1
$\begingroup$

The matrix in question is A = [(1,1,1),(1,2,k),(1,4,k^2)]. I know that I can row reduce the matrix to rref, which should in theory leave me with some k values in the matrix from which I can see what values will make the matrix invertible. However, Wolfram is telling me rref(A) is just the 3x3 identity, which would make me think A is invertible for all real k. However, I know from the determinant that k cannot be either 1 or 2. Is there some approach I am missing here?

$\endgroup$
1
  • $\begingroup$ It is rather obvious that $k=1,2$ give the third column equal to the first or second respectively. $\endgroup$ Commented Sep 6, 2014 at 19:15

1 Answer 1

2
$\begingroup$

$$\begin{pmatrix}1&1&1\\ 1&2&k\\ 1&4&k^2\end{pmatrix}\stackrel{R_2,R_3-R_1}\longrightarrow\begin{pmatrix}1&1&1\\ 0&1&k-1\\ 0&3&k^2-1\end{pmatrix}\stackrel{R_3-3R_2}\longrightarrow\begin{pmatrix}1&1&1\\ 0&1&k-1\\ 0&0&k^2-3k+2\end{pmatrix}$$

You need the last line is $\;\neq 0\;$ , and $\;k^2-3k+2=(k-1)(k-2)\;$ ...

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .