সৃজনশীলঃ ‘দৃশ্যকল্প-১ : A=[1322031−11]…’ ২০২৪ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $A=\left[\begin{array}{ccc}1& 3& 2\\ 2& 0& 3\\ 1& -1& 1\end{array}\right]$

এখানে, $I=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\therefore A^2=A\cdot A=\left[\begin{array}{ccc}1& 3& 2\\ 2& 0& 3\\ 1& -1& 1\end{array}\right].\left[\begin{array}{ccc}1& 3& 2\\ 2& 0& 3\\ 1& -1& 1\end{array}\right]$

$=\left[\begin{array}{ccc}1+6+2& 3+0-2& 2+9+2\\ 2+0+3& 6+0-3& 4+0+3\\ 1-2+1& 3-0-1& 2-3+1\end{array}\right]=\left[\begin{array}{ccc}9& 1& 13\\ 5& 3& 7\\ 0& 2& 0\end{array}\right]$

এবং $A^3=A^2\cdot A=\left[\begin{array}{ccc}9& 1& 13\\ 5& 3& 7\\ 0& 2& 0\end{array}\right]\left[\begin{array}{ccc}1& 3& 2\\ 2& 0& 3\\ 1& -1& 1\end{array}\right]$

$=\left[\begin{array}{ccc}9+2+13& 27+0-13& 18+3+13\\ 5+6+7& 15+0-7& 10+9+7\\ 0+4+0& 0+0+0& 0+6+0\end{array}\right]$

$=\left[\begin{array}{ccc}24& 14& 34\\ 18& 8& 26\\ 4& 0& 6\end{array}\right]$

প্রদত্ত রাশি $=A^3-2A^2+A-2I$

$=\left[\begin{array}{ccc}24& 14& 34\\ 18& 8& 26\\ 4& 0& 6\end{array}\right]-2\left[\begin{array}{ccc}9& 1& 13\\ 5& 3& 7\\ 0& 2& 0\end{array}\right]+\left[\begin{array}{ccc}1& 3& 2\\ 2& 0& 3\\ 1& -1& 1\end{array}\right]-2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$=\left[\begin{array}{ccc}24& 14& 34\\ 18& 8& 26\\ 4& 0& 6\end{array}\right]-\left[\begin{array}{ccc}18& 2& 26\\ 10& 6& 14\\ 0& 4& 0\end{array}\right]+\left[\begin{array}{ccc}1& 3& 2\\ 2& 0& 3\\ 1& -1& 1\end{array}\right]-\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$

$=\left[\begin{array}{ccc}24-18& 14-2& 34-26\\ 18-10& 8-6& 26-14\\ 4-0& 0-4& 6-0\end{array}\right]+\left[\begin{array}{ccc}1-2& 3-0& 2-0\\ 2-0& 0-2& 3-0\\ 1-0& -1-0& 1-2\end{array}\right]$

$=\left[\begin{array}{ccc}6& 12& 8\\ 8& -2& 12\\ 4& -4& 6\end{array}\right]+\left[\begin{array}{ccc}-1& 3& 2\\ 2& -2& 3\\ 1& -1& -1\end{array}\right]$

$=\left[\begin{array}{ccc}6-1& 12+3& 8+2\\ 8+2& 2-2& 12+3\\ 4+1& -4-1& 6-1\end{array}\right]=\left[\begin{array}{ccc}5& 15& 10\\ 10& 0& 15\\ 5& -5& 5\end{array}\right].$

দেওয়া আছে, $B=\left|\begin{array}{ccc}(b+c{)}^2& a^2& a^2\\ b^2& (c+a{)}^2& b^2\\ c^2& c^2& (a+b{)}^2\end{array}\right|$

$=\left|\begin{array}{ccc}(b+c{)}^2& a^2-(b+c{)}^2& a^2-(b+c{)}^2\\ b^2& (c+a{)}^2-b^2& b^2-b^2\\ c^2& c^2-c^2& (a+b{)}^2-c^2\end{array}\right|\left[\begin{array}{c}{c}_2{}^{\prime }=c_2-c_1\\ c_3^{\prime }=c_3-c_1\end{array}\right]$

$=\left|\begin{array}{ccc}(b+c{)}^2& (a+b+c)(a-b-c)& (a+b+c)(a-b-c)\\ b^2& (a+b+c)(c+a-b)& 0\\ c^2& 0& (a+b+c)(a+b-c)\end{array}\right|$

$=(a+b+c{)}^2\left|\begin{array}{ccc}{b}^2+2bc+c^2& a-b-c& a-b-c\\ b^2& c+a-b& 0\\ c^2& 0& a+b-c\end{array}\right|$

$=(a+b+c{)}^2\left|\begin{array}{ccc}2bc& -2c& -2b\\ b^2& c+a-b& 0\\ c^2& 0& a+b-c\end{array}\right|[r_1^{\prime }=r_1-r_2-r_3]$

$=2(a+b+c{)}^2\left|\begin{array}{ccc}bc& -c& -b\\ b^2& c+a-b& 0\\ c^2& 0& a+b-c\end{array}\right|$

$=2(a+b+c{)}^2\left|\begin{array}{ccc}bc& 0& 0\\ b^2& c+a& \frac{b^2}{c}\\ c^2& \frac{c^2}{b}& a+b\end{array}\right|\left[\begin{array}{c}{c}_2{}^{\prime }=c_2+\frac{1}{b}{c}_1\\ c_3^{\prime }=c_3+\frac{1}{c}{c}_1\end{array}\right]$

$=\frac{2}{bc}(a+b+c{)}^2\left|\begin{array}{ccc}bc& 0& 0\\ b^2& bc+ab& b^2\\ c^2& c^2& ac+bc\end{array}\right|$

$=\frac{2}{bc}(a+b+c{)}^{2}bc\left|\overline{)\begin{array}{ccc}bc& 0& 0\\ b& c+a& b\\ c& c& a+b\end{array}}\right|$

$=2(a+b+c{)}^2\left\{bc(ca+bc+a^2+ab-bc)\right\}$

$=2(a+b+c{)}^2\left\{bc(a^2+ab+ca)\right\}$

$=2(a+b+c{)}^2[bc\left\{a\right(a+b+c\left)\right\}]$

$=2(a+b+c{)}^2\{abc(a+b+c)\}=2abc(a+b+c{)}^2$

$\therefore B=2abc(a+b+c{)}^3$ (প্রমাণিত)