সৃজনশীলঃ ‘A=|3−42−210−1−11|,B=|x2−2y5−4z7−5|…’ সিলেট বোর্ড ২০২২ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে,

$A=|\begin{array}{ccc}3& -4& 2\\ -2& 1& 0\\ -1& -1& 1\end{array}\mid |$

$|A|=\left|\begin{array}{ccc}-1& -1& 1\\ -2& 1& 0\\ -1& -1& 1\end{array}\right|$

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

= 3 - 8 + 8

= 1

$\therefore A^{-1}$ নির্ণয়যোগ্য

$A_{11}=\left|\begin{array}{cc}1& 0\\ -1& 1\end{array}\right|$

$=1+0$

$=1$

$A_{12}=-\left|\begin{array}{cc}-2& 0\\ -1& 1\end{array}\right|$

$=-(-2+0)$

$=2$

$A_{13}=\left|\begin{array}{cc}-2& 1\\ -1& -1\end{array}\right|$

$=2+1$

$=3$

$\begin{array}{cc}& A_{21}=-\left|\begin{array}{cc}-4& 2\\ -1& 1\end{array}\right|A_{21}=-\left|\begin{array}{cc}-4& 2\\ -1& 1\end{array}\right|\\ & =-(-4+2)\\ & =2\\ & A_{22}=\left|\begin{array}{cc}3& 2\\ -1& 1\end{array}\right|\\ & =3+2\\ & =5\\ & A_{z3}=-\left|\begin{array}{cc}3& -4\\ -1& -1\end{array}\right|\\ & =-(-3-4)\\ & 7\\ & A_{31}=\left|\begin{array}{cc}-4& 2\\ 1& 0\end{array}\right|\\ & =0-2\\ & =-2\\ & A_{32}=\left|\begin{array}{cc}3& 2\\ -2& 0\end{array}\right|\\ & =-(0+4)\\ & =-4\\ & A_{33}=\left|\begin{array}{cc}3& -4\\ & 16\\ -2& 1\end{array}\right|\\ & =3-8\\ & =-5\\ & \therefore \text{ Adj }A={\left[\begin{array}{ccc}1& 2& 3\\ 2& 5& 7\\ -2& -4& -5\end{array}\right]}^{\top }\\ & =\left[\begin{array}{ccc}1& 2& -2\\ 2& 5& -4\\ 3& 7& -5\end{array}\right]\\ & \therefore A^{-1}=\frac{1}{\left|A\right|}\text{ Adj }A\\ & =\frac{1}{1}\left[\begin{array}{ccc}1& 2& -2\\ 2& 5& -4\\ 3& 7& -5\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 2& -2\\ 2& 5& -4\\ 3& 7& -5\end{array}\right]\\ & \end{array}$

$\begin{array}{cc}& \therefore \left[\begin{array}{ccc}3& -4& 2\\ -2& 1& 0\\ -1& -1& 1\end{array}\right]\left[\begin{array}{ccc}x& 2& -2\\ y& 5& -4\\ z& 7& -5\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ & =\left[\begin{array}{ccc}3x-4y+2z& 2\cdot 3+(-4)\cdot 5+2\cdot 7& 3(-2)+(-4)\cdot (-4)+2(-5)\\ -2x+y+0\cdot z& -2\cdot 2+1\cdot 5+0\cdot 7& -2(-2)+(-4)\cdot 1+0(-5)\\ -x-y+z& -1\cdot 2+(-1)\cdot 5+1\cdot 7& -1(-2)+(-1)\cdot (-4)+\left(1\right)(-5)\end{array}\right]\\ & =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ & =\left[\begin{array}{ccc}3x-4y+zz& 0& 0\\ -2x+y+0.z& 1& 0\\ -x-y+z& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array}$

ম্যাট্রিক্সর সমতা অনুসারে,

$\begin{array}{cc}3x-4y+2z& =1\\ -2x+y+0.z& =0\\ -x-y+z& =0\end{array}$

x, y, z এর সহগ নিয়ে গঠিত নির্ণায়ক

$\begin{array}{cc}& D=\left|\begin{array}{ccc}3& -4& 2\\ -2& 1& 0\\ -1& -1& 1\end{array}\right|\\ & =3(1+0)+4(-2+0)+2(2+1)\\ & =3-8+6\\ & =1\\ & D_x=\left|\begin{array}{ccc}1& -4& 2\\ 0& 1& 0\\ 0& -1& 1\end{array}\right|=1\left|\begin{array}{cc}1& 0\\ -1& 1\end{array}\right|\\ & =1(1+0)\\ & =1\\ & \text{ Dy }=\left|\begin{array}{ccc}3& 1& 2\\ -2& 0& 0\\ -1& 0& 1\end{array}\right|\\ & =3(0-0)-1(-2-0)+2(0-0)\\ & =0+2+0=2\\ & Dz=\left|\begin{array}{ccc}3& -4& 1\\ -2& 1& 0\\ -1& -1& 0\end{array}\right|\\ & =3(0-0)+4(0-0)+1(2+1)\\ & =3\\ & \therefore x=\frac{\mathrm{Dx}}{x}=\frac{1}{1}=1\\ & \therefore y=\frac{Dy}{y}=\frac{2}{7}=2\\ & \therefore z=\frac{\mathrm{Dz}}{z}=\frac{3}{1}=3\\ & \therefore (x,y,z)=(1,2,3)\\ & \end{array}$