দৃশ্যকল্প-২ হতে প্রমাণ কর যে. [গণিত]
উত্তর: দেওয়া আছে, [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
দৃশ্যকল্প-২ হতে প্রমাণ কর যে.
উদ্দীপক
দৃশ্যকল্প-১ : \mathrm{A}=\left[\[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1\end{array}\]\right]
দৃশ্যকল্প-২ : B=\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
দেওয়া আছে, A=\left[\[\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\]\right], B=\left[\[\begin{array}{r}4 \\ 6 \\ -1\end{array}\]\right] এবং C=\left[\[\begin{array}{llll}1 & 2 & -5 & 6\end{array}\]\right]
\therefore \quad A B=\left[\[\begin{array}{lll}1 & 2 & 3 \\4 & 5 & 6\end{array}\]\right]\left[\[\begin{array}{r} 4 \\6 \\-1\end{array}\]\right]=\left[\[\begin{array}{c}4+12-3 \\16+30-6\end{array}\]\right]=\left[\[\begin{array}{l} 13 \\40\end{array}\]\right]
(A B) C=\left[\[\begin{array}{l}13 \\ 40\end{array}\]\right]\left[\[\begin{array}{llll}1 & 2 & -5 & 6\end{array}\]\right]=\left[\[\begin{array}{cccc}13 & 26 & -65 & 78 \\ 40 & 80 & -200 & 240\end{array}\]\right]
উত্তর
দেওয়া আছে, A=\left[\[\begin{array}{rrr}1 & 3 & 2 \\2 & 0 & 3 \\1 & -1 & 1\end{array}\]\right]
এখানে, I=\left[\[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\]\right]
\therefore A^{2}=A \cdot A=\left[\[\begin{array}{rrr}1 & 3 & 2 \\2 & 0 & 3 \\1 & -1 & 1\end{array}\]\right] .\left[\[\begin{array}{rrr}1 & 3 & 2 \\2 & 0 & 3 \\1 & -1 & 1\end{array}\]\right]
=\left[\[\begin{array}{lll}1+6+2 & 3+0-2 & 2+9+2 \\2+0+3 & 6+0-3 & 4+0+3 \\1-2+1 & 3-0-1 & 2-3+1 \end{array}\]\right] =\left[\[\begin{array}{lll}9 & 1 & 13 \\5 & 3 & 7 \\0 & 2 & 0\end{array}\]\right]
এবং A^3 =A^2 \cdot A=\left[\[\begin{array}{lll}9 & 1 & 13 \\ 5 & 3 & 7 \\ 0 & 2 & 0\end{array}\]\right]\left[\[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1\end{array}\]\right]
=\left[\[\begin{array}{ccc}9+2+13 & 27+0-13 & 18+3+13 \\ 5+6+7 & 15+0-7 & 10+9+7 \\ 0+4+0 & 0+0+0 & 0+6+0\end{array}\]\right]
=\left[\[\begin{array}{ccc}24 & 14 & 34 \\ 18 & 8 & 26 \\ 4 & 0 & 6\end{array}\]\right]
প্রদত্ত রাশি
=\left[\[\begin{array}{ccc}24 & 14 & 34 \\ 18 & 8 & 26 \\ 4 & 0 & 6\end{array}\]\right]-2\left[\[\begin{array}{ccc}9 & 1 & 13 \\ 5 & 3 & 7 \\ 0 & 2 & 0\end{array}\]\right] +\left[\[\begin{array}{rrr}1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1\end{array}\]\right]-2\left[\[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\]\right]
=\left[\[\begin{array}{ccc}24 & 14 & 34 \\ 18 & 8 & 26 \\ 4 & 0 & 6\end{array}\]\right]-\left[\[\begin{array}{ccc}18 & 2 & 26 \\ 10 & 6 & 14 \\ 0 & 4 & 0\end{array}\]\right]+\left[\[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 0 & 3 \\ 1 & -1 & 1\end{array}\]\right]-\left[\[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\]\right]
=\left[\[\begin{array}{ccc}24-18 & 14-2 & 34-26 \\ 18-10 & 8-6 & 26-14 \\ 4-0 & 0-4 & 6-0\end{array}\]\right]+\left[\[\begin{array}{rrr}1-2 & 3-0 & 2-0 \\ 2-0 & 0-2 & 3-0 \\ 1-0 & -1-0 & 1-2\end{array}\]\right]
=\left[\[\begin{array}{ccc}6 & 12 & 8 \\ 8 & -2 & 12 \\ 4 & -4 & 6\end{array}\]\right]+\left[\[\begin{array}{ccc}-1 & 3 & 2 \\ 2 & -2 & 3 \\ 1 & -1 & -1\end{array}\]\right]
=\left[\[\begin{array}{ccc}6-1 & 12+3 & 8+2 \\ 8+2 & 2-2 & 12+3 \\ 4+1 & -4-1 & 6-1\end{array}\]\right]=\left[\[\begin{array}{ccc}5 & 15 & 10 \\ 10 & 0 & 15 \\ 5 & -5 & 5\end{array}\]\right] .
উত্তর
দেওয়া আছে, \mathrm{B}=\left|\[\begin{array}{ccc}(b+c)^{2} & \mathrm{a}^{2} & \mathrm{a}^{2} \\ \mathrm{~b}^{2} & (c+a)^{2} & \mathrm{~b}^{2} \\ \mathrm{c}^{2} & c^{2} & (a+b)^{2}\end{array}\]\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[\[\begin{array}{c}c_{2}{ }^{\prime}=c_{2}-c_{1} \c_{3}^{\prime}=c_{3}-c_{1}\end{array}\]\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[\[\begin{array}{cc}c_{2}{ }^{\prime}=c_{2}+\frac{1}{b} c_{1} \c_{3}^{\prime}=c_{3}+\frac{1}{c} c_{1}\end{array}\]\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|
=\frac{2}{b c}(a+b+c)^{2} b c\left|\[\begin{array}{|ccc}b c & 0 & 0 \b & c+a & b \c & c & a+b\end{array}\]\right|
(প্রমাণিত)