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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!

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    $\begingroup$ Replace $C_3$ by $C_3+C_2$, factor $3-\lambda$ in $C_3$, replace $C_2$ by $C_2-C_1-C_3$, factor $3-\lambda$ in $C_2$, replace $C_1$ by $C_1-C_2$, observe that the resulting matrix is upper triangular with diagonal $(3-\lambda,1,1)$, qed. $\endgroup$ – Did Sep 1 '16 at 11:49
  • $\begingroup$ @Did Please add this as answer. Thank you! $\endgroup$ – MM PP Sep 1 '16 at 11:50
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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.

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  • $\begingroup$ Thank you Kaligule, I tried to make some operations with the rows and/or the columns, but I was unable to compute this determinant, using this method. I was able to compute the determinant using Sarrus. $\endgroup$ – MM PP Sep 1 '16 at 11:43

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