সৃজনশীলঃ ‘দৃশ্যকল্প-১: B =(13−564−252−1);C=(−2−12−4−32−1−46)দৃশ্যকল্প-২:…’ ২০২৬ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

$A=B+C=\left[\begin{array}{ccc}1& 3& -5\\ 6& 4& -2\\ 5& 2& -1\end{array}\right]+\left[\begin{array}{ccc}-2& -1& 2\\ -4& -3& 2\\ -1& -4& 6\end{array}\right]=\left[\begin{array}{ccc}-1& 2& -3\\ 2& 1& 0\\ 4& -2& 5\end{array}\right]$

$\therefore |A|=\left|\begin{array}{ccc}-1& 2& -3\\ 2& 1& 0\\ 4& -2& 5\end{array}\right|=-1(5+0)-2(10-0)+(-3)(-4-4)$

$=-5-20+24=-1\ne 0$

$\therefore A^{-1}$ নির্ণয়যোগ্য।

$\therefore \mathrm{adj}A=\left[\begin{array}{ccc}\left|\begin{array}{cc}1& 0\\ -2& 5\end{array}\right|& -\left|\begin{array}{cc}2& 0\\ 4& 5\end{array}\right|& \left|\begin{array}{cc}2& 1\\ 4& -2\end{array}\right|\\ -\left|\begin{array}{cc}2& -3\\ -2& 5\end{array}\right|& \left|\begin{array}{cc}-1& -3\\ 4& 5\end{array}\right|& -\left|\begin{array}{cc}-1& 2\\ 4& -2\end{array}\right|\\ \left|\begin{array}{cc}2& -3\\ 1& 0\end{array}\right|& -\left|\begin{array}{cc}-1& -3\\ 2& 0\end{array}\right|& \left|\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right|\end{array}\right]$

$={\left[\begin{array}{ccc}5& -10& -8\\ -4& 7& 6\\ 3& -6& -5\end{array}\right]}^t=\left[\begin{array}{ccc}5& -4& 3\\ -10& 7& -6\\ -8& 6& -5\end{array}\right]$

$\therefore A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A=\frac{1}{-1}\left[\begin{array}{ccc}5& -4& 3\\ -10& 7& -6\\ -8& 6& -5\end{array}\right]$

$=\left[\begin{array}{ccc}-5& 4& -3\\ 10& -7& 6\\ 8& -6& 5\end{array}\right]$ (Ans.)

দৃশ্যকল্প অনুসারে, $\frac{2}{7}x+\frac{3}{7}y-\frac{5}{7}z=1$

$\frac{3x}{5}-\frac{y}{5}-\frac{2z}{5}=1$

$\therefore 3x-y-2z=5$

(i), (ii) ও (iii) নং হতে $x,y$ ও $z$ এর সহগগুচ্ছ নিয়ে গঠিত নির্ণায়ক,

$D=\left|\begin{array}{ccc}2& 3& -5\\ 1& -4& 1\\ 3& -1& -2\end{array}\right|=2(8+1)-3(-2-3)-5(-1+12)$

$=18+15-55=-22\ne 0$

$D_x=\left|\begin{array}{ccc}7& 3& -5\\ 4& -4& 1\\ 5& -1& -2\end{array}\right|=7(8+1)-3(-8-5)-5(-4+20)$

$=63+39-80=22$

$D_y=\left|\begin{array}{ccc}2& 7& -5\\ 1& 4& 1\\ 3& 5& -2\end{array}\right|=2(-8-5)-7(-2-3)-5(5-12)$

$=-26+35+35=44$

$D_z=\left|\begin{array}{ccc}2& 3& 7\\ 1& -4& 4\\ 3& -1& 5\end{array}\right|=2(-20+4)-3(5-12)+7(-1+12)$

$=-32+21+77=66$

$\therefore x=\frac{D_x}{D}=\frac{22}{-22}=-1$

$y=\frac{D_y}{D}=\frac{44}{-22}=-2,$

$z=\frac{D_z}{D}=\frac{66}{-22}=-3$

$\therefore$ নির্ণেয় সমাধান: $(x,y,z)=(-1,-2,-3)$ (Ans.)