# 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. Jul 26, 2012 at 18:07
• There is no obvious pattern from the eigenvalue/eigenvectors or the svd. Jul 26, 2012 at 18:13
• "Clearly"?? $\$
– user856
Jul 26, 2012 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!" Jul 26, 2012 at 18:43
• It's clearly 1664606914601. Jul 27, 2012 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 $n \in \mathbb{N}$? Jul 27, 2012 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 $n$. Jul 27, 2012 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.) Mar 26, 2016 at 11:52