# Gaussian elimination with 4 variables

After this step however, I end up with 2 simultanious equations but still with four unknowns, I'm not sure if my gaussian elimination is wrong or not but how will i find the exact values of x,y,z and w?

• Do you know about determinants? What is the determinant of the original $4 \times 4$ matrix? Do you know what it means to the system $Ax =b$ if $\det(A) = 0$? Oct 13, 2018 at 2:48
• It is the product of all the numbers diagonally from the top left to the bottom right? Oct 13, 2018 at 2:51
• No it's not the product of the diagonal. The sum of the diagonal is called the matrix trace, but the determinant is different altogether. In your case since $\det(A) = 0$, then $Ax = b$ will not have a unique solution. Oct 13, 2018 at 3:18

If the solution you reach is correct, then, $$y$$ and $$w$$ can take any value, and $$x$$ and $$z$$ are equal to $$x=100-3y+96w$$ and $$z=54+52w$$.
$$y$$ and $$w$$ can take any value because the equations 3 and 4 are equivalent to: $$0x+0y+0z+0w=0$$ and from here, because those equations are pivotal: $$0z=0$$ and $$0w=0$$.
$$y$$ and $$w$$ are the 2 free variables parametrizing that solution space.
Actually, you can say that, i.e. $$x$$ and $$w$$ are your 2 free variables. The problem is the same in this case, if you reinterpret the problem by swapping the columns and proceeding as usually.