যদি [গণিত] হয়, তাহলে [গণিত] নির্ণয় করো ।
উত্তর: দেওয়া আছে, [গণিত] হলে, [গণিত] বিদ্যমান। Adj [গণিত] M এর সহগুণকগুলি হচ্ছে : [গণিত] নির্ণেয় [গণিত] . [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
যদি হয়, তাহলে নির্ণয় করো ।
উদ্দীপক
M=\left[\[\begin{array}{ccc}(b+c)^2 & a^2 & a^2 \b^2 & (c+a)^2 & b^2 \c^2 & c^2 & (a+b)^2\end{array}\]\right].
উত্তর
\left(\[\begin{array}{lll}3 & 1 & 9 \\ 2 x & 2 & 6 \x^2 & 3 & 3\end{array}\]\right) ম্যাট্রিক্সটি ব্যাত্রিমী হলে, \left|\[\begin{array}{ccc}3 & 1 & 9 \\ 2 x & 2 & 6 \x^2 & 3 & 3\end{array}\]\right|=0
বা, 3\left|\[\begin{array}{ll}2 & 6 \\ 3 & 3\end{array}\]\right|-1\left|\[\begin{array}{ll}2 x & 6 \x^2 & 3\end{array}\]\right|+9\left|\[\begin{array}{ll}2 x & 2 \x^2 & 3\end{array}\]\right|=0
বা,
বা,
বা,
বা,
বা,
বা,
বা,
বা,
হয়,
অথবা,
নির্ণেয় মান : .
উত্তর
দেওয়া আছে, M=\left[\[\begin{array}{ccc}(b+c)^2 & a^2 & a^2 \b^2 & (c+a)^2 & b^2 \c^2 & c^2 & (a+b)^2\end{array}\]\right]
বামপক্ষ
\begin{aligned} & =\left|\begin{array}{ccc}(b+c)^2 & a^2 & a^2 \b^2 & (c+a)^2 & b^2 \c^2 & c^2 & (a+b)^2\end{aligned}\right| \\ & =\left|\[\begin{array}{ccc}(b+c)^2 & a^2-(b+c)^2 & a^2-(b+c)^2 \b^2 & (c+a)^2-b^2 & b^2-b^2 \c^2 & c^2-c^2 & (a+b)^2-c^2\end{array}\]\right|\left[\[\begin{array}{l}c_2{ }^{\prime}=c_2-c_1 \c_3{ }^{\prime}=c_3-c_1\end{array}\]\right] \\ & =\left|\[\begin{array}{ccc}(b+c)^2 & (a+b+c)(a-b-c) & (a+b+c)(a-b-c) \b^2 & (a+b+c)(c+a-b) & 0 \c^2 & 0 & (a+b+c)(a+b-c)\end{array}\]\right| \\ & =(a+b+c)^2\left|\[\begin{array}{ccc}b^2+2 b c+c^2 & a-b-c & a-b-c \b^2 & c+a-b & 0 \c^2 & 0 & a+b-c\end{array}\]\right| \\ & =(a+b+c)^2\left|\[\begin{array}{ccc}2 b c & -2 c & -2 b \b^2 & c+a-b & 0 \c^2 & 0 & a+b-c\end{array}\]\right|\left[r_1{ }^{\prime}=r_1-r_2-r_3\right] \\ & =2(a+b+c)^2\left|\[\begin{array}{ccc}b c & -c & -b \b^2 & c+a-b & 0 \c^2 & 0 & a+b-c\end{array}\]\right| \\ & =2(a+b+c)^2\left|\[\begin{array}{ccc}b c & 0 & 0 \b^2 & c+a & \frac{b^2}{c} \c^2 & \frac{c^2}{b} & a+b\end{array}\]\right|\left[\[\begin{array}{ccc}c_2{ }^{\prime}=c_2+\frac{1}{b} c_1 \c_3{ }^{\prime}=c_3+\frac{1}{c} c_1\end{array}\]\right] \\ & =\frac{2}{b c}(a+b+c)^2\left|\[\begin{array}{ccc}b c & 0 & 0 \b^2 & b c+a b & b^2 \c^2 & c^2 & a c+b c\end{array}\]\right|\end{aligned}
\end{array}
\begin{aligned} & =\frac{2}{b c}(a+b+c)^2 b c\left|\begin{array}{ccc}b c & 0 & 0 \b & c+a & b \c & c & a+b\end{aligned}\right| \\ & =2(a+b+c)^2\left\{b c\left(c a+b c+a^2+a b-b c\right)\right\} \\ & =2(a+b+c)^2\left\{b c\left(a^2+a b+c a\right)\right\} \\ & =2(a+b+c)^2[b c\{a(a+b+c)\}]\end{aligned}
