সৃজনশীলঃ ‘দৃশ্যকল্প-১: x5+3y10+z10=x4+y4=3y7+4z7=1.দৃশ্যকল্প-২ :…’ ঢাকা বোর্ড ২০২১ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $\frac{x}{5}+\frac{3y}{10}+\frac{7}{10}=1$

বা, $2x+3y+z=10..........\left(i\right)$

এবং $\frac{x}{4}+\frac{y}{4}=1$

বা, $x+y=4$

বা, $x+y+0.z=4.........\left(ii\right)$

এবং, $\frac{3y}{7}+\frac{4z}{7}=1$

বা, $3y+4z=7$

বা, $0\cdot x+3y+4z=7..........\left(iii\right)$

সমীকরণ (i), (ii) ও (iii) হতে পাই,

$D=\left|\begin{array}{ccc}2& 3& 1\\ 1& 1& 0\\ 0& 3& 4\end{array}\right|$

$=2(4-0)-3(4-0)+1(3-0)$

$=8-12+3=-1$

$D_x=\left|\begin{array}{ccc}10& 3& 1\\ 4& 1& 0\\ 7& 3& 4\end{array}\right|$

$=10(4-0)-3(16-0)+1(12-7)=40-48+5=-3$

$D_y=\left|\begin{array}{ccc}2& 10& 1\\ 1& 4& 0\\ 0& 7& 4\end{array}\right|$

$=2(16-0)-10(4-0)+1(7-0)=32-40+7=-1$

$D_z=\left|\begin{array}{ccc}2& 3& 10\\ 1& 1& 4\\ 0& 3& 7\end{array}\right|$

$=2(7-12)-3(7-0)+10(3-0)=-10-21+30=-1$

$\therefore x=\frac{D_x}{D}=\frac{-3}{-1}=3;y=\frac{D_y}{D}=\frac{-1}{-1}=1;z=\frac{D_z}{D}=\frac{-1}{-1}=1$

নির্ণেয় সমাধান, $(x,y,z)=(3,1,1)$.

দেওয়া আছে,

$\Delta =\left|\begin{array}{ccc}(s-x{)}^2& x^2& x^2\\ y^2& (s-y{)}^2& y^2\\ z^2& z^2& (s-z{)}^2\end{array}\right|,s=x+y+z$

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

এবংপ্রয়োগকরে$=\left|\begin{array}{ccc}(y+z{)}^2& x^2-(y+z{)}^2& x^2-(y+z{)}^2\\ y^2& (x+z{)}^2-y^2& y^2-y^2\\ z^2& z^2-z^2& (x+y{)}^2-z^2\end{array}\right|\left[\begin{array}{c}{c}_2^{\prime }=c_2-c_1\text{ এবং }\\ c_3^{\prime }=c_3-c_1\text{ প্রয়োগ করে }\end{array}\right]$

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

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

$=(x+y+z{)}^2\left|\begin{array}{ccc}2yz& -2z& -2y\\ y^2& z+x-y& 0\\ z^2& 0& x+y-z\end{array}\right|[r_1=r_1-r_2-r_3$ প্রয়োগ করে $]$

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

এবংপ্রয়োগকরে$=2(x+y+z{)}^2\left|\begin{array}{ccc}\mathrm{yz}& 0& 0\\ y^2& z+x& \frac{y^2}{z}\\ z^2& \frac{z^2}{y}& x+y\end{array}\right|\mid \begin{array}{c}{c}_2^{\prime }=c_2+\frac{1}{y}{c}_1\text{ এবং }\\ c_3^{\prime }=c_3+\frac{1}{y}{c}_1\text{ প্রয়োগ করে }\end{array}]$

$=2(x+y+z{)}^{2}yz\left|\begin{array}{cc}z+x& y^2\\ \frac{z^2}{y}& x+y\end{array}\right|$

$=2(x+y+z{)}^{2}yz(zx+x^2+yz+xy-yz)$

$=2\mathrm{yz}(x+y+z{)}^2\cdot x(x+y+z)$

$=2xyz(x+y+z{)}^3=2xyz\cdot s^3$

$\therefore \Delta =2x^{2}y{z}^3$ (প্রমাণিত)