Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to find the eigenvalues of a matrix and I cannot remember how to find the determinant of $A-\lambda I$:

\begin{equation} \pmatrix{1-\lambda& 2& 1 \\ 2& -\lambda & -2 \\ -1& 2& 3-\lambda} \end{equation}

I know the rules for finding the determinant of an $m \times n$ matrix.

  1. $\det A = -\det A'$ when switching a row
  2. $\det A = \det A'$ when performing the replacement row operating
  3. $\det A = \dfrac{1}{\lambda}\det A'$ when scaling a row

Then I row reduce until I have a diagonal matrix...

Do I have to use the above rules or is there a simpler way? I am not looking to use the formula for a $3\times 3$ matrix. I am looking for a more general solution so that I will be able to solve this problem for a $4 \times 4$ matrix or $5 \times 5$ matrix.

share|cite|improve this question
Does this help Det of $nxn$ Matrix and Examples? – Amzoti Dec 16 '12 at 2:46
For a $4 \times 4$ or larger matrix, unless there is special structure that makes it easy, finding the determinant by hand is really a waste of time. Use a computer algebra system or Wolfram Alpha: e.g.… – Robert Israel Dec 16 '12 at 3:35
up vote 4 down vote accepted

The standard way to calculate the determinant of small matrices manually is the Laplace expansion:

share|cite|improve this answer

My favorite method for finding $3\times3$ determinants is shown here. For larger matrices, you'll have to resort to the more general formula given in the Wikipedia article on determinants.

share|cite|improve this answer
He wanted a general approach, not just $3\times3$. – Amzoti Dec 16 '12 at 2:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.