সৃজনশীলঃ ‘সমীকরণ জোট: px+qy+rz=5p2x+q2y+r2z=5(p3−1)x+(q3−1)y+(r3−1)z=−5…’ সিলেট বোর্ড ২০২১ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

সমীকরণণ জোটঃ $px+qy+rz=5$

$p^{2}x+q^{2}y+r^{2}z=5$

$(p^3-1)x+(q^3-1)y+(r^3-1)z=-5$

$p,q,r$এর মান বসিয়ে পাই,

x+2 y+3 z=5 ......... (i)

x+4 y+9 z=5 ......... (ii)

7 y+26 z=-5 ......... (iii)

প্রদত্ত সমীকরণগুলোর $x,y,z$ এর সহগগুচছ নিয়ে পাই,

$\begin{array}{cc}& D=\left[\begin{array}{ccc}1& 2& 3\\ 1& 4& 9\\ 0& 7& 26\end{array}\right]\\ & |D|=1(104-63)-2(26-0)+3(7-0)\\ & =41-52+21=10\end{array}$

$\begin{array}{cc}& D_x=\left[\begin{array}{ccc}5& 2& 3\\ 5& 4& 9\\ -5& 7& 26\end{array}\right]\\ & =5(104-63)-2(130+45)+3(35+20)\\ & =205-350+165=20\end{array}$

$\begin{array}{cc}& D_y=\left[\begin{array}{ccc}1& 5& 3\\ 1& 5& 9\\ 0& -5& 26\end{array}\right]\\ & =1(130+45)-5(26-0)+3(-5-0)\\ & =175-130-15=30\end{array}$

$\begin{array}{cc}& D_z=\left[\begin{array}{ccc}1& 2& 5\\ 1& 4& 5\\ 0& 7& -5\end{array}\right]\\ & =1(-20-35)-2(-5-0)+5(7-0)\\ & =-55+10+35\\ & =-10\end{array}$

$\begin{array}{cc}\therefore x& =\frac{Dx}{D}\\ & =\frac{20}{10}\\ & =2\end{array}$

$\begin{array}{cc}& y=\frac{Dy}{D}\\ & =\frac{30}{10}\\ & =3\end{array}$

$\begin{array}{cc}& Z=\frac{Dz}{D}\\ & =\frac{-10}{10}\\ & =-1\end{array}$

$\therefore (x,y,z)=(2,3,-1)$

ইহাই নির্ণেয় সমাধান।।

$D=\left[\begin{array}{ccc}p& q& r\\ p^2& q^2& r^2\\ p^3-1& q^3-1& r^3-1\end{array}\right]$

$=\left[\begin{array}{ccc}p& q& r\\ p^2& q^2& r^2\\ p^3& q^3& r^3\end{array}\right]-\left[\begin{array}{ccc}p& q& r\\ p^2& q^2& r^2\\ 1& 1& 1\end{array}\right]$

$=\left(pqr\right)\left[\begin{array}{ccc}1& 1& 1\\ p& q& r\\ p^2& q^2& r^2\end{array}\right]+\left[\begin{array}{ccc}p& q& r\\ 1& 1& 1\\ p^2& q^2& r^2\end{array}\right]$

$=\left(pqr\right)\left[\begin{array}{ccc}1& 1& 1\\ p& q& r\\ p^2& q^2& r^2\end{array}\right]-\left[\begin{array}{ccc}1& 1& 1\\ p& q& r\\ p^2& q^2& r^2\end{array}\right]$

$=(pqr-1)\left[\begin{array}{ccc}1& 1& 1\\ p& q& r\\ p^2& q^2& r^2\end{array}\right]$

$=(pqr-1)\left[\begin{array}{ccc}0& 0& 1\\ p-q& q-r& r\\ p^2-q^2& q^2-r^2& r^2\end{array}\right]$

$=(pqr-1)\left[\begin{array}{cc}p-q& q-r\\ (p+q)(p-q)& (q+r)(q-r)\end{array}\right]$

$=(pqr-1)(p-q)(q-r)\left[\begin{array}{cc}1& 1\\ p+q& q+r\end{array}\right]$

$=(pqr-1)(p-q)(q-r)(q+r-p-q)$

$=(pqr-1)(p-q)(q-r)(r-p)$

$D=(pqr-1)(p-q)(q-r)(r-p)$ [Proved]