A school wants to allocate students into three clubs: Sports, Music and Drama, under following conditions: - The number of students in Sports club…

Let $\displaystyle \mathrm{x}, \mathrm{y}$ and z be the no. of students allocated to Sports, Music and Drama clubs respectively. Here, $\displaystyle \mathrm{x}=\mathrm{y}+z, \mathrm{y}=\frac{\mathrm{x}}{2}+20, \mathrm{x}+\mathrm{y}+z=180$ \[\Rightarrow \mathrm{x}-\mathrm{y}-z=0, \mathrm{x}-2 \mathrm{y}=-40, \mathrm{x}+\mathrm{y}+z=180 \] Given equations can be written as $\displaystyle \boldsymbol{A} \boldsymbol{X} \boldsymbol{=} \boldsymbol{B}$ \[\text { where, } A=\left[\begin{array}{ccc} 1 & -1 & -1 \\ 1 & -2 & 0 \\ 1 & 1 & 1 \end{array}\right], B=\left[\begin{array}{c} 0 \\ -40 \\ 180 \end{array}\right], X=\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ z \end{array}\right] \] \[\begin{aligned} & |A|=-4 \neq 0 \Rightarrow A^{-1} \mathrm{ex} \\ & \operatorname{adj} A=\left[\begin{array}{ccc} -2 & 0 & -2 \\ -1 & 2 & -1 \\ 3 & -2 & -1 \end{array}\right] \end{aligned} \] \[A^{-1}=\frac{1}{|A|} \times \operatorname{adj} A=\frac{1}{4}\left[\begin{array}{ccc} 2 & 0 & 2 \\ 1 & -2 & 1 \\ -3 & 2 & 1 \end{array}\right] \] Number of students allocated in sports, music and drama are $\displaystyle 90,65$ and $\displaystyle 25$ respectively.