(a) Solve the following system of equations, using matrices: 2/x+3/y+10/z=4, 4/x-6/y+5/z=1, 6/x+9/y-20/z=2 where x, y, z ≠ 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 $$\begin{aligned} & \mathrm{A}_{11}=75, \mathrm{A}_{12}=110, \mathrm{A}_{13}=72 \\ & \mathrm{A}_{21}=150, \mathrm{A}_{22}=-100, \mathrm{A}_{23}=0 \\ & \mathrm{A}_{31}=75, \mathrm{A}_{32}=30, \mathrm{A}_{33}=-24 \end{aligned} $$ $$\operatorname{adj} \mathrm{A}=\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right] $$ $$\mathrm{A}^{-1}=\frac{\operatorname{adjA}}{|\mathrm{A}|}=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right] $$ $$X=\mathrm{A}^{-1} \mathrm{B}=\frac{1}{1200}\left[\begin{array}{ccc} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{array}\right]\left[\begin{array}{l} 4 \\ 1 \\ \end{array}\right] $$ $$\therefore \mathrm{x}=2, \mathrm{y}=3, \mathrm{z}=5 $$ $$\begin{aligned} & |\mathrm{A}|=1+\cot ^{2} \mathrm{x}=\operatorname{cosec}^{2} \mathrm{x} \\ & \operatorname{adj} \mathrm{A}=\left[\begin{array}{cc} 1 & -\cot \mathrm{x} \\ \cot \mathrm{x} & 1 \end{array}\right] \end{aligned} $$