simpler way to calculate a determinant?

Simpler way to calculate this?

$$A = \begin{bmatrix}\lambda -2 & 2 & 0 \\ 2 & \lambda -1 & 2 \\ 0 & 2 & \lambda \end{bmatrix}$$

my method:

\begin{align*} \det A &= \det \begin{bmatrix}\lambda -2 & 2 & 0 \\ 0 & \lambda -1 & 2 \\ -\lambda & 2 & \lambda \end{bmatrix} \\ &= \det \begin{bmatrix} \lambda -2 & 2 & 0 \\ 0 & \lambda -1 & 2 \\ -2 & 4 & \lambda \end{bmatrix} \\ &= \det \begin{bmatrix} \lambda -2 & 2 & 0 \\ 0 & \lambda -1 & 2 \\ 0 & 4 & \lambda \end{bmatrix}\end{align*}

After stuck for a long time, I decided to calculate by diagonal rule,

$$(\lambda -1)^2\lambda -8(\lambda -1)=(\lambda -1)(\lambda ^2-\lambda -8)$$

The rule of Sarrus hinted by @Dietrich below: \begin{align} (\lambda -2)(\lambda -1)\lambda -4\lambda -4(\lambda -2) &= (\lambda -2)(\lambda -1)\lambda -4(2\lambda -2) \\ &= (\lambda -1)((\lambda -2)\lambda -8) \\ &= (\lambda-1)\left(\lambda ^2-2\lambda -8\right) \\ &= (\lambda -1)(\lambda -4)(\lambda +2)\end{align}

• Could you elaborate on what your method is? I don't see what you're doing. – DBFdalwayse Sep 6 '13 at 8:32
• @DBF seems I've made an error. wait... – realmatrix Sep 6 '13 at 8:33
• @DBF, the determinant is invariant under the first kind of row/column operation. I don't see what operation we did in the last line, though. – Jonathan Y. Sep 6 '13 at 8:34
• There. Finally fixed the formatting so it's possible to see what's going on. Now whether the computations are correct... – kahen Sep 6 '13 at 8:36
• You may also use the Laplace's formula (since your matrix has two zero terms). More details can be found here : link. – user37238 Sep 6 '13 at 8:42

The rule of Sarrus (see http://en.wikipedia.org/wiki/Rule_of_Sarrus) is quite convenient here. You should obtain $$\det (A)=(\lambda-2)(\lambda-1)\lambda-4\lambda-4(\lambda-2)=(\lambda -1)(\lambda+2)(\lambda-4).$$
• +1 I always urge students using this way for a $3\times 3$ matrix. – mrs Sep 6 '13 at 8:48