[Math] General Solution of System Of Equations (with 3 variables)

linear algebramatricesmatrix-rank

A system of equations is given as 

              x + 4y +2z = 0
              3x     -2z = 4
              3x -3y -4z = 5

The task is to find the general solution of the system.

I wrote down the augmented matrix as follows and went on to reduce it down to row-echelon form. But, then I realized the equations are inconsistent because the rank of the system alone is 2 while the rank of the augmented matrix is 1.

Can someone give me a few pointers as to where I might have gone wrong with this and what I can do to obtain an answer?

$$
\left[\begin{array}{rrr|r}
1 & 4 & 2 & 0 \\
3 & 0 & -2 & 4 \\
3 & -3 & -4 & 5
\end{array}\right]
$$

$$
\left[\begin{array}{rrr|r}
1 & 4 & 2 & 0 \\
0 & -12 & -8 & 4 \\
0 & -15 & -10 & 5
\end{array}\right]
$$

$$
\left[\begin{array}{rrr|r}
1 & 4 & 2 & 0 \\
0 & -3 & -2 & 1 \\
0 & -3 & -2 & 1
\end{array}\right]
$$

$$
\left[\begin{array}{rrr|r}
1 & 4 & 2 & 0 \\
0 & -3 & -2 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]
$$

EDIT : Okay, so I worked it out using elimination instead of the augmented matrix approach and got x = 4/3, y = -1/3 and z = 0, they are valid solutions but I need to find the general solution. How can I go about to find this?

Best Answer

You can try to continue with Gaussian elimination, but taking $z$ as a constant and reducing the system to a two variable problem: $$ \left[\begin{array}{rrr|r} \color{green}{1} & \color{green}{4} & \color{red}{2} & 0 \\ \color{green}{0} & \color{green}{-3} & \color{red}{-2} & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \longrightarrow \left[\begin{array}{rr|r} \color{green}{1} & \color{green}{4} & 0\color{red}{-2z} \\ \color{green}{0} & \color{green}{-3} & 1\color{red}{+2z} \\ \end{array}\right]$$ Note that you need to change the sign of the column corresponding to $z$, since you are moving it to the right-hand side of the equations.

You will obtain the general solution in terms of $z$.

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