Solution of System of linear equations

I have three equations: $$\begin{cases} 4y + z = 2\\ 2x + 6y - 2z = 3\\ 4x + 8y - 5z = 4 \end{cases}$$

Applying Gauss elimination I get:

$$\left[ \begin{array}{ccc|c} 1&0&-\frac{7}{4} & 0\\ 0 & 1 & \frac{1}{4} & \frac{1}{2} \\ 0 & 0 & 0 & 0 \end{array} \right]$$

Now how can I get the solution as there one row is eliminated leaving 2 equations with 3 unknowns?

• shows that there is an infinite number of solutions. You should choose one of the unknown like z as parameter and compute the other unknowns x and y based on z – Sepideh Abadpour Aug 14 '15 at 18:01

You can arbitrarily choose $z$. Then, $x$ and $y$ can be computed from that: $$x - \frac{7}{4}z = 0 \;\Rightarrow\; x = \frac{7}{4} z\\ y + \frac{1}{4}z = \frac{1}{2} \;\Rightarrow\; y=\frac{1}{2} - \frac{1}{4}z$$
The solution of your equations would be the vector $$\begin{bmatrix}x\\y\\z\\\end{bmatrix} = \begin{bmatrix}7/4z\\1/2 - 1/4z\\z\\ \end{bmatrix} = \begin{bmatrix}0\\1/2\\0 \end{bmatrix}+\begin{bmatrix}7/4\\-1/4\\1\\ \end{bmatrix}z$$ for some $z$.