সৃজনশীলঃ ‘দৃশ্যকল্প-১ : P=[1+x2−y22xy2y2xy1−x2+y2−2x−2y2x1−x2−y2]…’ ২০২৪ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দৃশ্যকল্প-১ হতে পাই,

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

এখন,

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

$=\left|\begin{array}{ccc}1+x^2-y^2+2y^2& 2xy& 1-x^2-y^2\\ 2xy-2xy& 1-x^2+y^2& 2y\\ -2y+y-x^{2}y-y^3& 2x& 1-x^2-y^2\end{array}\right|$

$[c_1^{\prime }=c_1+y\times c_3$ প্রয়োগ করে]

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

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

$=(1+x^2+y^2)\left|\begin{array}{ccc}1& 2xy-2xy& 2y\\ 0& 1-x^2+y^2+2x^2& -2x\\ -y& 2x-x+x^3+x{y}^2& 1-x^2-y^2\end{array}\right|$

$c_2^{\prime }=c_2-x\times c_3$ প্রয়োগ করে]

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

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

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

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

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

দেওয়া আছে,

$det\left(P\right)=0$

$\Rightarrow {(1+x^2+y^2)}^3=0$

$\Rightarrow 1+x^2+y^2=0$

$\therefore x^2+y^2=-1$ (প্রমাণিত)

দৃশ্যকল্প-২ হতে পাই,

$R={\left[r_{ij}\right]}_{3\times 3}$

$=\left[\begin{array}{ccc}{r}_{11}& r_{12}& r_{13}\\ r_{21}& r_{22}& r_{23}\\ r_{31}& r_{32}& r_{33}\end{array}\right]$

এখন, $r_{ij}=i+2j$

$\therefore r_{11}=1+2\cdot 1=3$

$r_{12}=1+2\cdot 2=5$

$r_{13}=1+2\cdot 3=7$

$r_{21}=2+2\cdot 1=4$

$r_{22}=2+2\cdot 2=6$

$r_{23}=2+2\cdot 3=8$

$r_{31}=3+2\cdot 1=5$

$r_{32}=3+2\cdot 2=87$

$r_{33}=3+2\cdot 3=9$

$\therefore R=\left[\begin{array}{ccc}3& 5& 7\\ 4& 6& 8\\ 5& 7& 9\end{array}\right]$

$\therefore R^T=\left[\begin{array}{ccc}3& 4& 5\\ 5& 6& 7\\ 7& 8& 9\end{array}\right]$

আবার,

$S=\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 3& 1& 2\end{array}\right]$

$\therefore |S|=\left|\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 3& 1& 2\end{array}\right|$

$=1(6-1)-2(4-3)+3(2-9)$

$=5-2-21$

$=-18$

|s| -এর সহ গুণক গুলো নিম্নরূপ :

$S_{11}=(6-1)=5,S_{12}=-(4-3)=-1;S_{13}=(2-9)=-7$

$S_{21}=-(4-3)=-1,S_{22}=(2-9)=-7,S_{23}=-(1-6)=5$

$S_{31}=(2-9)=-7,S_{32}=-(1-6)=5,S_{33}=(3-4)=-1$

$\therefore \text{ Adj }S={\left[\begin{array}{ccc}5& -1& -7\\ -1& -7& 5\\ -7& 5& -1\end{array}\right]}^T=\left[\begin{array}{ccc}5& -1& -7\\ -1& -7& 5\\ -7& 5& -1\end{array}\right]$

$\therefore S^{-1}=\frac{\text{ Adj }S}{\mid Aj}=\frac{1}{-18}\left[\begin{array}{ccc}5& -1& -7\\ -1& -7& 5\\ -7& 5& -1\end{array}\right]$

$\therefore R^T+54S^{-1}=\left[\begin{array}{ccc}3& 4& 5\\ 5& 6& 7\\ 7& 8& 9\end{array}\right]+54\cdot \left(\frac{1}{-18}\right)\left[\begin{array}{ccc}5& -1& -7\\ -1& -7& 5\\ -7& 5& -1\end{array}\right]$

$=\left[\begin{array}{ccc}3& 4& 5\\ 5& 6& 7\\ 7& 8& 9\end{array}\right]+(-3)\left[\begin{array}{ccc}5& -1& -7\\ -1& -7& 5\\ 7& 5& -1\end{array}\right]$

$=\left[\begin{array}{ccc}3& 4& 5\\ 5& 6& 7\\ 7& 8& 9\end{array}\right]+\left[\begin{array}{ccc}-15& 3& 21\\ 3& 21& -15\\ 21& -15& 3\end{array}\right]$

$=\left[\begin{array}{ccc}3-15& 4+3& 5+21\\ 5+3& 6+21& 7-15\\ 7+21& 8-15& 9+3\end{array}\right]$

$=\left[\begin{array}{ccc}-12& 7& 26\\ 8& 27& -8\\ 28& -7& 12\end{array}\right]$