সৃজনশীলঃ ‘A=[1−236],B=[3721],C=[0128]M=[x+2y+3z2x+y+4z3x+2y+z]এবং N=[−123]…’ চট্টগ্রাম বোর্ড ২০১৯ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$A=\left[\begin{array}{cc}1& -2\\ 3& 6\end{array}\right],B=\left[\begin{array}{cc}3& 7\\ 2& 1\end{array}\right],C=\left[\begin{array}{cc}0& 1\\ 2& 8\end{array}\right]$

$\therefore AB=\left[\begin{array}{cc}1& -2\\ 3& 6\end{array}\right]\left[\begin{array}{cc}3& 7\\ 2& 1\end{array}\right]$

$=\left[\begin{array}{cc}3-4& 7-2\\ 9+12& 21+6\end{array}\right]=\left[\begin{array}{cc}-1& 5\\ 21& 27\end{array}\right]$

$B^t={\left[\begin{array}{cc}3& 7\\ 2& 1\end{array}\right]}^t=\left[\begin{array}{cc}3& 2\\ 7& 1\end{array}\right]$

এবং $2C=2\left[\begin{array}{cc}0& 1\\ 2& 8\end{array}\right]=\left[\begin{array}{cc}0& 2\\ 4& 16\end{array}\right]$

$\therefore AB+B^t-2C=\left[\begin{array}{cc}-1& 5\\ 21& 27\end{array}\right]+\left[\begin{array}{cc}3& 2\\ 7& 1\end{array}\right]-\left[\begin{array}{cc}0& 2\\ 4& 16\end{array}\right]$

$=\left[\begin{array}{cc}-1+3-0& 5+2-2\\ 21+7-4& 27+1-16\end{array}\right]=\left[\begin{array}{cc}2& 5\\ 24& 12\end{array}\right]$

দেওয়া আছে,

$M=\left[\begin{array}{ccc}x+2y+3z& 2x+y+4z& 3x+2y+z\end{array}\right]$

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

$M=N$ হলে,

$[x+2y+3z2x+y+4z3x+2y+z]=\left[\begin{array}{ccc}-1& 2& 3\end{array}\right]$

$\therefore x+2y+3z$

$2x+y+4z$

$3x+2y+z$

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

$=1(1-8)-2(2-12)+3(4-3)$

$=-7+20+3=16$

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

$=-1(1-8)-2(2-12)+3(4-3)$

$=7+20+3=30$

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

$=1(2-12)+1(2-12)+3(6-6)$

$=-10-10+0=-20$

এবং $D_z=\left|\begin{array}{ccc}1& 2& -1\\ 2& 1& 2\\ 3& 2& 3\end{array}\right|$

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

$=-1-0-1=-2$

$\therefore x=\frac{D_x}{D}=\frac{30}{16}=\frac{15}{8},y=\frac{D_y}{D}=\frac{-20}{16}=-\frac{5}{4}$

এবং $z=\frac{D_z}{D}=\frac{-2}{16}=-\frac{1}{8}$

নির্ণেয় সমাধান, $(x,y,z)=(\frac{15}{8},-\frac{5}{4},-\frac{1}{8})$