\right)
B=\left(\[\begin{array}{rrr}1 & 3 & 1 \\ -1 & -1 & 1 \\ 2 & 17 & 4\end{array}\]\right)
এবং C=\left(\[\begin{array}{lll}1 & 2 & 5 \\ 7 & 0 & 9 \\ 3 & 4 & 6\end{array}\]\right) .
চট্টগ্রাম কলেজ · 2025
উত্তর
দেওয়া আছে, C=\left(\[\begin{array}{lll}1 & 2 & 5 \\ 7 & 0 & 9 \\ 3 & 4 & 6\end{array}\]\right)
সমঘাতী ম্যাট্রিক্স হলে, হবে।
এখন,
\begin{aligned} & =\left(\begin{array}{lll}1 & 2 & 5 \\ 7 & 0 & 9 \\ 3 & 4 & 6\end{aligned}\right)\left(\[\begin{array}{ccc}1 & 2 & 5 \\ 7 & 0 & 9 \\ 3 & 4 & 6\end{array}\]\right) \\ & =\left(\[\begin{array}{lll}1.1+2.7+5.3 & 1.2+2.0+5.4 & 1.5+2.9+5.6 \\ 7.1+0.7+9.3 & 7.2+0.0+9.4 & 7.5+0.9+9.6 \\ 3.1+4.7+6.3 & 3.2+4.0+6.4 & 3.5+4.9+6.6\end{array}\]\right) \\ & =\left(\[\begin{array}{lll}1+14+15 & 2+0+20 & 5+18+30 \\ 7+0+27 & 14+0+36 & 35+0+54 \\ 3+28+18 & 6+0+24 & 15+36+36\end{array}\]\right) \\ & = \left(\[\begin{array}{lll}30 & 22 & 53 \\ 34 & 50 & 89 \\ 49 & 30 & 87\end{array}\]\right) \neq C\end{aligned}
\end{array}
যেহেতু
সুতরাং, ম্যাট্রিক্সটি সমঘাতী ম্যাট্রিক্স নয়।\end{array}
উত্তর
দেওয়া আছে, A =\left(\[\begin{array}{ccc}(q+r)^2 & (r+p)^2 & (p+q)^2 \p^2 & q^2 & r^2 \\ 1 & 1 & 1\end{array}\]\right)
\therefore|A|=\left|\[\begin{array}{ccc}(q+r)^{2} & (r+p)^{2} & (p+q)^{2} \p^{2} & q^{2} & r^{2} \\ 1 & 1 & 1\end{array}\]\right|
=\left|\[\begin{array}{ccc}(q+r)^{2}-(r+p)^{2} & (r+p)^{2}-(p+q)^{2} & (p+q)^{2} \p^{2}-q^{2} & q^{2}-r^{2} & r^{2} \\ 0 & 0 & 1\end{array}\]\right|
=\left|\[\begin{array}{cc}(p+q+2 r)(q-p) & (2 p+q+r)(r-q) \\ -\left(q^{2}-p^{2}\right) & -\left(r^{2}-q^{2}\right)\end{array}\]\right|
=(q-p)(r-q)\left|\[\begin{array}{cc}p+q+2 r & 2 p+q+r \\ -(q+p) & -(r+q)\end{array}\]\right|
=-(p-q)(q-r)\left|\[\begin{array}{cc}p+q+2 r & 2 p+q+r \p+q & q+r\end{array}\]\right|
=-(p-q)(q-r)\left|\[\begin{array}{cc}2 p+2 q+2 r & 2 p+2 q+2 r \p+q & q+r\end{array}\]\right|
=-(2 p+2 q+2 r)(p-q)(q-r)\left|\[\begin{array}{cc}1 & 1 \p+q & q+r\end{array}\]\right|
(দেখানো হলো)
উত্তর
দেওয়া আছে,, \mathrm{B}=\left(\[\begin{array}{rrr}1 & 3 & 1 \\ -1 & -1 & 1 \\ 2 & 17 & 4\end{array}\]\right)
এখন, হলে,
\mathrm{D}=\left[\[\begin{array}{rrr}1 & 3 & 1 \\ -1 & -1 & 1 \\ 2 & 17 & 4\end{array}\]\right]^{\mathrm{T}}=\left[\[\begin{array}{rrr}1 & -1 & 2 \\ 3 & -1 & 17 \\ 1 & 1 & 4\end{array}\]\right]
আবার,
এখন,
|D| =\left|\[\begin{array}{rrr} 1 & -1 & 2 \\ 3 & -1 & 17 \\ 1 & 1 & 4 \end{array}\]\right|
বিদ্যমান।
D এর সহগুণকসমূহ :
D_{11}=\left|\[\begin{array}{rr} -1 & 17 \\ 1 & 4 \end{array}\]\right|
D_{12}=-\left|\[\begin{array}{rr} 3 & 17 \\ 1 & 4 \end{array}\]\right|
D_{13}=\left|\[\begin{array}{rr} 3 & -1 \\ 1 & 1 \end{array}\]\right|
D_{21}=-\left|\[\begin{array}{rr} -1 & 2 \\ 1 & 4 \end{array}\]\right|
D_{22}=\left|\[\begin{array}{rr} 1 & 2 \\ 2 & 4 \end{array}\]\right|
D_{23}=-\left|\[\begin{array}{rr} 1 & -1 \\ 1 & 1 \end{array}\]\right|
D_{31}=\left|\[\begin{array}{cc} -1 & 2 \\ -1 & 17 \end{array}\]\right|
D_{32}=-\left|\[\begin{array}{cc} 1 & 2 \\ 3 & 17 \end{array}\]\right|
D_{33}=\left|\[\begin{array}{ll} 1 & -1 \\ 3 & -1 \end{array}\]\right|
Adj \mathbf{D} =\left[\[\begin{array}{lll} D_{11} & D_{12} & D_{13} \D_{21} & D_{22} & D_{23} \D_{31} & D_{32} & D_{33} \end{array}\]\right]^{\mathrm{T}}
=\left[\[\begin{array}{rrr} -21 & 5 & 4 \\ 6 & 2 & -2 \\ -15 & -11 & 2 \end{array}\]\right]^{\mathrm{T}}
=\left[\[\begin{array}{rrr} -21 & 6 & -15 \\ 5 & 2 & -11 \\ 4 & -2 & 2 \end{array}\]\right]
=\frac{1}{-18}\left[\[\begin{array}{rrr} -21 & 6 & -15 \\ 5 & 2 & -11 \\ 4 & -2 & 2 \end{array}\]\right]
=\frac{1}{18}\left[\[\begin{array}{rrr} 21 & -6 & 15 \\ -5 & -2 & 11 \\ -4 & 2 & -2 \end{array}\]\right]
নির্ণেয় E=\frac{1}{18}\left[\[\begin{array}{rrr} 21 & -6 & 15 \\ -5 & -2 & 11 \\ -4 & 2 & -2 \end{array}\]\right]