Let we are given the matrices as $X=\left[ \begin{matrix}
   {{x}_{1}} \\
   {{x}_{2}} \\
   {{x}_{3}} \\
\end{matrix} \right]$, $A=\left[ \begin{matrix}
   1 & -1 & 2 \\
   2 & 0 & 1 \\
   3 & 2 & 1 \\
\end{matrix} \right]$ and $B=\left[ \begin{matrix}
   3 \\
   1 \\
   4 \\
\end{matrix} \right]$. If $AX=B$ then $X$ is equal to \[\]
A. $\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]$\[\]
B. $\left[ \begin{matrix}
   -1 \\
   -2 \\
   -3 \\
\end{matrix} \right]$\[\]
C. $\left[ \begin{matrix}
   -1 \\
   2 \\
   3 \\
\end{matrix} \right]$\[\]
D. $\left[ \begin{matrix}
   0 \\
   2 \\
   1 \\
\end{matrix} \right]$\[\]

Hint: We multiply ${{A}^{-1}}$ at both sides of the given equation on the left and get $X={{A}^{-1}}B$. We find the inverse matrix ${{A}^{-1}}=\dfrac{\text{adj}\text{.}A}{\det \left( A \right)}$ where $\text{adj}\text{.}A$ is the adjoint matrix and $\det \left( A \right)$ is the determinant value of $A$. We find the adjoint by taking the transpose of the cofactor matrix of $A$ which means $ text{adj}.A={{\left( \text{cof}.A \right)}^{T}}$ , the determinant value and then ${{A}^{-1}}$. We multiply ${{A}^{-1}}B$ to get $X$.

