I see that the linear equations $\left\{\begin{array}{l} 2 x+y-t=-2 \\ 3 y+z+2 t=3 \end{array}\right.$ can be solved in the following way:
I have uploaded a picture to give the details of the algorithm for the mathematical solution. It is obvious that this method is quite different from the calculation method commonly used in linear algebra textbooks. I want to explain the principle of this algorithm clearly, but I don't even understand the principle at present.
The above similar contents are from page 206 of this textbook:
$$\left(\begin{array}{cccc|c} 2 & 1 & 0 & -1 & 2 \\ 0 & 3 & 1 & 2 & -3 \\ \hline 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right) \rightarrow\left(\begin{array}{cccc|c} 1 & 0 & 0 & 0 & 2 \\ 3 & -6 & 1 & 5 & -3 \\ \hline 0 & 1 & 0 & 0 & 0 \\ 1 & -2 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right)\rightarrow\left(\begin{array}{cc|cc|c} 1 & 0 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 & -9 \\ \hline 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & -2 & 1 & -2 \\ 0 & 1 & 6 & -5 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right)\rightarrow\left(\begin{array}{cc|cc|c} 1 & 0 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 & 0 \\ \hline 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & -2 & 1 & -2 \\ 0 & 1 & 6 & -5 & -9 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right)$$
Then we can get the following relationship:
$$\left(\begin{array}{l} x \\ y \\ z \\ t \end{array}\right)=\left(\begin{array}{c} 0 \\ -2 \\ 9 \\ 0 \end{array}\right)+\left(\begin{array}{cc} 1 & 0 \\ -2 & 1 \\ 6 & -5 \\ 0 & 1 \end{array}\right)\left(\begin{array}{l} c_{1} \\ c_{2} \end{array}\right)$$
But I want to use matrix language to describe and prove the solution process:
$$\begin{array}{c} A.X = B\\ M.A.X = M.B\\ M.A.X.N = M.B.N\\ \text{If }M.B.N = 0\\ \text{then }M.A.X.N = 0\\ \end{array}$$
How can I simply describe and prove the solution process?
I don't understand your approach in the picture for to solve this problem. However, this problem in the matricix language is easy. Note that if $$(S.E.L): \left\{\begin{aligned} 2x+y+0z-t=-2 \\ 0x+3y+z+2t=3 \end{aligned} \right.$$ so, we can write the S.E.L in the form $$AX=B \quad \text{and} \quad (A|B)$$ Taking the coefficients of the system of linear equations, we construct the matrix $A$, this's $$A=\begin{pmatrix} 2 & 1 & 0 & -1 \\ 0 & 2 & 1 & 2 \end{pmatrix}$$ and taking the unknows of the S.E.L, we have $$X=\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}$$ and finally, $$B=\begin{pmatrix} -2 \\ 3 \end{pmatrix}$$ Therefore, you have $$\left\{\begin{aligned} 2x+y+0z-t=-2 \\ 0x+3y+z+2t=3 \end{aligned} \right. \iff \underbrace{\begin{pmatrix} 2 & 1 & 0 & -1 \\ 0 & 2 & 1 & 2 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}}_{X}=\underbrace{\begin{pmatrix} -2 \\ 3 \end{pmatrix}}_{B}$$ Now, you can use Gaussian elimination here $$(A|B) \iff \begin{pmatrix} 2 & 1 & 0 & -1 & | & -2 \\ 0 & 2 & 1 & 2 & | & 3\end{pmatrix}$$