(a)OR(b)

Solve the following system of equations, using matrices : \[\frac{2}{\mathrm{x}}+\frac{3}{\mathrm{y}}+\frac{10}{\mathrm{z}}=4, \quad \frac{4}{\mathrm{x}}-\frac{6}{\mathrm{y}}+\frac{5}{\mathrm{z}}=1, \quad \frac{6}{\mathrm{x}}+\frac{9}{\mathrm{y}}-\frac{20}{\mathrm{z}}=2 \] where $\displaystyle \mathrm{x}, \mathrm{y}, \mathrm{z} \neq 0$

Given system of linear equations is equivalent to $\displaystyle \mathrm{AX}=\mathrm{B}$, where\begin{aligned} & \mathrm{A}=\left[\begin{array}{ccc} 2 & 3 & 10 4 & -6 & 5 6 & 9 & -20 \end{array}\right], X=\left[\begin{array}{c} \frac{1}{\mathrm{x}} \frac{1}{\mathrm{y}} \frac{1}{\mathrm{z}} \end{array}\right], \mathrm{B}=\left[\begin{array}{l} 4 1 \end{array}\right] & |\mathrm{A}|=1200 \neq 0 \end{aligned}Cofactors of the elements of A are\begin{aligned} & \mathrm{A}_{11}=75, \mathrm{A}_{12}=110, \mathrm{A}_{13}=72 & \mathrm{A}_{21}=150, \mathrm{A}_{22}=-100, \mathrm{A}_{23}=0 & \mathrm{A}_{31}=75, \mathrm{A}_{32}=30, \mathrm{A}_{33}=-24 \end{aligned} \operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc} 75 & 150 & 75 110 & -100 & 30 72 & 0 & -24 \end{array}\right] \mathrm{A}^{-1}=\frac{\operatorname{adjA}}{|\mathrm{A}|}=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 110 & -100 & 30 72 & 0 & -24 \end{array}\right] X=\mathrm{A}^{-1} \mathrm{B}=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 110 & -100 & 30 72 & 0 & -24 \end{array}\right]\left[\begin{array}{l} 4 1 \end{array}\right] \therefore \mathrm{x}=2, \mathrm{y}=3, \mathrm{z}=5 \begin{aligned} & |\mathrm{A}|=1+\cot ^{2} \mathrm{x}=\operatorname{cosec}^{2} \mathrm{x} & \operatorname{adj} \mathrm{A}=\left[\begin{array}{cc} 1 & -\cot \mathrm{x} \cot \mathrm{x} & 1 \end{array}\right] \end{aligned} $$