Finding values of a matrix to make it consistent

Question

Determine the values of $k \in \mathbb{R}$ so that the following system $$ \left\{\begin{aligned} x+y-z &=2 \\ x+y &=1 \\ x+y+k z &=0 \end{aligned}\right. $$ consistent and compute the solution.

So I have a system of linear equations as seen on the image I attached up here.

$x+y-z=2$

$x+y=1$

$x+y+kz=0$

I want to find the values of $k$ in order to make the system consistent. As the question asks.

I was thinking , as $x+y=1$, to plug this in the first equation and get $z=-1$. Then I plug the $z=-1$ into the third equation to get $k =1$. So is the system consistent as long as $k$ is not equal to $1$?

Can somebody please help me ?

Answer 1

The system is consistent as long as $k$ is equal to $1$. When $k = 1$, the equations admit solutions. In fact, the equations admit an infinite family of solutions: any $(x, y, z)$ with $x + y = 1$ and $z = -1$ is a solution, i.e. the set of solutions when $k = 1$ is the set $\{(t, 1 - t, -1) : t \in \mathbb R \}$, which looks like a straight line in $\mathbb R^3$.

When $k \neq 1$, the equations are inconsistent, and have no solutions. Indeed, the first two equations tell us that $x + y = 1$ and $z = -1$, which implies that $x + y + kz = 1 - k$. But the third equation states that $x + y + kz = 0$. The conclusion is that the third equation cannot possibly be satisfied while the first two equations are satisfied if $k \neq 1$.

Answer 2

The key word here is "simultaneously." For the linear system to be consistent, all three of the equations must be satisfied simultaneously for at least one fixed triplet $(x,y,z)$. That is, there must be at least one set of values $x = a, y = b, z = c, (a, b,$ and $c$ real numbers), for which ALL of the equations hold. By choosing $k = 1$, we guarantee this is the case.