How to compute the determinant of some 3x3 matrix more quickly than by Sarrus method? I know that the determinant of this matrix: $ \begin{bmatrix} 2-\lambda & -1 & 1\\1& 5-\lambda & -2 \\ 0 & 1 & 2-\lambda \end{bmatrix} $ is $(3-\lambda)^3$ by simply computing the determinant (using Sarrus method), but I want to know if I can compute this determinant faster, by doing some operations with the rows and columns.
Thank you for your help!
 A: For $3 \times 3$ Matrices, Sarrus is the fastest way to go. You might find matrices that have certain features (a coloumn or a row with no or only one nonzero element), which can give you a boost, but in general (as in this case), you should use Sarrus.

Example where you can go faster using Laplace's formula:
$$M=\begin{bmatrix} 2-\lambda & -1 & 1\\0& 5-\lambda & -2 \\ 0 & 1 & 2-\lambda \end{bmatrix}$$
As you can see, this is very similar to your matrix, but the first row has only one element different from zero ($2-\lambda$). If this is not the case, Laplace's won't save you work on $3\times 3$ or smaller.
So we go down the first row with Laplace:
\begin{align}
\det(M) &= \det\begin{pmatrix} 2-\lambda & -1 & 1\\0& 5-\lambda & -2 \\ 0 & 1 & 2-\lambda \end{pmatrix}\\
 &= (2-\lambda) \cdot \det\begin{pmatrix} 5-\lambda & -2 \\ 1 & 2-\lambda \end{pmatrix}
-0  \cdot \det\begin{pmatrix} -1 & 1\\1 & 2-\lambda \end{pmatrix}
+0  \cdot \det\begin{pmatrix} -1 & 1\\5-\lambda & -2 \end{pmatrix}\\
&=(2-\lambda) \cdot \det\begin{pmatrix} 5-\lambda & -2 \\ 1 & 2-\lambda \end{pmatrix}
\end{align}
... and this is much easier to calculate.
For more detail, look here.
