Using matrix method, solve the following system of equations:
$x-2y=10;2x+y+3z=8;-2y+z=7$

Hint: We first try to form the matrices of the equations. We have three matrices coefficient, variable, solution matrices. We use the inverse of the coefficient matrix to multiply with the solution matrix to form the relation $X={{A}^{-1}}B$ from $AX=B$. We equate corresponding elements to find the variables.

Complete step-by-step answer:
We express this given three equations of three unknowns x, y, z in the form of matrices.
There will be three matrices. Matrix A being the coefficient matrix, matrix X is the variable matrix and B is the solution matrix.
We first form every given equation for all variables.
$x-2y+0z=10;2x+y+3z=8;0x-2y+z=7$.
We now form the coefficient matrix as $A=\left[ \begin{matrix}
   1 & -2 & 0 \\
   2 & 1 & 3 \\
   0 & -2 & 1 \\
\end{matrix} \right]$.
The variable and the solution matrices are $X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right];B=\left[ \begin{matrix}
   10 \\
   8 \\
   7 \\
\end{matrix} \right]$.
The matrix multiplication form for the matrices is $AX=B$.
Now we perform matrix inverse to get the value of matrix X as $X={{A}^{-1}}B$.
We need to find the inverse of matrix A. we define $\left| A \right|$ as the determinant of the matrix A.
We know that ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}$, where $adj\left( A \right)$ defines the matrix $adj\left( A \right)=\left[ {{A}_{ji}} \right]$ for the matrix $A=\left[ {{a}_{ij}} \right]$. ${{a}_{ij}}$ are the elements of ${{i}^{th}}$ row and ${{j}^{th}}$ column. ${{A}_{ji}}$ represents the cofactors of ${{a}_{ij}}$. Basically, we are taking the cofactors and taking transpose of that matrix.
For ${{a}_{11}}=1$, the cofactor will be $\left| \begin{matrix}
   1 & 3 \\
   -2 & 1 \\
\end{matrix} \right|=1+6=7$.
For ${{a}_{12}}=-2$, the cofactor will be $-\left| \begin{matrix}
   2 & 3 \\
   0 & 1 \\
\end{matrix} \right|=-\left( 2-0 \right)=-2$.
For ${{a}_{13}}=0$, the cofactor will be $\left| \begin{matrix}
   2 & 1 \\
   0 & -2 \\
\end{matrix} \right|=-4-0=-4$.
For ${{a}_{21}}=2$, the cofactor will be $-\left| \begin{matrix}
   -2 & 0 \\
   -2 & 1 \\
\end{matrix} \right|=-\left( -2-0 \right)=2$.
For ${{a}_{22}}=1$, the cofactor will be $\left| \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right|=1-0=1$.
For ${{a}_{23}}=3$, the cofactor will be $\left| \begin{matrix}
   1 & -2 \\
   0 & -2 \\
\end{matrix} \right|=-\left( -2-0 \right)=2$.
For ${{a}_{31}}=0$, the cofactor will be $\left| \begin{matrix}
   -2 & 0 \\
   1 & 3 \\
\end{matrix} \right|=-6-0=-6$.
For ${{a}_{32}}=-2$, the cofactor will be $\left| \begin{matrix}
   1 & 0 \\
   2 & 3 \\
\end{matrix} \right|=-\left( 3-0 \right)=-3$.
For ${{a}_{33}}=1$, the cofactor will be $\left| \begin{matrix}
   1 & -2 \\
   2 & 1 \\
\end{matrix} \right|=1+4=5$.
So, $adj\left( A \right)=\left[ {{A}_{ji}} \right]={{\left[ \begin{matrix}
   7 & -2 & -4 \\
   2 & 1 & 2 \\
   -6 & -3 & 5 \\
\end{matrix} \right]}^{T}}=\left[ \begin{matrix}
   7 & 2 & -6 \\
   -2 & 1 & -3 \\
   -4 & 2 & 5 \\
\end{matrix} \right]$.
Now expanding through the first row, we get the determinant value as $\left| A \right|=\left( 1+6 \right)-2\left( 0-2 \right)+0\left( -4-0 \right)=11$.
Therefore, ${{A}^{-1}}=\dfrac{adj\left( A \right)}{\left| A \right|}=\dfrac{1}{11}\left[ \begin{matrix}
   7 & 2 & -6 \\
   -2 & 1 & -3 \\
   -4 & 2 & 5 \\
\end{matrix} \right]$. We multiply it with B to get the variables.
$X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]={{A}^{-1}}B=\dfrac{1}{11}\left[ \begin{matrix}
   7 & 2 & -6 \\
   -2 & 1 & -3 \\
   -4 & 2 & 5 \\
\end{matrix} \right]\left[ \begin{matrix}
   10 \\
   8 \\
   7 \\
\end{matrix} \right]=\dfrac{1}{11}\left[ \begin{matrix}
   70+16-42 \\
   -20+8-21 \\
   -40+16+35 \\
\end{matrix} \right]=\dfrac{1}{11}\left[ \begin{matrix}
   44 \\
   -33 \\
   11 \\
\end{matrix} \right]=\left[ \begin{matrix}
   4 \\
   -3 \\
   1 \\
\end{matrix} \right]$.
We used row-column multiplication and now we equate corresponding elements to get
$x=4,y=-3,z=1$. This is the solution of the system of equations.

Note: This type of solution is only possible for matrices. If we want to solve in a determinant way then we need to apply the Cramer’s rule. We also need to remember that the inverse will exist for only non-singular matrices, which means the determinant value has to be non-zero.