সৃজনশীলঃ ‘A=[1213−3−121−1] এবং B=[xyz]…’ ২০২৪ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $A_{\overline{T}}=\left[\begin{array}{ccc}1& 2& 1\\ 3& -3& -1\\ 2& 1& -1\end{array}\right]$

$\therefore$|A| =1(3+1)-2(-3+2)+1(3+6) 

$=4-2(-1)+9$

$=4+2+9$

$=15$

$\begin{array}{cc}& A_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}-3& -1\\ 1& -1\end{array}\right|=4\\ & A_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}3& -1\\ 2& -1\end{array}\right|=-(-3+2)=1\\ & A_{12}=(-1{)}^{1+3}\left|\begin{array}{cc}3& -3\\ 2& 1\end{array}\right|=3+6=9\\ & A_{21}=(-1{)}^{1+2}\left|\begin{array}{cc}2& 1\\ 1& -1\end{array}\right|=-(-2-1)=3\\ & A_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& 1\\ 2& -1\end{array}\right|=-1-2=-3\\ & A_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& 2\\ 2& 1\end{array}\right|=-(1-4)=3\\ & A_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}2& 1\\ -3& -1\end{array}\right|=-2+3=1\\ & A_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& 1\\ 3& -1\end{array}\right|=-(-1-3)=4\\ & A_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& 2\\ 3& -3\end{array}\right|=-3-6=-9\end{array}$

$\begin{array}{cc}{A}^{-1}& =\frac{1}{\left|A\right|}\mathrm{adj}A\\ & =\frac{1}{15}{\left[\begin{array}{ccc}4& 1& 9\\ 3& -3& 3\\ 1& 4& -9\end{array}\right]}^{\top }\\ & =\frac{1}{15}\left[\begin{array}{ccc}4& 3& 1\\ 1& -3& 9\\ 9& 3& -9\end{array}\right]\end{array}$

দেওয়া আছে, $AB=\left[\begin{array}{c}1\\ 3\\ 2\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}1& 2& 1\\ 3& -3& -1\\ 2& 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 3\\ 2\end{array}\right]$

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

x, y, z এর সহগ দ্বারা গঠিত ম্যাট্রিক্স D হলে,

$|D|=\left|\begin{array}{ccc}1& 2& 1\\ 3& -3& -1\\ 2& 1& -1\end{array}\right|=15$

এখন,

$D_x=\left|\begin{array}{ccc}1& 2& 1\\ 3& -3& -1\\ 2& 1& -1\end{array}\right|=1(3+1)-2(-3+2)+1(3+6)$

$=15$

$D_y=\left|\begin{array}{ccc}1& 1& 1\\ 3& 3& -1\\ 2& 2& -1\end{array}\right|=1(-3+2)-1(-3+2)+1(6-6)$

$=-1+1+10$

= 0

$D_z=\left|\begin{array}{ccc}31& 2& 1\\ 3& -3& 3\\ 2& 1& 2\end{array}\right|=1(-6-3)-2(6-6)+1(3+6)$

= 0

$\begin{array}{cc}\therefore x& =\frac{D_x}{\mid D}=\frac{15}{15}=1\\ y& =\frac{D_y}{\left|D\right|}=\frac{0}{15}=0\\ z& =\frac{D_z}{\left|D\right|}=\frac{0}{15}=0\end{array}$

$\therefore \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

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