The given system of equation is $\frac{2}{\mathit{x}}+\frac{3}{\mathit{y}}+\frac{10}{\mathit{z}}$ = 4 , $\frac{4}{\mathit{x}}-\frac{6}{\mathit{y}}+\frac{5}{\mathit{z}}$ = 1 , $\frac{6}{\mathit{x}}+\frac{9}{\mathit{y}}-\frac{20}{\mathit{z}}$ = 2
The given system of equation can be written as
$\left[\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right]\left[\begin{array}{c}\frac{1}{\mathit{x}}\\ \frac{1}{\mathit{y}}\\ \frac{1}{\mathit{z}}\end{array}\right]$ = $\left[\begin{array}{c}4\\ 1\\ 2\end{array}\right]$
or AX B,Where A = $\left[\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right]$ , X = $\left[\begin{array}{c}\frac{1}{\mathit{x}}\\ \frac{1}{\mathit{y}}\\ \frac{1}{\mathit{z}}\end{array}\right]$ and B = $\left[\begin{array}{c}4\\ 1\\ 2\end{array}\right]$
Now, |A| = $\left[\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right]$
= 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36)
= 1200 ≠ 0
Hence, the unique solution of the system of equation is given by X = ${\mathit{A}}^{-1}\mathit{B}$
Now, the cofactors of A are computed as :
${\mathit{C}}_{11}$ = ${\left(-1\right)}^2$ (120 - 45) = 75, ${\mathit{C}}_{12}$ = ${\left(-1\right)}^3$ (- 80 - 30) = 110, ${\mathit{C}}_{13}$ = ${\left(-1\right)}^4$ (36 + 36) = 72
${\mathit{C}}_{21}$ = ${\left(-1\right)}^3$ (- 60 - 90) = 150, ${\mathit{C}}_{22}$ = ${\left(-1\right)}^4$ (- 40 - 60) = - 100, ${\mathit{C}}_{23}$ = ${\left(-1\right)}^5$ (18 - 18) = 0
${\mathit{C}}_{31}$ = ${\left(-1\right)}^4$ (15 + 60) = 75, ${\mathit{C}}_{12}$ = ${\left(-1\right)}^5$ (10 - 40) = 30, ${\mathit{C}}_{33}$ = ${\left(-1\right)}^6$ (- 12 - 12) = - 24
∴ Adj A = ${\left[\begin{array}{ccc}75& 110& 72\\ 150& -100& 0\\ 75& 30& -24\end{array}\right]}^{\mathit{T}}$ = $\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]$
⇒ ${\mathit{S}}^{-1}$ = $\frac{\mathit{A}\mathit{d}\mathit{j}\mathit{A}}{\left|\mathit{A}\right|}$ = $\frac{1}{1200}\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]$
X = ${\mathit{A}}^{-1}\mathit{B}$
=
= $\frac{1}{1200}\left[\begin{array}{c}400+150+150\\ 440-100+60\\ 288+0-48\end{array}\right]$ = $\frac{1}{1200}\left[\begin{array}{c}600\\ 400\\ 240\end{array}\right]$
X = $\left[\begin{array}{c}\frac{600}{1200}\\ \frac{400}{1200}\\ \frac{240}{1200}\end{array}\right]$ = $\left[\begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array}\right]$ ⇒ $\left[\begin{array}{c}\frac{1}{\mathit{x}}\\ \frac{1}{\mathit{y}}\\ \frac{1}{\mathit{z}}\end{array}\right]$ = $\left[\begin{array}{c}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array}\right]$
⇒ $\frac{1}{\mathit{x}}$ = $\frac{1}{2},\frac{1}{\mathit{y}}$ = $\frac{1}{3}$ and $\frac{1}{\mathit{z}}$ = $\frac{1}{5}$
⇒ x = 2 , y = 3 and z = 5
Thus, solution of given system of equation is given by x = 2, y = 3 and z = 5.
OR
The given matrix is A = $\left[\begin{array}{ccc}1& 3& -2\\ -3& 0& -1\\ 2& 1& 0\end{array}\right]$
We have $\mathit{A}{\mathit{A}}^{-1}$ = I
Thus, A = IA
Or, $\left[\begin{array}{ccc}1& 3& -2\\ -3& 0& -1\\ 2& 1& 0\end{array}\right]$ = $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ A
Applying ${\mathit{R}}_2$ → ${\mathit{R}}_2+3{\mathit{R}}_1$ and ${\mathit{R}}_3$ → ${\mathit{R}}_3-2{\mathit{R}}_1$
$\left[\begin{array}{ccc}1& 3& -2\\ 0& 9& -7\\ 0& -5& 4\end{array}\right]$ = $\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ -2& 0& 1\end{array}\right]$ A
Now,applying ${\mathit{R}}_2$ → $\frac{1}{9}{\mathit{R}}_2$
$\left[\begin{array}{ccc}1& 3& -2\\ 0& 1& -\frac{7}{9}\\ 0& -5& 4\end{array}\right]$ = $\left[\begin{array}{ccc}1& 0& 0\\ \frac{1}{3}& \frac{1}{9}& 0\\ -2& 0& 1\end{array}\right]$ A
Applying ${\mathit{R}}_1$ → ${\mathit{R}}_1-3{\mathit{R}}_2$ and ${\mathit{R}}_3$ → ${\mathit{R}}_3+5{\mathit{R}}_2$
$\left[\begin{array}{ccc}1& 0& \frac{1}{3}\\ 0& 1& -\frac{7}{9}\\ 0& 0& \frac{1}{9}\end{array}\right]$ = $\left[\begin{array}{ccc}0& -\frac{1}{3}& 0\\ \frac{1}{3}& \frac{1}{9}& 0\\ -\frac{1}{3}& \frac{5}{9}& 1\end{array}\right]$ A
Applying ${\mathit{R}}_3$ → $9{\mathit{R}}_3$
$\left[\begin{array}{ccc}1& 0& \frac{1}{3}\\ 0& 1& -\frac{7}{9}\\ 0& 0& 1\end{array}\right]$ = $\left[\begin{array}{ccc}0& -\frac{1}{3}& 0\\ \frac{1}{3}& \frac{1}{9}& 0\\ -3& 5& 9\end{array}\right]$ A
Applying ${\mathit{R}}_1$ → ${\mathit{R}}_1-\frac{1}{3}{\mathit{R}}_3$ and ${\mathit{R}}_2$ → ${\mathit{R}}_2+\frac{7}{9}{\mathit{R}}_3$
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ = $\left[\begin{array}{ccc}1& -2& -3\\ -2& 4& 7\\ -3& 5& 9\end{array}\right]$ A ⇒ I = $\left[\begin{array}{ccc}1& -2& -3\\ -2& 4& 7\\ -3& 5& 9\end{array}\right]$ A
∴ ${\mathit{A}}^{-1}$ = $\left[\begin{array}{ccc}1& -2& -3\\ -2& 4& 7\\ -3& 5& 9\end{array}\right]$
Hence, inverse of the matrix A is $\left[\begin{array}{ccc}1& -2& -3\\ -2& 4& 7\\ -3& 5& 9\end{array}\right]$