If $\mathrm{A}=\left[\begin{array}{ccc}1& 2& -3\\ 3& 2& -2\\ 2& -1& 1\end{array}\right]$, then find A^-1 and use it to solve following system of the equation:
$\begin{array}{r}\mathrm{x}+2\mathrm{y}-3\mathrm{z}=6\\ 3\mathrm{x}+2\mathrm{y}-2\mathrm{z}=3\\ 2\mathrm{x}-\mathrm{y}+\mathrm{z}=2\end{array}$

OR

Using properties of determinants, prove that
$\left|\begin{array}{l}(\mathrm{b}+\mathrm{c}{)}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{bc}\\ (\mathrm{c}+\mathrm{a}{)}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{b}}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{ca}\\ (\mathrm{a}+\mathrm{b}{)}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{c}}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{ab}\end{array}\right|=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})(\mathrm{a}+\mathrm{b}+\mathrm{c})\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)$

First, we will find,
$\left|\mathrm{A}\right|=\det \mathrm{A}=\left|\begin{array}{ccc}1& 2& -3\\ 3& 2& -2\\ 2& -1& 1\end{array}\right|=1(2-2)-2(3+4)-3(-3-4)=0-14+21=7$
Now, We will find the cofactors of the matrix A:
$\begin{array}{l}{\mathrm{A}}_{11}=\left|\begin{array}{lc}2& -2\\ -1& 1\end{array}\right|=0,\\ {\mathrm{A}}_{12}=-\left|\begin{array}{cc}3& -2\\ 2& 1\end{array}\right|=-7,\\ {\mathrm{A}}_{13}=\left|\begin{array}{cc}3& 2\\ 2& -1\end{array}\right|=-7,\\ {\mathrm{A}}_{21}=-\left|\begin{array}{cc}2& -3\\ -1& 1\end{array}\right|=1,\\ {\mathrm{A}}_{22}=\left|\begin{array}{cc}1& -3\\ 2& 1\end{array}\right|=7,\\ {\mathrm{A}}_{23}=-\left|\begin{array}{cc}1& 2\\ 2& -1\end{array}\right|=5,\\ {\mathrm{A}}_{31}=\left|\begin{array}{ll}2& -3\\ 2& -2\end{array}\right|=2,\\ {\mathrm{A}}_{32}=-\left|\begin{array}{ll}1& -3\\ 3& -2\end{array}\right|=-7,\\ {\mathrm{A}}_{33}=\left|\begin{array}{ll}1& 2\\ 3& 2\end{array}\right|=-4\end{array}$
Therefore, adjoint of matrix A will be given by:
$\begin{array}{r}\mathrm{adjA}={\left[\begin{array}{lcc}0& -7& -7\\ 1& 7& 5\\ 2& -7& -4\end{array}\right]}^{\mathrm{T}}=\left[\begin{array}{ccc}0& 1& 2\\ -7& 7& -7\\ -7& 5& -4\end{array}\right]\\ \therefore {\mathrm{A}}^{-1}=\frac{1}{\left|\mathrm{A}\right|}\text{ adj }\mathrm{A}=\frac{1}{7}\left[\begin{array}{ccc}0& 1& 2\\ -7& 7& -7\\ -7& 5& -4\end{array}\right]\end{array}$
Now, the system of equations can be reduced in the matrix form AX=B as
Where, $\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right],\mathrm{B}=\left[\begin{array}{l}6\\ 3\\ 2\end{array}\right]$
$\begin{array}{l}\mathrm{AX}=\mathrm{B}\Rightarrow \mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\\ \left[\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{7}\left[\begin{array}{ccl}0& 1& 2\\ -7& 7& -7\\ -7& 5& -4\end{array}\right]\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}7\\ -35\\ -35\end{array}\right]=\left[\begin{array}{c}1\\ -5\\ -5\end{array}\right]\\ \mathrm{x}=1,\mathrm{y}=-5\text{ and }\mathrm{z}=-5\end{array}$
Therefore, $A^{-1}=\frac{1}{7}\left[\begin{array}{ccc}0& 1& 2\\ -7& 7& -7\\ -7& 5& -4\end{array}\right]\text{ and }(\mathrm{x},\mathrm{y},\mathrm{z})=(1,-5,-5)$

OR

Given: 
$\begin{array}{l}\text{ L.H.S. }=\left|\begin{array}{lll}(\mathrm{b}+\mathrm{c}{)}^2& {\mathrm{a}}^2& \mathrm{bc}\\ (\mathrm{c}+\mathrm{a}{)}^2& {\mathrm{b}}^2& \mathrm{ca}\\ (\mathrm{a}+\mathrm{b}{)}^2& {\mathrm{c}}^2& \mathrm{ab}\end{array}\right|\\ =\left|\begin{array}{ccc}{\mathrm{b}}^2+{\mathrm{c}}^2& {\mathrm{a}}^2& \mathrm{bc}\\ {\mathrm{c}}^2+{\mathrm{a}}^2& {\mathrm{b}}^2& \mathrm{ca}\\ {\mathrm{a}}^2+{\mathrm{b}}^2& {\mathrm{c}}^2& \mathrm{ab}\end{array}\right|\left(\because {\mathrm{C}}_1\to {\mathrm{C}}_1-2{\mathrm{C}}_3\right)\\ =\left|\begin{array}{ccc}{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2& {\mathrm{a}}^2& \mathrm{bc}\\ {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2& {\mathrm{b}}^2& \mathrm{ca}\\ {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2& {\mathrm{c}}^2& \mathrm{ab}\end{array}\right|\left(\because {\mathrm{c}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2\right)\\ =\left|\begin{array}{ccc}{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2& {\mathrm{a}}^2& \mathrm{bc}\\ 0& {\mathrm{b}}^2-{\mathrm{a}}^2& \mathrm{ca}-\mathrm{bc}\\ 0& {\mathrm{c}}^2-{\mathrm{a}}^2& \mathrm{ab}-\mathrm{bc}\end{array}\right|\left(\because {\mathrm{R}}_1\to {\mathrm{R}}_2-{\mathrm{R}}_1,{\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_1\right)\end{array}$
$=(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{a})\left|\begin{array}{ccc}{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2& {\mathrm{a}}^2& \mathrm{bc}\\ 0& \mathrm{b}+\mathrm{a}& -\mathrm{c}\\ 0& \mathrm{c}+\mathrm{a}& -\mathrm{b}\end{array}\right|(\because$get common (b-a) from $R_2$and (c-a) from R$R_3$)
Expand along column 1
$\begin{array}{l}=\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{a})\left(-{\mathrm{b}}^2-\mathrm{ab}+{\mathrm{c}}^2+\mathrm{ac}\right)\\ =(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{a})(\mathrm{a}+\mathrm{b}+\mathrm{c})\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)\\ =\text{ R.H.S }\\ \therefore \text{ L.H.S. }=\text{ R.H.S }\end{array}$
Therefore, which has proven.