# Quod Erat Demonstrandum

## 2017/04/16

### 行列式特性

Filed under: NSS,Pure Mathematics — johnmayhk @ 5:05 下午
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Factorize $\left|\begin{array}{rcl}a &b &c\\b+c &c+a &a+b \\a^2 &b^2 &c^2\\\end{array}\right|$.

$\left|\begin{array}{rcl}a+b+c &a+b+c &a+b+c\\b+c &c+a &a+b \\a^2 &b^2 &c^2\\\end{array}\right|$

$=(a+b+c)\left|\begin{array}{rcl}1 &1 &1\\b+c &c+a &a+b \\a^2 &b^2 &c^2\\\end{array}\right|$

$\left|\begin{array}{rcl}a &b-a &c-a\\b+c &a-b &a-c \\a^2 &b^2-a^2 &c^2-a^2\\\end{array}\right|$

$\left|\begin{array}{rcl}a &b-a &c-a\\b+c &a-b &a-c \\a^2 &b^2-a^2 &c^2-a^2\\\end{array}\right|$

$=(a-b)(c-a)\left|\begin{array}{rcl}a &-1 &1\\b+c &1 &-1 \\a^2 &-b-a &c+a\\\end{array}\right|$

$=(a-b)(c-a)\left|\begin{array}{rcl}a &-1 &0\\b+c &1 &0 \\a^2 &-b-a &c-b\\\end{array}\right|$

$=(a-b)(c-a)(c-b)\left|\begin{array}{rcl}a &-1 \\b+c &1 \\\end{array}\right|$

$=(a-b)(c-a)(c-b)(a+b+c)$

Factorize $\left|\begin{array}{rcl}(a-x)^2 &(a-y)^2 &(a-z)^2\\(b-x)^2 &(b-y)^2 &(b-z)^2 \\(c-x)^2 &(c-y)^2 &(c-z)^2\\\end{array}\right|$.

$|MN|=|M||N|$

$\left|\begin{array}{rcl}(a-x)^2 &(a-y)^2 &(a-z)^2\\(b-x)^2 &(b-y)^2 &(b-z)^2 \\(c-x)^2 &(c-y)^2 &(c-z)^2\\\end{array}\right|$

$=\left|\begin{array}{rcl}a^2 &-2a &1\\b^2 &-2b &1 \\c^2 &-2c &1\\\end{array}\right|$　$\left|\begin{array}{rcl}1 &1 &1\\x &y &z \\x^2 &y^2 &z^2\\\end{array}\right|$

$|MN|=|M||N|$

Factorize $\left|\begin{array}{rcl}2 &a+b &a^2+b^2\\a+b &a^2+b^2 &a^3+b^3 \\1 &c &c^2\\\end{array}\right|$.

$\left(\begin{array}{rcl}2 &a+b &\\a+b &a^2+b^2\\\end{array}\right)=\left(\begin{array}{rcl}1 &1 &\\a &b\\\end{array}\right)\left(\begin{array}{rcl}1 &a &\\1 &b\\\end{array}\right)$

Factorize $\Delta =\left|\begin{array}{rcl}(b+c)^2 &ab &ac\\ab &(c+a)^2 &bc \\ac &bc &(a+b)^2\\\end{array}\right|$.

$\left|\begin{array}{rcl}(b+c)^2 &0 &0 \\0 &c^2 &bc \\ 0 &bc &b^2\\\end{array}\right|=c^2b^2-(bc)(bc)=0$

$\left|\begin{array}{rcl}(b+c)^2 &(-b-c)b &(-b-c)c \\(-b-c)b &(-b)^2 &bc \\ (-b-c)c &bc &(-c)^2\\\end{array}\right|=(b+c)bc\left|\begin{array}{rcl}b+c &-(b+c) &-(b+c) \\-b & b & b \\ -c &c &c\\\end{array}\right|$

$=0$

$a+b+c$ 也是因子。

$\Delta$ 可表為

$abc(a+b+c)Q(a,b,c)$

$\left|\begin{array}{rcl}(a+c)^2 &ba &bc\\ba &(c+b)^2 &ac \\bc &ac &(b+a)^2\\\end{array}\right|$

$=-\left|\begin{array}{rcl}ab &(c+a)^2 &bc\\(b+c)^2 &ab &ac \\ac &bc &(a+b)^2\\\end{array}\right|$

$=\left|\begin{array}{rcl}(b+c)^2 &ab &ac\\ab &(c+a)^2 &bc \\ac &bc &(a+b)^2\\\end{array}\right|$

$Q(a,b,c)=\alpha(a^2+b^2+c^2)+\beta(ab+bc+ca)$

$\left|\begin{array}{rcl}(b+c)^2 &ab &ac\\ab &(c+a)^2 &bc \\ac &bc &(a+b)^2\\\end{array}\right|=abc(a+b+c)(\alpha(a^2+b^2+c^2)+\beta(ab+bc+ca))$

$\left|\begin{array}{rcl}4 &1 &1\\1 &4 &1 \\1 &1 &4\\\end{array}\right|=9(\alpha+\beta)$

$\Rightarrow \alpha+\beta=6$……………….. (1)

$\left|\begin{array}{rcl}0 &1 &-1\\1 &0 &-1 \\-1 &-1 &4\\\end{array}\right|=-3\alpha+\beta$

$\Rightarrow -3\alpha+\beta=-2$……………….. (2)

$\left|\begin{array}{rcl}(b+c)^2 &ab &ac\\ab &(c+a)^2 &bc \\ac &bc &(a+b)^2\\\end{array}\right|=abc(a+b+c)(2(a^2+b^2+c^2)+4(ab+bc+ca))$

$=abc(a+b+c)\times 2(a+b+c)^2$

$=2abc(a+b+c)^2$

https://www.wolframalpha.com/input/?i=factor+det(%7B%7B(b%2Bc)%5E2,ab,ac%7D,%7Bab,(c%2Ba)%5E2,bc%7D,%7Bac,bc,(a%2Bb)%5E2%7D%7D)