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

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?

  • 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, 2012 at 18:07
  • 1
    $\begingroup$ There is no obvious pattern from the eigenvalue/eigenvectors or the svd. $\endgroup$
    – copper.hat
    Jul 26, 2012 at 18:13
  • 153
    $\begingroup$ "Clearly"?? $\ $ $\endgroup$
    – user856
    Jul 26, 2012 at 18:21
  • 18
    $\begingroup$ 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, 2012 at 18:43
  • 133
    $\begingroup$ It's clearly 1664606914601. $\endgroup$ Jul 27, 2012 at 2:49

1 Answer 1


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.

  • 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, 2012 at 1:46
  • 33
    $\begingroup$ 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, 2012 at 2:33
  • 3
    $\begingroup$ @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, 2014 at 15:47
  • 2
    $\begingroup$ @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? $\endgroup$ Jun 1, 2014 at 16:12
  • 6
    $\begingroup$ 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.) $\endgroup$ Mar 26, 2016 at 11:52

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