সৃজনশীলঃ ‘A=[1332−22311] এবং f(x)=x2−4x−3…’ ২০২৪ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

এখানে,

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

$A^T=\left[\begin{array}{ccc}1& 2& 3\\ 3& -2& 1\\ 3& 2& 1\end{array}\right]$

$\therefore \left|A^T\right|=1(-2-2)-2(3-3)+3(6+6)$

$=-4-0+36$

= 32

$A_{11}^T=(-1{)}^{1+1}\left|\begin{array}{cc}-2& 1\\ 2& 1\end{array}\right|=(-2-2)=-4$

$A_{12}^T=(-1{)}^{1+2}\left|\begin{array}{cc}3& 1\\ 3& 1\end{array}\right|=(-1\left)\right(3-3)=0$

$A_{13}^T=(-1{)}^{1+3}\left|\begin{array}{cc}3& -2\\ 3& 2\end{array}\right|=(6+6)=12$

$A_{21}^T=(-1{)}^{2+1}\left|\begin{array}{cc}2& 3\\ 2& 1\end{array}\right|=(-1\left)\right(2-6)=4$

$A_{22}^T=(-1{)}^{2+2}\left|\begin{array}{cc}1& 3\\ 3& 1\end{array}\right|=(1-9)=-8$

$A_{23}^T=(-1{)}^{2+3}\left|\begin{array}{cc}1& 2\\ 3& 2\end{array}\right|=(-1\left)\right(2-6)=4$

$A_{31}^T=(-1{)}^{3+1}\left|\begin{array}{cc}2& 3\\ -2& 1\end{array}\right|=(2+6)=8$

$A_{32}^T=(-1{)}^{3+2}\left|\begin{array}{cc}1& 3\\ 3& 1\end{array}\right|=(-1\left)\right(1-9)=8$

$A_{33}^T=(-1{)}^{3+3}\left|\begin{array}{cc}1& 2\\ 3& -2\end{array}\right|=-2-6=-8$

$\therefore Adj{A}^{\top }={\left[\begin{array}{ccc}-4& 0& 12\\ 4& -8& 4\\ 8& 8& -8\end{array}\right]}^t$

$=\left[\begin{array}{ccc}-4& 4& 8\\ 0& -8& 8\\ 12& 4& -8\end{array}\right]$

$\therefore {\left(A^T\right)}^{-1}=\frac{1}{\left|A^{\top }\right|}\left[\begin{array}{ccc}-4& 4& 8\\ 0& -8& 8\\ 12& 4& -8\end{array}\right]$

$=\frac{1}{32}\left[\begin{array}{ccc}-4& 4& 8\\ 0& -8& 8\\ 12& 4& -8\end{array}\right]=\frac{1}{8}\left[\begin{array}{ccc}-1& 1& 2\\ 0& -2& 2\\ 3& 1& -2\end{array}\right]$

(Ans)

দেওয়া আছে,

$f\left(x\right)=x^2-4x-3$

$\therefore f\left(A\right)=A^2-4A-3$

এখন, $A=\left[\begin{array}{ccc}1& 3& 3\\ 2& -2& 2\\ 3& 1& 1\end{array}\right]$

$\therefore A^2=A\cdot A$

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

$=\left[\begin{array}{ccc}1+6+9& 3-6+3& 3+6+3\\ 2-4+6& 6+4+2& 6-4+2\\ 3+2+3& 9-2+1& 9+2+1\end{array}\right]$

$=\left[\begin{array}{ccc}16& 0& 12\\ 4& 12& 4\\ 8& 8& 12\end{array}\right]$

$\therefore A^2-4A-3=\left[\begin{array}{ccc}16& 0& 12\\ 4& 12& 4\\ 8& 8& 12\end{array}\right]-\left[\begin{array}{ccc}4& 12& 12\\ 8& -8& 8\\ 12& 4& 4\end{array}\right]-\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]$

$=\left[\begin{array}{ccc}16-4-3& 0-12-0& 12-12-0\\ 4-8-0& 12+8-3& 4-8+0\\ 8-12-0& 8-4-0& 12-9-3\end{array}\right]$

$=\left[\begin{array}{ccc}9& -12& 0\\ -4& 17& -4\\ -4& 4& 5\end{array}\right]$

Ans.