সৃজনশীলঃ ‘A=[2−1131−452−3],B=[p2qr2pq2rp2qr2pq2r].…’ দিনাজপুর বোর্ড ২০২৩ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

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

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

$=2(-3+8)+1(-9+20)+1(6-5)$

$=10+11+1=22\ne 0$

$\therefore A^{-1}$ বিদ্যামান।

$A_{11}=\left|\begin{array}{cc}1& -4\\ 2& -3\end{array}\right|$

$=-3+8=5$

$A_{12}=-\left|\begin{array}{cc}3& -4\\ 5& -3\end{array}\right|$

$=-(-9+20)=-11$

$A_{13}=\left|\begin{array}{cc}3& 1\\ 5& 2\end{array}\right|$

$=6-5=1$

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

$=-(3-2)=-1$

$A_{22}=\left|\begin{array}{cc}2& 1\\ 5& -3\end{array}\right|$

$=-6-5=-11$

$A_{23}=-\left|\begin{array}{cc}2& -1\\ 5& 2\end{array}\right|$

$=-(4+5)=-9$

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

$=4-1=3$

$A_{32}=-\left|\begin{array}{cc}2& 1\\ 3& -4\end{array}\right|$

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

$A_{33}=\left|\begin{array}{cc}2& -1\\ 3& 1\end{array}\right|$

$=2+3=5$

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

$={\left[\begin{array}{ccc}5& -11& 1\\ -1& -11& -9\\ 3& 11& 5\end{array}\right]}^T=\left[\begin{array}{ccc}5& -1& 3\\ -11& -11& 11\\ 1& -9& 5\end{array}\right]$

$A^{-1}=\frac{1}{\left|A\right|}\mathrm{Adj}A=\frac{1}{22}\left[\begin{array}{ccc}5& -1& 3\\ -11& -11& 11\\ 1& -9& 5\end{array}\right]$

নির্ণেয় $A^{-1}=\frac{1}{22}\left[\begin{array}{ccc}5& -1& 3\\ -11& -11& 11\\ 1& -9& 5\end{array}\right]$

দেওয়া আছে, $B=\left[\begin{array}{ccc}{p}^2& \mathrm{qr}& 2p\\ q^2& \mathrm{rp}& 2q\\ r^2& \mathrm{pq}& 2r\end{array}\right]$

$\therefore |B|=\left|\begin{array}{ccc}{p}^2& \mathrm{qr}& 2p\\ q^2& \mathrm{rp}& 2q\\ r^2& \mathrm{pq}& 2r\end{array}\right|$

$=2\left|\begin{array}{ccc}{p}^2& \mathrm{qr}& p\\ q^2& \mathrm{rp}& q\\ r^2& \mathrm{pq}& r\end{array}\right|$

$=2\left|\begin{array}{ccc}{p}^2-q^2& qr-rp& p-q\\ q^2-r^2& rp-pq& q-r\\ r^2& pq& r\end{array}\right|$

$[\because r_1^{\prime }=r_1-r_2$ এবং $r_2^{\prime }=r_2-r_3]$

$=2\left|\begin{array}{ccc}(p+q)(p-q)& -r(p-q)& p-q\\ (q+r)(q-r)& -p(q-r)& q-r\\ r^2& pq& r\end{array}\right|$

$=2(p-q)(q-r)\left|\begin{array}{ccc}p+q& -r& 1\\ q+r& -p& 1\\ r^2& pq& r\end{array}\right|$

$=2(p-q)(q-r)\left|\begin{array}{ccc}p-r& p-r& 0\\ q+r& -p& 1\\ r^2& pq& r\end{array}\right|[r_1^{\prime }=r_1-r_2]$

$=2(p-q)(q-r)(p-r)\left|\begin{array}{ccc}1& 1& 0\\ q+r& -p& 1\\ r^2& pq& r\end{array}\right|$

$=2(p-q)(q-r)(p-r)\left|\begin{array}{ccc}1& 0& 0\\ q+r& -p-q-r& 1\\ r^2& pq-r^2& r\end{array}\right|[c_2{}^{\prime }=c_2-c_1]$

$=2(p-q)(q-r)(p-r)\left|\begin{array}{cc}-p-q-r& 1\\ pq-r^2& r\end{array}\right|$

$=2(p-q)(q-r)(p-r)(-rp-qr-r^2-pq+r^2)$

$=2(p-q)(q-r)(p-r)(-pq-qr-rp)$

$=2(p-q)(q-r)(p-r)\{-(pq+qr+rp\left)\right\}$

$=-2(p-q)(q-r)(p-r)(pq+qr+rp)$

$\therefore |B|=-2(p-q)(q-r)(p-r)(pq+qr+rp)$ (প্রমাণিত)