# 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, the matrix is invertible.

Is there a more elegant way to do this?

Is there a pattern among these entries?

• 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
• 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.) – André Nicolas Mar 26 '16 at 11:52