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we have the determinant:

$$ \begin{vmatrix} 6&1&1\\ 4&-2&5\\ 2&8&7\\ \end{vmatrix} $$ A stumbled upon this question: Is the following matrix invertible?

If I apply the modulo 2 procedure to this matrix, I get a zero determinant, which is not true. Why doesn't that procedure work in this case?

VLC
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2 Answers2

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The modulo $2$ idea does not work because this determinant, albeit nonzero, is divisible by $2$. (In fact, it is $-306$.) If something is nonzero modulo $2$, it is certainly nonzero, but if something is zero modulo $2$, this only means that it is even but doesn't mean it is zero.

You can still apply the same technique, but first divide the first column by $2$: you get the determinant:

$$\begin{vmatrix}3&1&1\\2&-2&5\\1&8&7\end{vmatrix}$$

which is, modulo $2$:

$$\begin{vmatrix}1&1&1\\0&0&1\\1&0&1\end{vmatrix}=1$$

so this new determinant is nonzero, and so the original determinant (which is twice the new determinant) must be nonzero as well.

  • Could we also use modulo 1 without using matrix transformations? – VLC Dec 31 '20 at 13:43
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    No, every number modulo $1$ is zero because every number is divisible by $1$. Thus, your matrix $\pmod 1$ will always be a zero matrix and you will not be able to conclude anything. –  Dec 31 '20 at 13:44
  • Oh I see.... But I guess we can use any modulo that is an integer and larger than 1 ? – VLC Dec 31 '20 at 13:45
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    @BiliDebili You can use some modulus that happens not to divide the actual determinant of the matrix. so in this case $\pmod 4$ would probably work. However, the bigger modulus - the more work to calculate the determinant - until it gets as bad as calculating the original determinant.... –  Dec 31 '20 at 13:46
  • The original determinant is -306, not twice the new determinant, if I am not mistaking. – VLC Dec 31 '20 at 13:50
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    @BiliDebili The original determinant is -306, the new determinant is -153, and the original determinant is twice the new determinant. (Dividing a row or a column by a number divides the determinant by the same number.) –  Dec 31 '20 at 13:51
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In the given link, we can conclude that the matrix is invertible since its determinant is $1$ modulo $2$, which cannot be $0$, clearly. However, this procedure cannot be used to show that matrix is not invertible, since if it tells you that determinant is $0$ modulo $2$ the determinant still can be an even number other than $0$.

ArsenBerk
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  • So we can use let's say modulo 1 and get a non zero matrix? – VLC Dec 31 '20 at 13:42
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    @ArsenBerk Usually modulo $1$ refers to the equivalence relation, for $x,y\in \mathbb R$, that $x\sim y\iff x-y\in\mathbb Z$. – Kenta S Dec 31 '20 at 13:49
  • @KentaS Yes, you are right. That's just what happens when you struggle with algebra too much. Because in rings, we can say that a unit is a divisor of every element and $1$ is a unit of the ring $\mathbb{R}$. But for this context, I should have said integer instead of real number because of the modulo, as you said. Thank you for the correction :) – ArsenBerk Dec 31 '20 at 13:55
  • How about this example @ArsenBerk :

    $$ \begin{vmatrix} \lambda & 1 & 1\ 1 & \lambda & 1\ 1 & 1 & \lambda\ \end{vmatrix} $$ If $\lambda$ is even then the diagonal is zero. The determinant is non zero. But I tried for: $$ \begin{vmatrix} -2 & 1 & 1\ 1 & -2 & 1\ 1 & 1 & -2\ \end{vmatrix} $$ And the determinant is zero...why is that so ?

     – VLC Dec 31 '20 at 16:18
  • I am not sure if I understand your question correctly but in your first example, if we are considering the determinant in modulo $2$, then determinant also becomes $0$ modulo $2$. And your second example is a determinant whose value is $0$, which is $0$ modulo $2$. So I don't see any problem for your first example to be $0$ modulo $2$. – ArsenBerk Dec 31 '20 at 18:04