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\begin{vmatrix} \lambda & 1 & 1\\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{vmatrix}

If we want to see that the determinant is non zero we can do modulo 2, and we get if $\lambda$ is even: $$ \begin{vmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}$$

The determinant is in this case not zero.

But... if we input $\lambda = -2$, then the determinant is zero. Shouldn't the determinant be non zero for every even number in the diagonal since it's modulo 2 determinant is not zero.

EDIT: We want to see if determinant is non zero in integers and used modulo 2 only as a tool for quicker computation. Like shown here in the first answer Is the following matrix invertible?

VLC
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    the determinant is zero modulo 2 in both cases – Exodd Dec 31 '20 at 16:41
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    The matrix in the attached link is invertible because the determinant is $1$ modulo $2$ and hence cannot be $0$. However, if the determinant is $0$ modulo $2$, it might be $0$ or some non-zero even number in $\Bbb Z$. – Shubham Johri Dec 31 '20 at 16:52
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    The determinant is $\lambda^3-3\lambda+2=(\lambda-1)^2(\lambda+2).$ In particular, it is even when $\lambda$ is an integer. – Thomas Andrews Dec 31 '20 at 16:57

1 Answers1

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$$ \begin{vmatrix} 0 & 1 & 1\\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix}=2\equiv0\pmod2$$

J. W. Tanner
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  • Okay, but if we input -4 then it's non zero again. – VLC Dec 31 '20 at 16:43
  • With $\lambda=-4,$ the determinant is non-zero in $\mathbb Z$, but $0$ modulo $2$ – J. W. Tanner Dec 31 '20 at 16:44
  • I used modulo 2 as a tool to see if it is non-zero in integers. Does everything still imply anyways? I wanted to show like the first answer here: https://math.stackexchange.com/questions/175561/is-the-following-matrix-invertible – VLC Dec 31 '20 at 16:45
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    if the determinant mod $2$ is $1$, then the determinant is non-zero, but if it’s even, you can’t conclude it’s non-zero – J. W. Tanner Dec 31 '20 at 17:01