How do you solve using gaussian elimination or gauss Jordan elimination, x-2y-z=2 ,
2x – y + z = 4, -x + y -2z = -4 ?

Hint: While solving a system of linear equation by using gauss elimination method, we write all the coefficient and constants of equations in matrix form and then by doing row operation we try to make the coefficient of any rows 0 except one coefficient and then solve for unknowns.

Complete step by step solution:
Equations given in the question is x-2y-z=2 ,2x – y + z = 4, -x + y -2z = -4 , if we write in matrix form we get
$\left| \begin{matrix}
   1 & -2 & -1 & 2 \\
   2 & -1 & 1 & 4 \\
   -1 & 1 & -2 & -4 \\
\end{matrix} \right|$
Now we can perform row operation, let’s write ${{R}_{3}}$ as sum of ${{R}_{1}}$ and ${{R}_{3}}$ and ${{R}_{2}}$ as ${{R}_{2}}-2{{R}_{1}}$
$\left| \begin{matrix}
   1 & -2 & -1 & 2 \\
   0 & 3 & 3 & 0 \\
   0 & -1 & -3 & -2 \\
\end{matrix} \right|$
Now writing ${{R}_{3}}$ as $\dfrac{1}{3}{{R}_{2}}+{{R}_{3}}$
$\left| \begin{matrix}
   1 & -2 & -1 & 2 \\
   0 & 3 & 3 & 0 \\
   0 & 0 & -2 & -2 \\
\end{matrix} \right|$
From row 3, we get -2z is equal to -2, so z is equal to 1
From row 2, we get 3y + 3z =0 , so if we put z = 1, y comes out to be -1.
From row 1, we get x – 2y - z = 2, if we put z = 1 and y = -1 we get x equal to 1.

So x = 1, y = -1 and z = 1 is solution of the system of equation.

Note: The given 3 equations in the above question represent a plane in 3D coordinate system. If any 2 of the planes are parallel to each other then there will be no solution for the system. If one plane passes through the line formed by the intersection of another 2 planes then there will be an infinite no of solutions.