\end{array}
ডানপক্ষ
. (দেখানো হলো)\end{array}
উত্তর
দেওয়া আছে, \mathrm{M}=\left[\[\begin{array}{ccc}(b+c)^2 & a^2 & a^2 \b^2 & (c+a)^2 & b^2 \c^2 & c^2 & (a+b)^2\end{array}\]\right]
হলে, \mathbf{M}=\left[\[\begin{array}{ccc}(1+1)^2 & 1^2 & 2^2 \\ 1^2 & (1+1)^2 & 1^2 \\ 1^2 & 1^2 & (1+1)^2\end{array}\]\right]
\begin{aligned} \therefore & M \\ \therefore & =\left[\begin{array}{lll}4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4\end{aligned}\right] \\ \therefore & |M|=\left|\[\begin{array}{lll}4 & 1 & 1 \\ 1 & 4 & 1 \\ 1 & 1 & 4\end{array}\]\right| \\ & =4(16-1)-1(4-1)+1(1-4)=60-3-3=54 \neq 0\end{aligned}
\end{array}
বিদ্যমান।
Adj M=\left[\[\begin{array}{lll}M_{11} & M_{12} & M_{13} \M_{21} & M_{22} & M_{23} \M_{31} & M_{32} & M_{33}\end{array}\]\right]^T
M এর সহগুণকগুলি হচ্ছে :
\begin{aligned} & M_{11}=\left|\begin{array}{ll}4 & 1 \\ 1 & 4\end{aligned}\right|=16-1=15 \\ & M_{12}=-\left|\[\begin{array}{ll}1 & 1 \\ 1 & 4\end{array}\]\right|=-(4-1)=-3 \\ & M_{13}=\left|\[\begin{array}{ll}1 & 4 \\ 1 & 1\end{array}\]\right|=1-4=-3 \\ & M_{21}=-\left|\[\begin{array}{ll}1 & 1 \\ 1 & 4\end{array}\]\right|=-(4-1)=-3 \\ & M_{22}=\left|\[\begin{array}{ll}4 & 1 \\ 1 & 4\end{array}\]\right|=16-1=15 \\ & M_{23}=-\left|\[\begin{array}{ll}4 & 1 \\ 1 & 1\end{array}\]\right|=-(4-1)=-3 \\ & M_{31}=\left|\[\begin{array}{ll}1 & 1 \\ 4 & 1\end{array}\]\right|=1-4=-3 \\ & M_{32}=-\left|\[\begin{array}{ll}4 & 1 \\ 1 & 1\end{array}\]\right|=-(4-1)=-3 \\ & M_{33}=\left|\[\begin{array}{ll}4 & 1 \\ 1 & 4\end{array}\]\right|=16-1=15 \\ & \therefore A^{\mathrm{Adj}(M)}=\left[\[\begin{array}{lll}15 & -3 & -3 \\ -3 & 15 & -3 \\ -3 & -3 & 15\end{array}\]\right]^{\mathrm{T}}=\left[\[\begin{array}{ccc}15 & -3 & -3 \\ -3 & 15 & -3 \\ -3 & -3 & 15\end{array}\]\right] \\ & \therefore M^{-1}=\frac{1}{|M|} \text{ Adj (M) }=\frac{1}{54}\left[\[\begin{array}{ccc}15 & -3 & -3 \\ -3 & 15 & -3 \\ -3 & -3 & 15\end{array}\]\right]\end{aligned}
\end{array}
=\frac{1}{54} \times 3\left[\[\begin{array}{rrr}5 & -1 & -1 \\ -1 & 5 & -1 \\ -1 & -1 & 5\end{array}\]\right]=\frac{1}{18}\left[\[\begin{array}{rrr}5 & -1 & -1 \\ -1 & 5 & -1 \\ -1 & -1 & 5\end{array}\]\right]
নির্ণেয় \mathrm{M}^{-1}=\frac{1}{18}\left[\[\begin{array}{rrr}5 & -1 & -1 \\ -1 & 5 & -1 \\ -1 & -1 & 5\end{array}\]\right].\end{array}