135
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$$\begin{bmatrix} 1235 &2344 &1234 &1990\\ 2124 & 4123& 1990& 3026 \\ 1230 &1234 &9095 &1230\\ 1262 &2312& 2324 &3907 \end{bmatrix}$$

Clearly, its determinant is not zero and, hence, the matrix is invertible.

Is there a more elegant way to do this?

Is there a pattern among these entries?

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  • 3
    $\begingroup$ I don't know which one is easier, but you could try to row-reduce it. This way, it's invertible if and only if it is full-rank. $\endgroup$ – M Turgeon Jul 26 '12 at 18:07
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    $\begingroup$ There is no obvious pattern from the eigenvalue/eigenvectors or the svd. $\endgroup$ – copper.hat Jul 26 '12 at 18:13
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    $\begingroup$ "Clearly"?? $\ $ $\endgroup$ – Rahul Jul 26 '12 at 18:21
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    $\begingroup$ Yeah...that "clearly" is the way some have to say "Hey, I already calculated (or better: some programm did it for me) this ugly thing's determinant and found out it is not zero...so I'll show off and tell you that "cloearly" it is not zero!" $\endgroup$ – DonAntonio Jul 26 '12 at 18:43
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    $\begingroup$ It's clearly 1664606914601. $\endgroup$ – Matthew Crumley Jul 27 '12 at 2:49
473
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Find the determinant. To make calculations easier, work modulo $2$! The diagonal is $1$'s, the rest are $0$'s. The determinant is odd, and therefore non-zero.

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  • 17
    $\begingroup$ Since I'm not cool enough to ask in Andre' Nicolas' comments about his solution, I am asking Andre Nicolas' or anybody else in this answer body: why can we go from a what I assume is $M_4 \left( \mathbb{Z} \right)$ to $M_4 \left( \mathbb{Z}/2\mathbb{Z} \right)$? I like the idea for its simplicity but will this always work if we start with a matrix in $M_4 \left( \mathbb{Z} \right)$ or in general $M_n \left( \mathbb{Z} \right)$ for $n \in \mathbb{N}$? $\endgroup$ – torrho Jul 27 '12 at 1:46
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    $\begingroup$ Modulo "moves into" addition, multiplication, and the base of exponentiation, ie $a + bc^d \mod n = (a \mod n) + (b \mod n)(c\mod n)^d \mod n$. Since the determinant is a polynomial in the matrix entries, reducing the entries modulo $n$ does not change the value of the determinant module $n$. $\endgroup$ – user7530 Jul 27 '12 at 2:33
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    $\begingroup$ @MusséRedi You can see this straight from the permutation definition of the determinant: the determinant is a linear combination of products of the matrix entries. Or you can see by induction from Laplace's formula that the determinant is polynomial. $\endgroup$ – user7530 Jun 1 '14 at 15:47
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    $\begingroup$ @user7530 Okay, I was thinking about the usual expression of a polynomial of some variable but by polynomial you mean a linear combination of products of the matrix entries. Could you also answer my second question? $\endgroup$ – Mussé Redi Jun 1 '14 at 16:12
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    $\begingroup$ You are right, if the determinant modulo $2$ were $0$, we could draw no conclusion about whether the matrix is invertible. (But another modulus might yield useful information.) $\endgroup$ – André Nicolas Mar 26 '16 at 11:52

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