সৃজনশীলঃ ‘A−1=[2−1130−21−41]x5−−2y5+z=1x+y4+3z4=1×3−y+2z3=1…’ দিনাজপুর বোর্ড ২০২২ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

A^-1 ম্যাট্রিক্স এর বীপরতি ম্যাট্রিক্সটিই হবে A ম্যাট্রিক্স [বিপরীত ম্যাট্রিক্স বিদ্যামান]

এখন$,এখন\left|A^{-1}\right|=2(0-8)+(-1)(-2-3)+1(-12-0)=-23\ne 0$

এখানে, $A_{11}^{-1}=(0-8)=-8$

$A^{-1}12=(-2-3)=-5$

$A^{-1}13=(-12-0)=-12$

$A^{-1}21=(-4+1)=-3$

$A^{-1}22=2-1=1$

$A^{-1}23=-1+8=7$

$\therefore A=\frac{1}{\left|A^{-1}\right|}\mathrm{adj}\left(A^{-1}\right)$

$=\frac{1}{-23}\left|\begin{array}{ccc}-8& -3& 2\\ -5& 1& 7\\ -12& 7& 3\end{array}\right|$

$=\left|\begin{array}{ccc}\frac{8}{23}& \frac{3}{23}& \frac{-2}{23}\\ \frac{5}{23}& \frac{-1}{23}& \frac{-7}{23}\\ \frac{12}{23}& \frac{-7}{23}& \frac{-3}{23}\end{array}\right|$

দেওয়া আছে,

$\frac{x}{z}-\frac{zy}{3}+z=1\dots \dots \dots .....(\text{ (i) }$

$x+\frac{1}{4}y+\frac{z}{4}z=1\dots \dots \dots .....\text{ (ii) }$

$\frac{x}{z}-y+\frac{z}{z}z=1\dots \dots \dots \dots ......\text{ (iii) }$

সমীকরণ গুলিকে ম্যাট্রিক্স আকারে প্রকাশ করলে,

$\left|\begin{array}{ccc}\frac{1}{5}& -\frac{2}{5}& 1\\ 1& \frac{1}{4}& \frac{3}{4}\\ \frac{1}{3}& -1& \frac{2}{3}\end{array}\right]\left|\begin{array}{c}x\\ y\\ z\end{array}\right|=\left|\begin{array}{c}1\\ 1\\ 1\end{array}\right|$

ধরি,

$A\times =B$

$\therefore A\times =B$

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

$=\frac{1}{5}(\frac{2}{12}+\frac{3}{4})\left(\frac{-2}{5}\right)\times (\frac{3}{12}-\frac{2}{3})+1(-1\frac{-2}{12})$

$=\frac{1}{5}\times \frac{1}{12}+\frac{-2}{5}\times \frac{-5}{12}-\frac{13}{12}$

$=\frac{11+10-65}{60}$

$=\frac{-44}{60}=\frac{-11}{15}$

$D_x=\left|\begin{array}{ccc}1& \frac{-2}{5}& 1\\ 1& \frac{1}{4}& \frac{3}{4}\\ 1& -1& \frac{2}{3}\end{array}\right|=1(\frac{2}{12}+\frac{3}{4})+\left(\frac{-2}{5}\right)(\frac{3}{4}-\frac{2}{3})+1(-1-\frac{1}{4})$

$=\frac{2+8}{22}+\frac{-2}{3}\times \frac{1}{12}+\frac{-5}{4}$

$=\frac{11}{12}-\frac{2}{60}-\frac{5}{4}$

$=\frac{ss-2-7s}{40}$

$=\frac{-11}{20}$

$\begin{array}{cc}& D_y=\left|\begin{array}{ccc}1/5& 1& 1\\ 1& 1& \frac{3}{4}\\ 1/3& 1& 2/3\end{array}\right|=\frac{1}{5}(\frac{2}{3}-\frac{3}{4})+1(\frac{3}{12}-\frac{2}{3})+1(1-\frac{1}{3})\\ & =\frac{1}{5}\times \frac{-1}{12}+\frac{3-8}{12}+\frac{2}{3}\\ & =\frac{-1}{60}-\frac{5}{12}+\frac{2}{3}\\ & =\frac{-1-25+40}{60}=\frac{14}{60}=\frac{7}{30}\\ & D_z=\left|\begin{array}{ccc}1/5& -2/5& 1\\ 1& 1/4& 1\\ 1/3& -1& 1\end{array}\right|=\frac{1}{5}(\frac{1}{4}+1)+\left(\frac{-2}{5}\right)(\frac{1}{3}-1)+1(-1-\frac{1}{12})\\ & =\frac{15}{5}\frac{5}{4}+\frac{-2}{5}\times \frac{-2}{3}+\frac{-13}{12}\\ & =\frac{1}{4}+\frac{4}{15}-\frac{13}{12}\\ & =\frac{15+16-65}{60}=\frac{-17}{30}\\ & x=\frac{D_x}{D}=\frac{-11/30}{-11/15}=1/2\\ & y=\frac{D_y}{D}=\frac{7/30}{-11/15}=\frac{-7}{22}\\ & z=\frac{D_z}{D}=\frac{-17/30}{-11/15}=\frac{17}{22}\\ & \end{array}$