1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Either you continue reducing it, or you start converting to algebra form: $x+4y+2z=0$, $-3y-2z=1$. $\endgroup$ – Kenny Lau May 5 '16 at 15:02
  • $\begingroup$ $z$ 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. $\endgroup$ – ja72 May 5 '16 at 15:32
  • $\begingroup$ The system is, in fact, consistent, but it doesn’t have an unique solution. $\endgroup$ – amd May 5 '16 at 18:11
  • $\begingroup$ @KennyLau That is exactly how I obtained the solutions which I mentioned in the edit. $\endgroup$ – Abdullah Zameek May 6 '16 at 8:44
  • $\begingroup$ @ja72 That seems to work here pretty well $\endgroup$ – Abdullah Zameek May 6 '16 at 8:47
1
$\begingroup$

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 $z$.

$\endgroup$
  • $\begingroup$ Thank you for the solution but, to be honest, I haven't studied Gaussian Elimination to solve system of equations so I was kinda hoping to solve it in the augmented matrix approach (which is what I was expected to do) $\endgroup$ – Abdullah Zameek May 6 '16 at 8:50
  • $\begingroup$ 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 $$-3y=1+2z \Rightarrow y=\frac{1+2z}{-3}.$$ $\endgroup$ – AugSB May 6 '16 at 9:12
  • $\begingroup$ Oh...alright. I conpletely misinterpreted what you wrote above. Sorry about that. I never considered shifting z to the right hand side but it seems like a pretty neat way to go about this sum. Thanks! $\endgroup$ – Abdullah Zameek May 6 '16 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.