সৃজনশীলঃ ‘P=[1+x2−y22xy2y2xy1−x2+y2−2x−2y2x1−x2−y2] এবং f(x)=x3−3x+2I…’ ২০২৬ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

প্রয়োগকরেপ্রয়োগকরে$\begin{array}{cc}& \det \left(P\right)=\left|\begin{array}{ccc}1+x^2-y^2& 2xy& 2y\\ 2xy& 1-x^2+y^2& -2x\\ -2y& 2x& 1-x^2-y^2\end{array}\right|\\ & =\left|\begin{array}{ccc}1+x^2+y^2& 2\mathrm{xy}& 2y\\ 0& 1-x^2+y^2& -2x\\ -y^{\prime }-x^{2}y-y^3& 2x& 1-x^2-y^2\end{array}\right|[c_1^{\prime }=c_1+{\mathrm{yc}}_3\text{ প্রয়োগ করে }]\\ & =(1+x^2+y^2)\left|\begin{array}{ccc}1& 2xy& 2y\\ 0& 1-x^2+y^2& -2x\\ -y& 2x& 1-x^2-y^2\end{array}\right|\\ & =(1+x^2+y^2)\left|\begin{array}{ccc}1& 0& 2y\\ 0& 1+x^2+y^2& -2x\\ -y& x+x^3+x^2& 1-x^2-y^2\end{array}\right|[c_2^{\prime }=c_2-x{c}_3\text{ [প্রয়োগ করে ] }\\ & ={(1+x^2+y^2)}^2\left|\begin{array}{ccc}1& 0& 2y\\ 0& 1& -2x\\ -y& x& 1-x^2-y^2\end{array}\right|\\ & ={(1+x^2+y^2)}^2\{1(1-x^2-y^2+2x^2)-0+2y(0+y\left)\right\}\\ & ={(1+x^2+y^2)}^2(1+x^2-y^2+2y^2)\\ & ={(1+x^2+y^2)}^2(1+x^2+y^2)\\ & ={(1+x^2+y^2)}^3\end{array}$

প্রশ্নমতে det $P=0$ হলে, ${(1+x^2+y^2)}^3=0$

বা, $1+x^2+y^2=0$

$\therefore x^2+y^2=-1$ (প্রমাণিত)

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

এখন, $x=1,y=2$ হলে,

$\begin{array}{cc}& P=\left[\begin{array}{ccc}1+1^2-2^2& 2\cdot 1\cdot 2& 2\cdot 2\\ 2\cdot 1\cdot 2& 1-1^2+2^2& -2\cdot 1\\ -2\cdot 2& 2\cdot 1& 1-1^2-2^2\end{array}\right]\\ & =\left[\begin{array}{ccc}-2& 4& 4\\ 4& 4& -2\\ -4& 2& -4\end{array}\right]\\ & \therefore P^2=P\cdot P=\left[\begin{array}{ccc}-2& 4& 4\\ 4& 4& -2\\ -4& 2& -4\end{array}\right]\left[\begin{array}{ccc}-2& 4& 4\\ 4& 4& -2\\ -4& 2& -4\end{array}\right]\\ & =\left[\begin{array}{ccc}-2(-2)+4\cdot 4+4(-4)& -2\cdot 4+4\cdot 4+4\cdot 2& (-2)\cdot 4+4(-2)+\left(4\right)(-4)\\ 4(-2)+4\cdot 4+(-2)(-4)& 4\cdot 4+4\cdot 4+(-2)\cdot 2& 4\cdot 4+4(-2)+(-2)\cdot (-4)\\ -4\cdot (-2)+2\cdot 4+(-4)(-4)& -4\cdot 4+2\cdot 4+(-4)\cdot 2& -4\cdot 4+2\cdot (-2)+(-4)\cdot (-4)\end{array}\right]\end{array}$

এখन$\begin{array}{cc}& =\left[\begin{array}{ccc}4& 16& -32\\ 16& 28& 16\\ 32& -16& -4\end{array}\right]\\ & P^3=P^2\cdot P\text{. }\\ & =\left[\begin{array}{ccc}4& 16& -32\\ 16& 28& 16\\ 32& -16& -4\end{array}\right]\left[\begin{array}{ccc}-2& 4& 4\\ 4& 4& -2\\ -4& 2& -4\end{array}\right]\\ & =\left[\begin{array}{ccc}4(-2)+16\cdot 4+(-32)\cdot (-4)& 4\cdot 4+16\cdot 4+(-32)\cdot 2& 4\cdot 4+16(-2)+(-32)(-4)\\ 16\cdot (-2)+28\cdot 4+16(-4)& 16+4+28\cdot 4+16\cdot 2& 16\cdot 4+22(-2)+16\cdot (-4)\\ 32(-2)+(-16)4+(-4)(-4)& 32\cdot 4+1-16)\cdot 4+(-4)\cdot 2& 32\cdot 4+(-16)\cdot (-2)+(-4)\end{array}\right]\\ & =\left[\begin{array}{ccc}184& 16& 112\\ 16& 208& -56\\ -112& 56& 176\end{array}\right]\\ & \text{ এখन, }f\left(x\right)=x^3-3x+2I\\ & \therefore f\left(P\right)=P^3-3P+2I\\ & =\left[\begin{array}{ccc}184& 16& 112\\ 16& 208& -56\\ -112& 56& 176\end{array}\right]-3\left[\begin{array}{ccc}-2& 4& 4\\ 4& 4& -2\\ -4& 2& -4\end{array}\right]+2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ & =\left[\begin{array}{ccc}184& 16& 112\\ 16& 208& -56\\ -112& 56& 176\end{array}\right]-\left[\begin{array}{ccc}-6& 12& 12\\ 12& 12& -6\\ -12& 6& -12\end{array}\right]+\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]\\ & =\left[\begin{array}{ccc}184+6+2& 16-12-0& 112-12-0\\ 16-12+0& 208-12+2& -56+6+0\\ -112+12+0& 56-6+0& 176+12+2\end{array}\right]\\ & =\left[\begin{array}{ccc}192& 4& 100\\ 4& 198& -50\\ -100& 50& 190\end{array}\right]\\ & \therefore f\left(P\right)=\left[\begin{array}{ccc}192& 4& 100\\ 4& 198& -50\\ -100& 50& 190\end{array}\right]\\ & \end{array}$