Solve the system of the following equations
$\cfrac{2}{x} + \cfrac{3}{y} + \cfrac{{10}}{z} = 4,\cfrac{4}{x} - \cfrac{6}{y} + \cfrac{5}{z} = 1,\cfrac{6}{x} + \cfrac{9}{y} - \cfrac{{20}}{z} = 2$

The equations can be written in the form AX$=$B,
where,

$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3&{10}\\4&{ - 6}&5\\6&9&{ - 20}\end{array}} \right],X$

$= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{1/x}\\{1/y}\\{1/z}\end{array}} \right]$ and $B = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4\\1\\2\end{array}} \right]$

Now,$|A| = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3&{10}\\4&{ - 6}&5\\6&9&{ - 20}\end{array}} \right|$

$=$ $2\left( {120 - 45} \right) - 3\left( { - 80 - 30} \right) + 10\left( {36 + 36} \right)$

$=$ $2\left( {75} \right) - 3\left( { - 110} \right) + 10\left( {72} \right)150 + 330 + 720 = 1200 \ne 0$

$\therefore$ ${A^{ - 1}}$ exists.

Now, let ${A_{ij}}$ be the cofactor of the element in ${i^{th}}$ row and ${j^{th}}$ column.

The cofactors are :
${A_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 6}&5\\9&{ - 20}\end{array}} \right| = 120 - 45 = 75$

${A_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&5\\6&{ - 20}\end{array}} \right| = - ( - 80 - 30) = 110.$

${A_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4&{ - 6}\\6&9\end{array}} \right| = 36 + 36 = 72,$

${A_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{10}\\9&{ - 20}\end{array}} \right| = - ( - 60 - 90) = 150$

${A_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{10}\\6&{ - 20}\end{array}} \right| = - 40 - 60 = - 100,$

${A_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\6&9\end{array}} \right| = - (18 - 18) = 0,$

${A_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3&{10}\\{ - 6}&5\end{array}} \right| = 15 + 60 = 75,$

${A_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{10}\\4&5\end{array}} \right| = - (10 - 40) = 30,$

${A_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&3\\4&{ - 6}\end{array}} \right| = - 12 - 12 = - 24$

$\therefore$ $adj\;A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{75}&{110}&{72}\\{150}&{ - 100}&0\\{75}&{30}&{ - 24}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{75}&{150}&{75}\\{110}&{ - 100}&{30}\\{72}&0&{ - 24}\end{array}} \right]$

Hence, ${A^{ - 1}} = \cfrac{1}{{|A|}}(adj\;A) = \cfrac{1}{{1200}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{75}&{150}&{75}\\{110}&{ - 100}&{30}\\{72}&0&{ - 24}\end{array}} \right]$

As, $AX = B \Rightarrow X = {A^{ - 1}}B$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cfrac{1}{x}}\\{\cfrac{1}{y}}\\{\cfrac{1}{z}}\end{array}} \right] = \cfrac{1}{{1200}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{75}&{150}&{75}\\{110}&{ - 100}&{30}\\{72}&0&{ - 24}\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}4\\1\\2\end{array}} \right]$

$= \cfrac{1}{{1200}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{300 + 150 + 150}\\{440 - 100 + 60}\\{288 + 0 - 48}\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cfrac{1}{x}}\\{\cfrac{1}{y}}\\{\cfrac{1}{z}}\end{array}} \right]$

$= \cfrac{1}{{1200}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{600}\\{400}\\{240}\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\cfrac{1}{2}}\\{\cfrac{1}{3}}\\{\cfrac{1}{5}}\end{array}} \right]$

Thus, $\cfrac{1}{x} = \cfrac{1}{2},\cfrac{1}{y} = \cfrac{1}{3},\cfrac{1}{z} = \cfrac{1}{5}$

Hence, $x = 2,y = 3,z = 5$