Question

Solving a System of Equations in Three Variables Find the complete solution of the linear system, or show that it is inconsistent.

$\left\{\begin{array}{l}x+y-z=0\\ x+2y-3z=-3\\ 2x+3y-4z=-3\end{array}\right\}$

Short answer

The solution set of the system$\left\{\begin{array}{l}x+y-z=0\\ x+2y-3z=-3\\ 2x+3y-4z=-3\end{array}\right\}$has infinitely many solutions and the system is

Step-by-step solution

Step 1. Given information

The given system of equations is$\left\{\begin{array}{l}x+y-z=0\\ x+2y-3z=-3\\ 2x+3y-4z=-3\end{array}\right\}$

Step 2. Find the complete solution of the given system of linear equations and show if it is inconsistent.

We will convert the system into triangular form that has the same solutions as the original system using gauss elimination method.

We will solve triangular form by using back substitution and check the consistency of the system.

The given system of equations is $\left\{\begin{array}{ll}x+y-z=0& \dots \dots \left(1\right)\\ x+2y-3z=-3& \dots \dots \left(2\right)\\ 2x+3y-4z=-3& \dots \dots \left(3\right)\end{array}\right.$

We can see that each equation is written in standard form,

$Ax+By+Cz=D$

We will begin by eliminating xfrom equation (2),

$\left\{\begin{array}{ll}x+y-z=0& \dots \mathrm{..}\left(1\right)\\ -y+2z=3& \dots \mathrm{..}\left(2\right)\\ 2x+3y-4z=-3& \dots \mathrm{..}\left(3\right)\end{array}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}equation}1-\text{equation}2\right.$

Now we will eliminate x from equation (3),

$\left\{\begin{array}{ll}x+y-z=0& \dots \dots \left(1\right)\\ -y+2z=3& \dots \dots \left(2\right)\\ -y+2z=3& \dots \dots .\left(3\right)\end{array}2\times \text{equation}1-\text{equation}3\right.$

On doing, equation 2 - equation 3 our system reduces to,

$\left\{\begin{array}{ll}x+y-z=0& \dots \mathrm{..}\left(1\right)\\ -y+2z=3& \dots \mathrm{..}\left(2\right)\\ 0=0& \dots \mathrm{..}\left(3\right)\end{array}\right.$

Here, the third equation is true but it gives us no new information, so we can drop it from the system. We will use remaining two equations to solve for x, y in terms of z but z can take any value

Hence, there are infinitely many solutions.