# General Solution of System Of Equations (with 3 variables)

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?

• Either you continue reducing it, or you start converting to algebra form: $x+4y+2z=0$, $-3y-2z=1$. – Kenny Lau May 5 '16 at 15:02
• $z$ should be arbitrary because you essentially have $0z=0$. The solution should be \begin{align} x & = \frac{2 (z+2)}{3} & y & =-\frac{2 z+1}{3} \end{align} In general you have to pick one variable as arbitrary (any of x, y or z) and solve for the other two. – ja72 May 5 '16 at 15:32
• The system is, in fact, consistent, but it doesn’t have an unique solution. – amd May 5 '16 at 18:11
• @KennyLau That is exactly how I obtained the solutions which I mentioned in the edit. – Abdullah Zameek May 6 '16 at 8:44
• @ja72 That seems to work here pretty well – Abdullah Zameek May 6 '16 at 8:47

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 $z$.
• That is, precisely, what I wrote above ;) You have the $3\times 2$ augmented matrix of a system of two equations and two variables ($x$ and $y$). The difference now is that you have to keep $z$ when solving the system. For instances, from the second row you have $$-3y=1+2z \Rightarrow y=\frac{1+2z}{-3}.$$ – AugSB May 6 '16 at 9:12