সৃজনশীলঃ ‘2x+y+3z=4,x+2z=0,3x+4y−5z=2…’ ২০২৪ – ম্যাট্রিকস ও নির্ণায়ক, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

$2x+y+3z=4$

$x+2z=0$

$3x+4y-5z=2$

$x,y$ও $z$এর সহগগুচ্ছ নিয়ে গঠিত নির্ণায়ক,

$D=\left|\begin{array}{ccc}2& 1& 3\\ 1& 0& 2\\ 3& 4& -5\end{array}\right|$

$=2(0-8)+1(6+5)+3(4-0)$

$=-16+11+12=7$

$D_x=\left|\begin{array}{ccc}4& 1& 3\\ 0& 0& 2\\ 2& 4& -5\end{array}\right|$

$=4(0-8)-1(0-4)+3(0-0)$

$=-32+4=-28$

$D_y=\left|\begin{array}{ccc}2& 4& 3\\ 1& 0& 2\\ 3& 2& -5\end{array}\right|$

$=2(0-4)+4(6+5)+3(2-0)$

$=-8+44+6=42$

$D_z=\left|\begin{array}{ccc}2& 1& 4\\ 1& 0& 0\\ 3& 4& 2\end{array}\right|=2(0-0)-1(2-0)+4(4-0)$

$=-2+16=14$

$\therefore x=\frac{D_x}{D}=\frac{-28}{7}=-4,y=\frac{D_y}{D}=\frac{42}{7}=6$ এবং $z=\frac{D_z}{D}=\frac{14}{7}=2$

$\therefore$নির্ণেয় সমাধান, $(x,y,z)=(-4,6,2)$ (Ans.)

$2x+y+3z=4$ ...... (i)

$x+2z=0$ ...... (ii)

$3x+4y-5z=2$ ...... (iii)

ম্যাট্রিক্সআাকারে লিখে পাই,

$\left[\begin{array}{ccc}2& 1& 3\\ 1& 0& 2\\ 3& 4& -5\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}4\\ 0\\ 2\end{array}\right]$

বা, $\mathrm{BX}=C\Rightarrow X=B^{-1}C$ ...... (iv)

যেখানে, $B=\left[\begin{array}{ccc}2& 1& 3\\ 1& 0& 2\\ 3& 4& -5\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$এবং $C=\left[\begin{array}{c}4\\ 0\\ 2\end{array}\right]$

এখন, $|B|=\left|\begin{array}{ccc}2& 1& 3\\ 1& 0& 2\\ 3& 4& -5\end{array}\right|=2(0-8)+1(6+5)+3(4-0)$

$=-16+11+12=7\ne 0$

$\therefore B^{-1}$বিদ্যামন।

$\mathrm{adj}B={\left[\begin{array}{ccc}\left|\begin{array}{cc}0& 2\\ 4& -5\end{array}\right|& -\left|\begin{array}{cc}1& 2\\ 3& -5\end{array}\right|& \left|\begin{array}{cc}1& 0\\ 3& 4\end{array}\right|\\ -\left|\begin{array}{cc}1& 3\\ 4& -5\end{array}\right|& +\left|\begin{array}{cc}2& 3\\ 3& -5\end{array}\right|& -\left|\begin{array}{cc}2& 1\\ 3& 4\end{array}\right|\\ \left|\begin{array}{cc}1& 3\\ 0& 2\end{array}\right|& -\left|\begin{array}{cc}2& 3\\ 1& 2\end{array}\right|& \left|\begin{array}{cc}2& 1\\ 1& 0\end{array}\right|\end{array}\right]}^t$

$={\left[\begin{array}{ccc}-8& 11& 4\\ 17& -19& -5\\ 2& -1& -1\end{array}\right]}^t=\left[\begin{array}{ccc}-8& 17& 2\\ 11& -19& -1\\ 4& -5& -1\end{array}\right]$

$\therefore B^{-1}=\frac{1}{\left|B\right|}\text{ adj }B=\frac{1}{7}[\begin{array}{ccc}-8& 17& 2\\ 11& -19& -1\\ 4& -5& -1\end{array}$

 (iv) নং হতে,$X=\frac{1}{7}\left[\begin{array}{ccc}-8& 17& 2\\ 11& -19& -1\\ 4& -5& -1\end{array}\right]\left[\begin{array}{c}4\\ 0\\ 2\end{array}\right]$

$=\frac{1}{7}\left[\begin{array}{c}-32+0+4\\ 44+0-2\\ 16+0-2\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}-28\\ 42\\ 14\end{array}\right]=\left[\begin{array}{c}-4\\ 6\\ 2\end{array}\right]$

$\therefore$নির্ণেয় সমাধান: $(x,y,z)=(-4,6,2)$ (Ans.)