Elimination Method by Matrix and Gauss Elimination method are different methods. For instance, you can't solve system with Elimination Method by Matrix if determinant is equal 0, but you can solve that kind of system with Gauss Elimination method.
"\\begin{cases}\n x +2y+3z=4 \\\\\n 2x+3y+4z=5 \\\\\n 3x+4y+5z=6\n\\end{cases}"
Let's rewrite the system of equations in matrix form and solve it using the Gauss method:
"\\begin{pmatrix}\n1&2&3&\\vert4 \\\\\n 2&3&4&\\vert5 \\\\\n 3&4&5&\\vert6\n\\end{pmatrix}"
Subtract from the second row the first row multiplied by 2
"\\begin{pmatrix}\n1&2&3&\\vert\\space\\space\\space\\space\\space4 \\\\\n 0&-1&-2&|-3 \\\\\n 3&4&5&\\vert\\space\\space\\space\\space\\space6\n\\end{pmatrix}"
Subtract from the third row the first row multiplied by 3
"\\begin{pmatrix}\n1&2&3&\\vert\\space\\space\\space\\space\\space4 \\\\\n 0&-1&-2&|-3 \\\\\n 0&-2&-4&|-6\n\\end{pmatrix}"
Multiply the second row by -1
"\\begin{pmatrix}\n1&2&3&\\vert\\space\\space\\space\\space\\space4 \\\\\n 0&1&2&\\vert\\space\\space\\space\\space\\space3 \\\\\n 0&-2&-4&|-6\n\\end{pmatrix}"
Subtract from the first row the second row multiplied by 2
"\\begin{pmatrix}\n1&0&-1&|-2 \\\\\n 0&1&2&\\vert\\space\\space\\space\\space\\space3 \\\\\n 0&-2&-4&|-6\n\\end{pmatrix}"
Add the second line multiplied by 2 to the third row
"\\begin{pmatrix}\n1&0&-1&|-2 \\\\\n 0&1&2&\\vert\\space\\space\\space\\space\\space3 \\\\\n 0&0&0&\\vert\\space\\space\\space\\space\\space0\n\\end{pmatrix}"
We can remove last row and get system of equations:
"\\begin{cases}\n x -z=-2 \\\\\n y+2z=3 \n\\end{cases}"
We leave unknown variables x and y on the left side of the equations system, and we move the unknown variable z to the right side
"\\begin{cases}\n x =z-2 \\\\\n y=3-2z \n\\end{cases}"
Let's take "z=\\alpha" , where "\\alpha" is an arbitrary number
"\\begin{cases}\n x =\\alpha-2 \\\\\n y=3-2\\alpha \n\\end{cases}"
Answer: "x=\\alpha-2, y=3-2\\alpha, z=\\alpha," where "\\alpha" is an arbitrary number
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