সৃজনশীলঃ ‘নিচের উদ্দীপকটি লক্ষ…’ ২০২৬ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $A=\left[\begin{array}{ccc}x& y& z\\ x^2& y^2& z^2\\ x^3-5& y^3-5& z^3-5\end{array}\right]$

$\therefore |A|=\left|\begin{array}{ccc}x& y& z\\ x^2& y^2& z^2\\ x^3-5& y^2-5& z^3-5\end{array}\right|$

$=\left|\begin{array}{ccc}x& y& z\\ x^2& y^2& z^2\\ x^3& y^3& z^3\end{array}\right|+\left|\begin{array}{ccc}x& y& z\\ x^2& y^2& z^2\\ -5& -5& -5\end{array}\right|$

$=xyz\left|\begin{array}{ccc}1& 1& 1\\ x& y& z\\ x^2& y^2& z^2\end{array}\right|-5\left|\begin{array}{ccc}x& y& z\\ x^2& y^2& z^2\\ 1& 1& 1\end{array}\right|$

$=xyz\left|\begin{array}{ccc}1& 1& 1\\ x& y& z\\ x^2& y^2& z^2\end{array}\right|-5\left|\begin{array}{ccc}1& 1& 1\\ x& y& z\\ x^2& y^2& z^2\end{array}\right|$

$=(xyz-5)\left|\begin{array}{ccc}1& 1& 1\\ x& y& z\\ x^2& y^2& z^2\end{array}\right|$

$=(xyz-5)\left|\begin{array}{ccc}0& 0& 1\\ x-y& y-z& z\\ x^2-y^2& y^2-z^2& z^2\end{array}\right|$

$=(\mathrm{xyz}-5)1\left\{\right(x-y)(y^2-z^2)-(y-z\left)(x^2-y^2)\right\}$

$=(\mathrm{xyz}-5)\left\{\right(x-y\left)\right(y+z\left)\right(y-z)-(y-z\left)\right(x+y\left)\right(x-y\left)\right\}$

$=\left(\mathrm{xyz}\right)-5(x-y)(y-z)(y+z-x-y)$

$=(xyz-5)(x-y)(y-z)(z-x)$

= ডানপক্ষ

$\therefore |A|=(\mathrm{xyz}-5)(x-y)(y-z)(z-x).$ (প্রমাণিত)

দেওয়া আছে, $A=\left[\begin{array}{ccc}x& y& z\\ x^2& y^2& z^2\\ x^3-5& y^3-5& z^3-5\end{array}\right]$

$x=1,y=2,z=-1$ হলে,

$A=\left[\begin{array}{ccc}1& 2& -1\\ 1& 4& 1\\ 1-5& 8-5& -1-5\end{array}\right]=\left[\begin{array}{ccc}1& 2& -1\\ 1& 4& 1\\ -4& 3& -6\end{array}\right]$

$\therefore |A|=\left|\begin{array}{ccc}1& 2& -1\\ 1& 4& 1\\ -4& 3& -6\end{array}\right|$

$=1(-24-3)+2(-4+6)-1(3+16)$

$=-27+4-19$

$=4-46=-42\ne 0$

$\therefore A^{-1}$ বিদ্যমান।

A এর সহগুণকসমূহ :

$A_{11}=\left|\begin{array}{cc}4& 1\\ 3& -6\end{array}\right|$

$=-24-3=-27$

$A_{12}=-\left|\begin{array}{cc}1& 1\\ -4& -6\end{array}\right|$

$=-(-6+4)=2$

$A_{13}=\left|\begin{array}{cc}1& 4\\ -4& 3\end{array}\right|$

$=3+16=19$

$A_{21}=-\left|\begin{array}{cc}2& -1\\ 3& -6\end{array}\right|$

$=-(-12+3)=9$

$A_{22}=\left|\begin{array}{cc}1& -1\\ -4& -6\end{array}\right|$

$=-6-4=-10$

$A_{23}=-\left|\begin{array}{cc}1& 2\\ -4& 3\end{array}\right|$

$=-(3+8)=-11$

$A_{31}=\left|\begin{array}{cc}2& -1\\ 4& 1\end{array}\right|$

$=2+4=6$

$A_{32}=-\left|\begin{array}{cc}1& -1\\ 1& 1\end{array}\right|$

$=-(1+1)=-2$

$A_{33}=\left|\begin{array}{cc}1& 2\\ 1& 4\end{array}\right|$

$=4-2=2$

$\mathrm{Adj}A={\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}^T$

$={\left[\begin{array}{ccc}-27& 2& 19\\ 9& -10& -11\\ 6& -2& 2\end{array}\right]}^T$

$=\left[\begin{array}{ccc}-27& 9& 6\\ 2& -10& -2\\ 19& -11& 2\end{array}\right]$

$A^{-1}=\frac{1}{\left|A\right|}\mathrm{Adj}A$

$=\frac{1}{-42}\left[\begin{array}{ccc}-27& 9& 6\\ 2& -10& -2\\ 19& -11& 2\end{array}\right]$

$=\frac{1}{42}\left[\begin{array}{ccc}27& -9& -6\\ -2& 10& 2\\ -19& 11& -2\end{array}\right]$

নির্ণেয় $A^{-1}=\frac{1}{42}\left[\begin{array}{ccc}27& -9& -6\\ -2& 10& 2\\ -19& 11& -2\end{array}\right]$.