Finding $k$ to make a linear system inconsistent

Question

I've been trying to solve the following problem for quite a while now.

Given the linear system

$$\begin{bmatrix}1&1&1\\2&4&4\\3&7&k\end{bmatrix} x= \begin{bmatrix}1\\2\\3\end{bmatrix}$$

for what values of $k$ are there no solutions or infinitely many solutions?

I tried two separate ways of doing this, which I think are correct. First, by setting the determinant equal to $0$ and finding $k=7$, then by creating the echelon matrix

$$\begin{bmatrix}1&1&1&1\\0&2&2&0\\0&0&(k-7)&0\end{bmatrix}$$

This gives the same value. This matrix is difficult for me to interpret. If I choose $k=7$, I make the third row accurate which tends to point to infinitely many solutions, but I do not think the second row makes sense in that case. I can understand there could be unique solution(s) if $k \neq 7$, but it is hard to tell if that equation above is satisfied just by having $k=7$.

Answer 1

In general if you have a linear equation $Ax = b$ then you know it has a unique solution if $A$ is invertible, i.e. $\det(A) \neq 0$.

If you set $k = 7$ then the Matrix $A$ isn't invertible anymore. $\begin{bmatrix}1&1&1&1\\0&2&2&0\\0&0&(k-7)&0\end{bmatrix}$ is just another way of writing $Bx = s$ where $$ B = \begin{bmatrix}1&1&1\\0&2&2\\0&0&0\end{bmatrix}, s = \begin{bmatrix}1\\0\\0\end{bmatrix}, $$ if you write x = \begin{bmatrix}y\\z\\w\end{bmatrix} you get $Bx = \begin{bmatrix}y + z + w\\z + w\\0\end{bmatrix} = \begin{bmatrix}1\\0\\0\end{bmatrix}$, so you get two equations \begin{align*} y+z+w = 1 \\ z + w = 1 \end{align*} and this has infinitely many solutions.

Answer 2

Hint:

I suppose $x$ denotes a vector in $K^3$ ($K$ being the base field). A nonhomogeneous linear system has solution(s) if and only the matrix of the l.h.s. and the augmented matrix have the same rank.

When they have the same rank, this rank is the codimension of the affine subspace of solutions.

Thus it wall have a unique solution if and only if the matrix of the l.h.s. has full rank.