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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?

Rodrigo de Azevedo
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Yongkai
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1 Answers1

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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.

André Nicolas
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    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 01:46
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    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 02:33
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    @user7530 How is the determinant a polynomial in the matrix entries? Why does this (non-zero) result, for the determinant modulo $n$ agree with the result of original determinant? – Mussé Redi Jun 01 '14 at 14:23
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    @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. – user7530 Jun 01 '14 at 15:47
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    @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? – Mussé Redi Jun 01 '14 at 16:12
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    @MusséRedi For your second question, it doesn't. But $(\det M)\mod 2 = \det(M\mod 2) \mod 2$ so if the determinant is 1 mod 2, it must be odd and therefore nonzero mod nothing. – user7530 Jun 01 '14 at 16:35
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    @André Nicolas If the det(m mod 2) would have been even then what does it imply? what I got that, if its odd then only it implies that the original determinant is non zero. If its even it does not convey anything. Means in case of even result the original determinant may be zero or non zero. Am I right? – ViX28 Mar 26 '16 at 08:11
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    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