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Prove that: `|(a,b-c,c-b),(a-c,b,c-a),(a-b,b-a,c)|=(a+b-c)(b+c-a)(c+a-b)` |
Answer» `[[a+b+c-c-b,b-c,c-b],[a-c+b+c-a,b,c-a],[a-b+b-a+c,b-a,c]]` `C_2->C_2+C_3` `[[a,b-c+c-b,c-b],[b,b+c-a,c-a],[c,b-a+c,c]]` `[[a,0,c-b],[b,b+c-a,c-a],[c,b+c-a,c]]` `(b+c-a)[[a,0,c-b],[b,1,c-a],[c,1,c]]` `(b+c-a)[a(c-c+a)+(-b)(b-c)-(b-c)(b-c)]` `(b+c-a)(a-(b-c))(a+(b-c))` `(a+b-c)(b+c-a)(a+c-b)`. |
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