20.

Using matrix method, solve the following system of equations:

$\frac{2}{\mathrm{x}} + \frac{3}{\mathrm{y}} + \frac{10}{\mathrm{z}} = 4, \frac{4}{\mathrm{x}} - \frac{6}{\mathrm{y}} + \frac{5}{\mathrm{z}}, \frac{6}{\mathrm{x}} + \frac{9}{\mathrm{y}} - \frac{20}{\mathrm{z}}; \mathrm{x}, \mathrm{y}, \mathrm{z} \ne 0$

$\mathrm{The} \mathrm{given} \mathrm{system} \mathrm{of} \mathrm{equation} \mathrm{is} \frac{2}{\mathrm{x}} + \frac{3}{\mathrm{y}} + \frac{10}{\mathrm{z}} = 4, \frac{4}{\mathrm{x}} - \frac{6}{\mathrm{y}} + \frac{5}{\mathrm{z}} = 1, \frac{6}{\mathrm{x} } + \frac{9}{\mathrm{y}} - \frac{20}{\mathrm{z}} = 2$

The given system of equation can be written as 

$$\begin{gathered} \left[ \begin{array}{ccc}2& 3& 10\\ 4& - 6& 5\\ 6& 9& - 20\end{array} \right] \left[\begin{array}{c}\frac{1}{ \mathrm{x}} \\ \frac{1}{\mathrm{y}} \\ \frac{1}{\mathrm{y}}\end{array}\right] = \left[\begin{array}{c}4\\ 1\\ 2\end{array}\right] \\ \mathrm{Or} \mathrm{AX} = \mathrm{B}, \mathrm{where} \mathrm{A} = \left[ \begin{array}{ccc}2& 3& 10\\ 4& - 6& 5\\ 6& 9& - 20\end{array} \right], \mathrm{X} = \left[\begin{array}{c}\frac{1}{ \mathrm{x}} \\ \frac{1}{\mathrm{y}} \\ \frac{1}{\mathrm{y}}\end{array}\right], \mathrm{and} \mathrm{B} = \left[\begin{array}{c}4\\ 1\\ 2\end{array}\right] \\ \mathrm{Now}, \left| \mathrm{A} \right| = \left[ \begin{array}{ccc}2& 3& 10\\ 4& - 6& 5\\ 6& 9& - 20\end{array} \right] \\ = 2 ( 120 - 45 ) - 3 ( - 80 - 30 ) + 10 ( 36 + 36 ) \\ = 1200 \ne 0 \end{gathered}$$

Hence, the unique solution of the system of equation is given by  X = A^{- 1} B

Now, the cofactors of  A  are computed as:

$$\begin{gathered} {\mathrm{C}}_{11} = ( - 1 {)}^2 \left( 120 - 45 \right) = 75, {\mathrm{C}}_{12} = ( - 1 {)}^3 \left( - 80 - 30 \right) = 110, {\mathrm{C}}_{13} = ( - 1 {)}^{4 }\left( 36 + 36 \right) = 72 \\ {\mathrm{C}}_{21} = ( - 1 {)}^3 \left( - 60 - 90 \right) = 150, {\mathrm{C}}_{22} = ( - 1 {)}^4 \left( - 60 - 90 \right) = 150, {\mathrm{C}}_{23} = ( - 1 {)}^5 \left( 18 - 18 \right) = 0 \\ {\mathrm{C}}_{31} = ( - 1 {)}^4 \left( 15+ 60 \right) =75, {\mathrm{C}}_{32} = ( - 1 {)}^5 \left( 10 - 40 \right) =30, {\mathrm{C}}_{33} = ( - 1 {)}^6 \left( - 12 - 12 \right) = - 24 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Adj} \mathrm{A} = {\left[ \begin{array}{ccc}75 & 110& 72\\ 150 & -100 & 0\\ 75& 30& - 24\end{array} \right]}^{\mathrm{T}} = \left[ \begin{array}{ccc}75 & 150& 75\\ 110 & -100 & 30\\ 72& 0& - 24\end{array} \right] \\ \Rightarrow {\mathrm{A}}^{-1} = \frac{\mathrm{Adj} \mathrm{A}}{\left| \mathrm{A} \right|} = \frac{1}{1200} \left[ \begin{array}{ccc}75 & 150& 75\\ 110 & -100 & 30\\ 72& 0& - 24\end{array} \right] \\ \mathrm{X} = {\mathrm{A}}^{-1} \mathrm{B} \\ = \frac{1}{1200} \left[ \begin{array}{ccc}75 & 150& 75\\ 110 & -100 & 30\\ 72& 0& - 24\end{array} \right] \left[\begin{array}{c} 4\\ 1\\ 2\end{array} \right] \\ =\frac{1}{1200} \left[ \begin{array}{c}3000 + 150 + 150\\ 440 - 100 + 60\\ 288 + 0 - 48\end{array} \right] = \frac{1}{1200} \left[\begin{array}{c} 600\\ 400\\ 240\end{array} \right] \end{gathered}$$

$$\begin{gathered} \mathrm{X} = \left[ \begin{array}{c}\frac{600}{1200}\\ \frac{ 400}{1200}\\ \frac{240}{1200}\end{array} \right] = \left[ \begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array} \right] \Rightarrow \left[ \begin{array}{c}\frac{1}{\mathrm{x}}\\ \frac{1}{\mathrm{y}}\\ \frac{1}{\mathrm{z}}\end{array} \right] = \left[ \begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array} \right] \\ \Rightarrow \frac{1}{\mathrm{x}} = \frac{1}{2}, \frac{1}{\mathrm{y}} = \frac{1}{3}, \mathrm{and} \frac{1}{\mathrm{z}} = \frac{1}{5} \\ \Rightarrow \mathrm{x} = 2, \mathrm{y} = 3, \mathrm{and} \mathrm{z} = 5 \end{gathered}$$

Thus, solution of given system of equation is given by  x = 2,   y = 3,   z = 5.