For which choice(s) of the constant $k$ is the following matrix invertible?
\[A=\begin{bmatrix}
1 & 1 & 1 \\
1 &2 &k \\
1 & 4 & k^2
\end{bmatrix}.\]  
(Johns Hopkins University, Linear Algebra Exam)
 

Hint.

An $n\times n$ matrix is invertible if and only if its rank is $n$.
The rank of a matrix is the number of nonzero rows of a (reduced) row echelon form matrix that is row equivalent to the given matrix.

Solution.

We compute the rank of the matrix $A$.
Applying elementary row operations, we obtain
\begin{align*}
\begin{bmatrix}
1 & 1 & 1 \\
1 &2 &k \\
1 & 4 & k^2
\end{bmatrix}
\xrightarrow{\substack{R_2-R_1 \\ R_3-R_1}}
\begin{bmatrix}
1 & 1 & 1 \\
0 & 1 &k-1 \\
0 & 3 & k^2-1
\end{bmatrix}
\xrightarrow{\substack{R_1-R_2 \\ R_3-3R_2}}
\begin{bmatrix}
1 & 0 & 2-k \\
0 &1 &k-1 \\
0 & 0 & k^2-3k+2
\end{bmatrix}.
\end{align*}
The last matrix is in row echelon form.

Note that $A$ is an invertible matrix if and only if its rank is $3$.
Therefore the $(3,3)$-entry of the last matrix must be nonzero: $k^2-3k+2=(k-1)(k-2)\neq 0$.

It follows that the matrix $A$ is invertible for any $k$ except $k=1, 2$.

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