Question 895038
You have equations with the 3 variables  x, y, and z.
You want to write each equation in the form
{{{Ax+By+C=D}}}
and then you will use the coefficients A, B, C and D to build that matrix.
{{{y+2=3x}}} <---> {{{3*x-1*y+0*z=2}}}
{{{(1/4)z=z-1}}} <---> {{{0*x+0*y+(3/4)*z=1}}}
{{{z-8=x/2)}}} <---> {{{(1/2)*x+0*y-1*z=-8}}}
So,
{{{system(y+2=3x,(1/4)z=z-1,z-8=x/2)}}} ---> {{{system(3*x-1*y+0*z=2,0*x+0*y+(3/4)*z=1,(1/2)*x+0*y-1*z=-8)}}}
and your matrix could be
{{{(matrix(3,4,3,-1,0,2,0,0,3/4,1,1/2,0,-1,-8))}}}
 
NOTE: There are other options, because each equation has an infinite number of equivalent equations, and you could rearrange your equations in different order.
For example, you could have used {{{(-1/2)*x+0*y+1*z=8}}} or {{{1*x+0*y-2*x=-16}}} as an equation equivalent to {{{z-8=x/2)}}} ,
and you could have used the coefficients of that equation for the top line of the matrix instead of the bottom line.