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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?

  • $\begingroup$ 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 $\det(A) = 0$? $\endgroup$
    – JavaMan
    Oct 13, 2018 at 2:48
  • $\begingroup$ It is the product of all the numbers diagonally from the top left to the bottom right? $\endgroup$
    – Utsav
    Oct 13, 2018 at 2:51
  • $\begingroup$ 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. $\endgroup$
    – JavaMan
    Oct 13, 2018 at 3:18

1 Answer 1


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 $0w=0$.

As you see, because the gaussian elimination discarded 2 equations, we have 4 variables and 2 LI equations, thus the space of available solutions has dimension 4-2=2.

$y$ 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.


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