Complete step-by-step solution:
We are given in the question a column vector $X$ of order $3\times 1$ with three unknowns${{x}_{1}},{{x}_{2}},{{x}_{3}}$, a square matrix $A$ of order $3\times 3$ with integers as entries and a column vector $B$of order $3\times 1$ with integer as entries. We have
\[X=\left[ \begin{matrix}
   {{x}_{1}} \\
   {{x}_{2}} \\
   {{x}_{3}} \\
\end{matrix} \right],A=\left[ \begin{matrix}
   1 & -1 & 2 \\
   2 & 0 & 1 \\
   3 & 2 & 1 \\
\end{matrix} \right],B=\left[ \begin{matrix}
   3 \\
   1 \\
   4 \\
\end{matrix} \right]\]
We are also given an equation of matrices
\[AX=B\]
Let us multiply the inverse of the matrix $A$ that is ${{A}^{-1}}$ on both sides of the equation from the left. We have,
\[\Rightarrow {{A}^{-1}}\left( AX \right)={{A}^{-1}}B\]
We know that matrix multiplication is associative. So we have
\[\begin{align}
  & \Rightarrow \left( {{A}^{-1}}A \right)X={{A}^{-1}}B \\
 & \Rightarrow IX={{A}^{-1}}B \\
 & \Rightarrow X={{A}^{-1}}B......\left( 1 \right) \\
\end{align}\]
Here “I” is the identity matrix of order $3\times 3$. So we have to find ${{A}^{-1}}$. We know that
\[{{A}^{-1}}=\dfrac{1}{\det \left( A \right)}\text{adj}A\]
Here $\text{adj}A$ is the adjoint matrix of A and $\det \left( A \right)$ is the determinant value of $A$, Let us find the determinant value first by expanding by first row. We have,
\[\det \left( A \right)=\left| \begin{matrix}
   1 & -1 & 2 \\
   2 & 0 & 1 \\
   3 & 2 & 1 \\
\end{matrix} \right|=1\left( 0-2 \right)-\left( -1 \right)\left( 2-3 \right)+2\left( 4-0 \right)=5\]
We know that the adjoint matrix of $A$ is the transpose of cofactor matrix of which means
\[\text{adj}A={{\left( \text{cof}A \right)}^{T}}\]
We know that the cofactor matrix is made up of entries as the cofactors of each element from the original matrix. The cofactor of the entry ${{a}_{ij}}$ (where $i$ is row position and $j$ is the column position) is equal to the determinant value of the square matrix (order 1 less than original matrix) t formed by rows and columns excluding the ${{i}^{\text{th}}}$row and ${{j}^{\text{th}}}$column and then multiplied by ${{\left( -1 \right)}^{i+j}}$. The co-factors of the entries in first row are
\[\begin{align}
  & C\left( 1 \right)={{\left( -1 \right)}^{1+1}}\left| \begin{matrix}
   0 & 1 \\
   2 & 1 \\
\end{matrix} \right|=1\left( 0-2 \right)=-2 \\
 & C\left( -1 \right)={{\left( -1 \right)}^{1+2}}\left| \begin{matrix}
   2 & 1 \\
   3 & 1 \\
\end{matrix} \right|=-1\left( 2-3 \right)=1 \\
 & C\left( 2 \right)={{\left( -1 \right)}^{1+3}}\left| \begin{matrix}
   2 & 0 \\
   3 & 2 \\
\end{matrix} \right|=1\left( 4-0 \right)=4 \\
\end{align}\]
The cofactors of the elements in the second row are
\[\begin{align}
  & C\left( 2 \right)={{\left( -1 \right)}^{2+1}}\left| \begin{matrix}
   -1 & 2 \\
   2 & 1 \\
\end{matrix} \right|=-1\left( -1-4 \right)=5 \\
 & C\left( 0 \right)={{\left( -1 \right)}^{2+2}}\left| \begin{matrix}
   1 & 2 \\
   3 & 1 \\
\end{matrix} \right|=1\left( 1-6 \right)=-5 \\
 & C\left( 1 \right)={{\left( -1 \right)}^{2+3}}\left| \begin{matrix}
   1 & -1 \\
   3 & 2 \\
\end{matrix} \right|=-1\left( 2-\left( -3 \right) \right)=-5 \\
\end{align}\]
The cofactors of the elements in the third row are
\[\begin{align}
  & C\left( 3 \right)={{\left( -1 \right)}^{3+1}}\left| \begin{matrix}
   -1 & 2 \\
   0 & 1 \\
\end{matrix} \right|=1\left( -1-0 \right)=-1 \\
 & C\left( 2 \right)={{\left( -1 \right)}^{3+2}}\left| \begin{matrix}
   1 & 2 \\
   2 & 1 \\
\end{matrix} \right|=-1\left( 1-4 \right)=3 \\
 & C\left( 1 \right)={{\left( -1 \right)}^{3+3}}\left| \begin{matrix}
   1 & -1 \\
   2 & 0 \\
\end{matrix} \right|=1\left( 0-\left( -2 \right) \right)=2 \\
\end{align}\]
So the co-factor matrix $A$ is,
\[\text{cof}\text{.}A=\left[ \begin{matrix}
   C\left( 1 \right) & C\left( -1 \right) & C\left( 2 \right) \\
   C\left( 2 \right) & C\left( 0 \right) & C\left( 1 \right) \\
   C\left( 3 \right) & C\left( 2 \right) & C\left( 1 \right) \\
\end{matrix} \right]=\left[ \begin{matrix}
   -2 & 1 & 4 \\
   5 & -5 & 5 \\
   -1 & 3 & 2 \\
\end{matrix} \right]\]
Let us take the transpose of the cofactor matrix of $A$ by writing rows as column to get adjoint matrix of $A$ We have
\[\text{adj}\text{.}A={{\left( \text{cof}\text{.}A \right)}^{T}}={{\left[ \begin{matrix}
   -2 & 1 & 4 \\
   5 & -5 & -5 \\
   -1 & 3 & 2 \\
\end{matrix} \right]}^{T}}=\left[ \begin{matrix}
   -2 & 5 & -1 \\
   1 & -5 & 3 \\
   4 & -5 & 2 \\
\end{matrix} \right]\]
So the inverse of matrix of $A$ is,
\[{{A}^{-1}}=\dfrac{adj.A}{\det \left( A \right)}=\dfrac{\left[ \begin{matrix}
   -2 & 5 & -1 \\
   1 & -5 & 3 \\
   4 & -5 & 2 \\
\end{matrix} \right]}{5}=\left[ \begin{matrix}
   \dfrac{-2}{5} & 1 & \dfrac{-1}{5} \\
   \dfrac{1}{5} & -1 & \dfrac{3}{5} \\
   \dfrac{4}{5} & -1 & \dfrac{2}{5} \\
\end{matrix} \right]\]
We put ${{A}^{-1}}$ and $B$ in equation (1) and find the unknown vector $X$ as,
\[X={{A}^{-1}}B=\left[ \begin{matrix}
   \dfrac{-2}{5} & 1 & \dfrac{-1}{5} \\
   \dfrac{1}{5} & -1 & \dfrac{3}{5} \\
   \dfrac{4}{5} & -1 & \dfrac{2}{5} \\
\end{matrix} \right]\times \left[ \begin{matrix}
   3 \\
   1 \\
   4 \\
\end{matrix} \right]=\left[ \begin{matrix}
   -1 \\
   2 \\
   3 \\
\end{matrix} \right]\]
So the correct choice is C.\[\]


Note: We note that we could multiply ${{A}^{-1}}$ from the left only not from the right as multiplication in matrix is not commutative but ${{A}^{-1}}A=A{{A}^{-1}}=I$. We can alternatively find the inverse by using Gauss-Jordan elimination of the augmented matrix$\left( A, B \right)$. The matrix equation $AX=B$ is a system of equations which here in this problem has a unique solution. . We note that we cannot find an inverse if when$\det \left( A \right)=0$.