Determine the value of $k$ so that the columns in this matrix are linearly dependent:

$$\begin{bmatrix} k & -1/2 & -1/2\\ -1/2 & k & -1/2\\ -1/2 & -1/2 & k \end{bmatrix}$$

Well then, the columns are dependent if this system has infinite solutions, right? In that case, I can try to calculate the determinant - because if the determinant happens to be $0$, we'd have infinite solutions and we'd be done.

But how do I calculate the determinant in this case? I'm trying to reduce the matrix to a triangular form (so I can just multiply the diagonal), but am unable to get rid of either side to make it triangular.

  • $\begingroup$ What 5xum said. But if you really want to compute the determinant, there is a "somewhat nice" formula for 3x3 matrices en.wikipedia.org/wiki/Determinant (beginning of the page) $\endgroup$ – T_O Apr 9 '14 at 8:00

Add the second and the third column to the first one and then subtract the first row from the other rows and finally develop the determinant along the first column we find

$$\det\begin{bmatrix} k-1 & -1/2 & -1/2\\ 0 & k+1/2 & 0\\ 0 & 0 & k+1/2 \end{bmatrix}=(k-1)\left(k+\frac12\right)^2$$ hence the columns are linearly dependent iff the determinant is $0$ iff $k=1$ or $k=-\frac12$.


Hint: Don't use detrminants, try to directly see when the columns are linearly dependent. For example, there is one value of $k$ which makes them obviously linearly dependent.


Have you tried to calculate the determinant? It equals $k^3 - \frac34 k - \frac14$, so setting it to $0$ and calculating $k$ should not be too hard, considering you already know of one root ($-\frac12$).

  • $\begingroup$ $-1/2$ makes them linearly dependent, I think? But I guess there could be other values too. $\endgroup$ – Zol Tun Kul Apr 9 '14 at 7:58
  • $\begingroup$ @Omega Possibly, but you said "find the value of $k$". $\endgroup$ – Git Gud Apr 9 '14 at 7:59
  • $\begingroup$ @Omega I edited my answer. $\endgroup$ – 5xum Apr 9 '14 at 8:09

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