সৃজনশীলঃ ‘দৃশ্যকল্প: A=[1+x2−y22xy−2y2xy1−x2+y22x2y−2×1−x2−y2]…’ ২০২৪ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,
$P=\left[\begin{array}{ccc}1+x^2-y^2& 2xy& 2y\\ 2xy& 1-x^2+y^2& -2x\\ -2y& 2x& 1-x^2-y^2\end{array}\right]$

এবং $\det \left(P\right)=0$

$\det \left(P\right)=\left|\begin{array}{ccc}1+x^2-y^2& 2xy& 2y\\ 2xy& 1-x^2+y^2& -2x\\ -2y,& 2x& 1-x^2-y^2\end{array}\right|$

$=\left|\begin{array}{ccc}1+x^2+y^2& 0& 2y\\ 0& 1+x^2+y^2& -2x\\ -y(1+x^2+y^2)& x(1+x^2+y^2)& 1-x^2-y^2\end{array}\right|$

$[C_1^{\prime }=C_1+C_{3}y$ এবং $C_2^{\prime }=C_2-C_{3}x]$

$={(1+x^2+y^2)}^2\left|\begin{array}{ccc}1& 0& 2y\\ 0& 1& -2x\\ -y& x& 1-x^2-y^2\end{array}\right|$

$={(1+x^2+y^2)}^2\{1-x^2-y^2+2x^2-0+2y(0+y\left)\right\}$

$={(1+x^2+y^2)}^2(1+x^2+y^2)={(1+x^2+y^2)}^3$

প্রশ্নমতে, ${(1+x^2+y^2)}^3=0$

বা, $1+x^2+y^2=0\therefore x^2+y^2=-1$ (প্রমাণিত)

দেওয়া আছে, $A=\left[\begin{array}{ccc}1+x^2-y^2& 2xy& -2y\\ 2xy& 1-x^2+y^2& 2x\\ 2y& -2x& 1-x^2-y^2\end{array}\right]$

$x=y=2$ হলে,

$A=\left[\begin{array}{ccc}1+2^2-2^2& 2.2.2& -2.2\\ 2.2.2& 1-2^2+2^2& 2.2\\ 2.2& -2.2& 1-2^2-2^2\end{array}\right]$

$=\left[\begin{array}{ccc}1& 8& -4\\ 8& 1& 4\\ 4& -4& -7\end{array}\right]$

এখন, $|A|=1.(-7+16)-8(-56-16)-4(-32-4)$

$=9+576+144=729\ne 0.$

$\therefore A^{-1}$নির্ণয়যোগ্য।

$\therefore \mathrm{adj}A={\left[\begin{array}{ccc}\left|\begin{array}{cc}1& 4\\ -4& -7\end{array}\right|& -\left|\begin{array}{cc}8& 4\\ 4& -7\end{array}\right|& \left|\begin{array}{cc}8& 1\\ 4& -4\end{array}\right|\\ -\left|\begin{array}{cc}8& -4\\ -4& -7\end{array}\right|& \left|\begin{array}{cc}1& -4\\ 4& -7\end{array}\right|& -\left|\begin{array}{cc}1& 8\\ 4& -4\end{array}\right|\\ \left|\begin{array}{cc}8& -4\\ 1& 4\end{array}\right|& -\left|\begin{array}{cc}1& -4\\ 8& 4\end{array}\right|& |\begin{array}{cc}1& 8\\ 8& 1\end{array}\end{array}\right]}^t$

$={\left[\begin{array}{ccc}9& 72& -36\\ 72& 9& 36\\ 36& -36& -63\end{array}\right]}^t=\left[\begin{array}{ccc}9& 72& 36\\ 72& 9& -36\\ -36& 36& -63\end{array}\right]$

$\therefore A^{-1}=\frac{1}{\left|A\right|}$ adj $A=\frac{1}{729}\left[\begin{array}{ccc}9& 72& 36\\ 72& 9& -36\\ -36& 36& -63\end{array}\right]$

এখানে, $|A|A^{-1}=729\cdot \frac{1}{729}\left[\begin{array}{ccc}9& 72& 36\\ 72& 9& -36\\ -36& 36& -63\end{array}\right]$

$=\left[\begin{array}{ccc}9& 72& 36\\ 72& 9& -36\\ -36& 36& -63\end{array}\right]=\mathrm{adj}\left(A\right)$

$\therefore |A|A^{-1}=\mathrm{adj}\left(A\right)$ (প্রমাণিত)