সৃজনশীলঃ ‘A=[12422343−5],f(x)=x2−x+3.…’ বরিশাল বোর্ড ২০২১ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

এখানে, $f\left(x\right)=x^2-x+3$

বা, $f\left(A\right)=A^2-A+3I$

$A^2=A\cdot A$

$\begin{array}{cc}& =\left[\begin{array}{ccc}1& 2& 4\\ 2& 2& 3\\ 4& 3& -5\end{array}\right]\left[\begin{array}{ccc}1& 2& 4\\ 2& 2& 3\\ 4& 3& -5\end{array}\right]\\ & =\left[\begin{array}{ccc}1+4+16& 1+4+16& 1+4+16\\ 1+4+16& 1+4+16& 1+4+16\\ 1+4+16& 1+4+16& 1+4+16\end{array}\right]\end{array}$

$\begin{array}{cc}& f\left(A\right)=A^2-A+3I\\ & =\left[\begin{array}{ccc}21& 18& -10\\ 18& 17& -1\\ -10& -1& 50\end{array}\right]-\left[\begin{array}{ccc}1& 2& 4\\ 2& 2& 3\\ 4& 3& -5\end{array}\right]+3\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ & =\left[\begin{array}{ccc}21& 18& -10\\ 18& 17& -1\\ -10& -1& 50\end{array}\right]-\left[\begin{array}{ccc}1& 2& 4\\ 2& 2& 3\\ 4& 3& -5\end{array}\right]+\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]\\ & =\left[\begin{array}{ccc}23& 16& -14\\ 16& 18& -4\\ -14& -4& 58\end{array}\right]\text{ [Ans. }\end{array}$

$A=\left|\begin{array}{ccc}1& 2& 4\\ 2& 2& 3\\ 4& 3& -5\end{array}\right|$

$\begin{array}{cc}& |A|=\left|\begin{array}{ccc}1& 2& 4\\ 2& 2& 3\\ 4& 3& -5\end{array}\right|\\ & =1(-10-9)-2(-10-12)+4(6-8)\\ & =-19+44-8=17\ne 0\end{array}$

A এর সহগুনক গুলো বের করে পাই ,

$\begin{array}{cc}& A_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}2& 3\\ 3& -5\end{array}\right|=-10-9=-19\\ & A_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}2& 3\\ 4& -5\end{array}\right|=-(-10-12)=22\\ & A_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}2& 2\\ 4& 3\end{array}\right|=6-8=-2\\ & A_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}2& 4\\ 3& -5\end{array}\right|=-(-10-12)=22\\ & A_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& 4\\ 4& -5\end{array}\right|=-5-16=-21\\ & A_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& 2\\ 4& 3\end{array}\right|=-(3-8)=5\\ & A_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}2& 4\\ 2& 3\end{array}\right|=6-8=-2\\ & A_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& 4\\ 2& 3\end{array}\right|=-(3-8)=5\\ & A_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& 2\\ 2& 2\end{array}\right|=2-4=-2\end{array}$

অতএব$\begin{array}{cc}& \text{ অতএব, adj }A={\left[\begin{array}{ccc}-19& 22& -2\\ 22& -21& 5\\ -2& 5& -2\end{array}\right]}^T=\left[\begin{array}{ccc}-19& 22& -2\\ 22& -21& 5\\ -2& 5& -2\end{array}\right]\end{array}$

এখন, $A^{-1}=\frac{1}{\left|A\right|}$ adj $A$

$=\frac{1}{17}\left[\begin{array}{ccc}-19& 22& -2\\ 22& -21& 5\\ -2& 5& -2\end{array}\right]$

$=\left[\begin{array}{ccc}-\frac{19}{17}& \frac{22}{17}& -\frac{2}{17}\\ \frac{22}{17}& -\frac{21}{17}& \frac{5}{17}\\ -\frac{2}{17}& \frac{5}{17}& -\frac{2}{17}\end{array}\right]$

আবার , $A^{-1},A=\left[\begin{array}{ccc}-\frac{19}{17}& \frac{22}{17}& -\frac{2}{17}\\ \frac{22}{17}& -\frac{21}{17}& \frac{5}{17}\\ -\frac{2}{17}& \frac{5}{17}& -\frac{2}{17}\end{array}\right]$

$\lceil \begin{array}{ccc}1& 2& 4\\ 2& 2& 3\\ 4& 3& -5\end{array}\rceil$

$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I_3$

$A^{-1}\cdot A=I_3$