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Let the matrix of dimension 4 be:

$$A=\begin{bmatrix} a11 & a12 & a13 & a14\\ a21 & a22 & a23 & a24\\ a31 & a32 & a33 & a34\\ a41 & a42 & a43 & a44 \end{bmatrix}$$

Also, let the matrix B be:

$$B=\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\ 0 & a11 & a12 & a13 & a14\\ 0 & a21 & a22 & a23 & a24\\ 0 & a31 & a32 & a33 & a34\\ 0 & a41 & a42 & a43 & a44 \end{bmatrix}$$

Now, by Laplace Expansion (cofactor expansion), we can say that:

$det(A)=1\cdot det(B)$

But by Leibniz formula ("Sarru's rule") we have that:

$det(B) = (1 \cdot a11\cdot a22\cdot a33\cdot a44) - (a41\cdot a32\cdot a23\cdot a14\cdot 1)$

I have a strong feeling that I'm applying Leibniz formula wrongly but could someone explain how?

PauloB
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1 Answers1

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The problem is Sarru's rule only works when the dimension is $3$.

Indeed, one formal definition of the determinant is

$$\sum_{\sigma \in S_n} \varepsilon(\sigma) \prod_{i=1}^n a_{i, \sigma (i)}$$

which looks scary (and might be to some extent), but all it's doing is taking all possible permutations of the row and column indices and multiplying those elements together (multiplied by $1$ or $-1$, depending on some property of that permutation). It just so happens that this follows a "nicer-looking" rule when the dimension of the matrix is $3$, and so that is what is (in my experience) commonly taught in schools before advancing to this level.

Even if you don't yet understand this notation, one take away is that the sum contains $|S_n| = n!$ terms in the expansion. Sarru's rule only gives you $2n$ terms. Hence it could only possibly work when $n=3$.

MT_
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