# Find the general solution of the system

Given:

\begin{cases} x- z = 1 \\ y + 2z - w = 3 \\ x + 2y + 3z - w = 7 \end{cases}

My work: Use Gauss method Leading variables: x,y, and z Free variables: w Express the leading variables x,y,and z in terms the free variable w.

1) I perform $-2R_2+R_3$ and got $x-z+w=1$ which replaces the third row of the system.

2) Then I perform $-R_1+R_3$ and got $w = 0$ which replaces the third row of the system.

3) Finally I perform $R_2+R_3$ and got $y+2z=3$ which replaces the third row of the system.

The system at the end looks like this:

\begin{cases} x- z = 1 \\ y + 2z - w = 3 \\ y = 3 -2 z \end{cases}

I am stuck as to how to get the leading variables in terms of the free variables since I know the value of the free variable now. But plugging in the value of w to the equations causes the free variable to disappear...

It appears you wrote the system as: $x, y, z, w$, so I followed suit.

The RREF gives:

• $w = 0$
• $y = 3 -2z$
• $x = 1 + z$

$z$ is a free variable.

• Oh I see w = 0 plugging makes z a free variable. So I am suppose to write it now in terms of z. Commented Dec 17, 2013 at 19:13
• That is correct. Thank you for showing your work and great job! Recall to upvote and/accept answers that are helpful! Regards Commented Dec 17, 2013 at 19:15
• You were busy again today! +1 Commented Dec 19, 2013 at 0:06
• @amWhy: Thanks, just saw some things I could not resist. Review queue is under 100, cannot wait! Commented Dec 19, 2013 at 2:37

As you can see, you have a linear system of three equations in four unknowns. So at least one of the variables $x, y, w, z$ will be a free variable and in this case, since it turns out that $w$ is identically zero, its value is fixed, and so it cannot be a free variable.

What we end with, if you use an augmented coefficient matrix to row reduce, and with columns representing the coefficients of each variable, ordered as follows: $x\;y\;w \; z$, , is

$$\begin{pmatrix} 1 & 0 & 0 & -1 &\mid& 1\\ 0 & 1 & 0 & 2 &\mid& 3 \\ 0 & 0 & 1 & 0 &\mid& 0 \\ 0 & 0 & 0 & 0 &\mid& 0\end{pmatrix}$$

corresponding to the system of equations: \begin{align} x & = 1 + z \\ \\ y & = 3 - 2z \\ \\ w & = 0 \\ \\ z &= t\end{align} where $z = t$ simply means that we have $z$ as the free variable, and we denote its value by the parameter $t$.

Then, the general solution is given by $$\begin{pmatrix} x \\ y \\ w \\ z \end{pmatrix}=\begin{pmatrix} 1 \\ 3 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ -2 \\ 0 \\ 1\end{pmatrix}$$

For a particular solution, just let the parameter $z = t = 0$.

• I think the coefficient matrix for the first row should be 1 0 -1 0 | 1 right? Commented Dec 17, 2013 at 19:20
• No, the fourth column I've designated as representing the coefficient of $z$. See the column vector: ordered x, y, w, z (just so we can put the zero row at the bottom in the augmented matrix). Commented Dec 17, 2013 at 19:37
• But z is free not w. Even with your ordered x,y,w,z. Should it not be that the third row of the coefficient matrix be 0 0 0 0 and the fourth row should be 0 0 0 1? Commented Dec 17, 2013 at 19:51
• The $1$ in the third row represents $1 \cdot w = 0$. The whole row of zeros at the bottom mean that z can be anything, since $0\cdot z = 0 \implies$ z can be any value whatsoever. Commented Dec 17, 2013 at 19:58
• The first row of the matrix means $1\cdot x - 1\cdot z = 1\iff x = 1 + z$. The second row of the matrix means $1\cdot y + 2z = 3\iff y = 3 - 2z$. Commented Dec 17, 2013 at 20:00

Note that $w$ is necessarily zero, since $(***) - 2(**) - (*) = w = 7 - 2\cdot 3 - 1 = 0$, where $(*)$ denotes the first equation, $(**)$ denotes the second and $(***)$ the third. Then look at the matrix over which you have to compute the inverse. It has rank $2$. Why? Do you see a problem? Hence, you can't simplify it beyond the parametric form you have it in with $x-z=1$ and $y=3-2z$.