# Is the following matrix invertible?

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

Clearly its determinant is not zero and hence it is invertible.

Any more elegant way to do this?

Is there a pattern among these entries?

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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. –  M Turgeon Jul 26 '12 at 18:07
There is no obvious pattern from the eigenvalue/eigenvectors or the svd. –  copper.hat Jul 26 '12 at 18:13
"Clearly"?? $\$ –  Rahul Jul 26 '12 at 18:21
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!" –  DonAntonio Jul 26 '12 at 18:43
It's clearly 1664606914601. –  Matthew Crumley Jul 27 '12 at 2:49

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.
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 $n \in \mathbb{N}$? –  torrho Jul 27 '12 at 1:46
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 $n$. –  user7530 Jul 27 '12 at 2:33