Timeline for Why does this matrix have zero determinant?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Oct 27, 2019 at 11:29 | vote | accept | Hailong Dao | ||
| Oct 27, 2019 at 8:47 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Oct 27, 2019 at 8:40 | comment | added | Zach Teitler | So in other words you are saying that the vector $(\alpha^2,\alpha\beta,\beta^2,-\gamma^2,-\gamma\delta,-\delta^2)$ is in the kernel of the matrix. Since the $\alpha,\beta,\gamma,\delta$ are cubic in the vectors $v_1,v_2,v_3$, this explains the degree $6$ polynomials. | |
| Oct 27, 2019 at 6:45 | history | answered | Fedor Petrov | CC BY-SA 4.0 |