If $AX=B$, where $A=\left[ \begin{matrix}
   3 & 1 \\
   -1 & 2 \\
\end{matrix} \right]$, $B=\left[ \begin{matrix}
   7 & 3 \\
   0 & 6 \\
\end{matrix} \right]$. Then $X=$?
A. $\left[ \begin{matrix}
   1 & 0 \\
   2 & 3 \\
\end{matrix} \right]$
B. $\left[ \begin{matrix}
   0 & 3 \\
   1 & 2 \\
\end{matrix} \right]$
C. $\left[ \begin{matrix}
   3 & 2 \\
   0 & 1 \\
\end{matrix} \right]$
D. $\left[ \begin{matrix}
   2 & 0 \\
   1 & 3 \\
\end{matrix} \right]$

Hint: We first form the multiplication form of matrices to find the coefficient matrix $AX=B$. Then we use the inverse matrix to find the variable matrix with the form of $X={{A}^{-1}}B$. We find the inverse after finding the matrix being singular or not.

Complete step-by-step solution:
We multiply the equation $AX=B$ with ${{A}^{-1}}$ to get
$\begin{align}
  & {{A}^{-1}}.AX={{A}^{-1}}.B \\
 & \Rightarrow IX=X={{A}^{-1}}B \\
\end{align}$
The variables matrix will be in the form of $X={{A}^{-1}}B$. We have to multiply the inverse matrix of A with the solution matrix.
The inverse of any matrix $A=\left[ {{a}_{ij}} \right]$ will be ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$. Here ${{a}_{ij}}$ denotes the element of ${{i}^{th}}$ row and ${{j}^{th}}$ column.
$\left| A \right|$ is defined as the determinant value of the matrix A.
The $adj\left( A \right)$ is defined by the $adj\left( A \right)={{\left[ {{A}_{ij}} \right]}^{T}}=\left[ {{A}_{ji}} \right]$. Here, ${{A}_{ij}}$ denotes the cofactor of the element ${{a}_{ij}}$. The term ‘T’ denotes the transpose of the matrix.
We find the cofactors of the matrix A and get $\left[ {{A}_{ij}} \right]=\left[ \begin{matrix}
   2 & 1 \\
   -1 & 3 \\
\end{matrix} \right]$ which gives
$adj\left( A \right)={{\left[ {{A}_{ij}} \right]}^{T}}=\left[ {{A}_{ij}} \right]=\left[ \begin{matrix}
   2 & -1 \\
   1 & 3 \\
\end{matrix} \right]$.
Now we find the determinant which is $\left| A \right|=3\times 2-1\times \left( -1 \right)=6+1=7$.
As the matrix A is a non-singular matrix the inverse of the matrix exists.
We have ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}=\dfrac{1}{7}\left[ \begin{matrix}
   2 & -1 \\
   1 & 3 \\
\end{matrix} \right]$.
We get $X={{A}^{-1}}B$.
Therefore, $X={{A}^{-1}}B=\dfrac{1}{7}\left[ \begin{matrix}
   2 & -1 \\
   1 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
   7 & 3 \\
   0 & 6 \\
\end{matrix} \right]=\dfrac{1}{7}\left[ \begin{matrix}
   14 & 0 \\
   7 & 21 \\
\end{matrix} \right]=\left[ \begin{matrix}
   2 & 0 \\
   1 & 3 \\
\end{matrix} \right]$.
The correct option is D.

Note: If the determinant value of the matrix of which we are finding the inverse is 0 then the matrix is non-invertible. Those types of matrices are called singular matrices.
Also, we need to remember that the multiplications $X={{A}^{-1}}B$ and $X=B{{A}^{-1}}$ are different.