Solve, $x + 2y -z= 5,3x - y + 3z = 7, 2x + 3y + z =11$

Solve: দেওয়া আছে,

$\left[\begin{array}{ccc}1 & 2 & -1 \\ 3 & -1 & 3 \\ 2 & 3 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}5 \\ 7 \\ 11\end{array}\right]$

$\begin{aligned} \Rightarrow & {\left[\begin{array}{c} x+2 y-z \\ 3 x-y+3 z \\ 2 x+3 y+z \end{array}\right]=\left[\begin{array}{c} 5 \\ 7 \\ 11 \end{array}\right] } \\ \therefore \quad & x+2 y-z=5 \\ & 3 x-y+3 z=7 \\ & 2 x+3 y+z=11 \end{aligned}$

এখখন, ক্রেমারের নিয়ম ব্যবহার করে আামরা পাই ,

$\begin{array}{l} D=\left|\begin{array}{ccc} 1 & 2 & -1 \\ 3 & -1 & 3 \\ 2 & 3 & 1 \end{array}\right| \\ =1(-1-9)-2(3-6)-1(9+2) \\ =-10+6-11=-15 \\ D_{x}=\left|\begin{array}{ccc} 5 & 2 & -1 \\ 7 & -1 & 3 \\ 11 & 3 & 1 \end{array}\right| \\ =5(-1-9)-2(7-33)-1(21+11) \\ =-50+52-32=-30 \\ D_{y}=\left|\begin{array}{lll} 1 & 5 & -1 \\ 3 & 7 & 3 \\ 2 & 11 & 1 \end{array}\right| \\ =1(7-33)-5(3-6)-1(33-14) \end{array}$

$\begin{aligned} & =-26+15-19=-30 \\ & D_{z}=\left|\begin{array}{ccc} 1 & 2 & 5 \\ 3 & -1 & 7 \\ 2 & 3 & 11 \end{array}\right| \\ & =1(-11-21)-2(33-14)+5(9+2) \\ & =-32-38+55=-15 \\ \therefore \quad & x=\frac{D_{x}}{D}=\frac{-30}{-15}=2, y=\frac{D_{y}}{D}=\frac{-30}{-15}=2 \\ & z=\frac{D_{z}}{D}=\frac{-15}{-15}=1 \end{aligned}$