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Factor the determinant.

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TOPIC 01

2022 JM02

The derivation does not divide by a+b+c, so it also covers a+b+c=0.

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01 / Standard#Your turn

Factor the determinant.

D=∣ab+cb2+c2ba+ca2+c2ca+ba2+b2∣D=\begin{vmatrix}a&b+c&b^2+c^2\\b&a+c&a^2+c^2\\c&a+b&a^2+b^2\end{vmatrix}

Official paper · jm02-2022 · 5(a) · PDF 7

Official original and suggested answers ↗ · Suggested answer PDF page 12

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Let s=a+b+c and t=a²+b²+c².
Hint 2
Use column operations to reduce to a Vandermonde determinant.
Worked solution
  1. Let v=(a,b,c)ᵀ and v^[2]=(a²,b²,c²)ᵀ. Add column 1 to column 2 and use multilinearity to remove the repeated constant column, without division.

    D=det⁡[ v, s1, t1−v[2] ]=−sdet⁡[ v,1,v[2] ]=sdet⁡[ 1,v,v[2] ]D=\det[\,v,\ s\mathbf1,\ t\mathbf1-v^{[2]}\,]=-s\det[\,v,\mathbf1,v^{[2]}\,]=s\det[\,\mathbf1,v,v^{[2]}\,]
  2. Evaluate the Vandermonde determinant and reorder factors.

    D=(a+b+c)(b−a)(c−a)(c−b)=(a−b)(b−c)(c−a)(a+b+c)D=(a+b+c)(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)(a+b+c)

(a−b)(b−c)(c−a)(a+b+c).

Checks and common pitfalls: The derivation does not divide by a+b+c, so it also covers a+b+c=0.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Let v=(a,b,c)ᵀ and v^[2]=(a²,b²,c²)ᵀ. Add column 1 to column 2 and use multilinearity to remove the repeated constant column, without division.
    D=det⁡[ v, s1, t1−v[2] ]=−sdet⁡[ v,1,v[2] ]=sdet⁡[ 1,v,v[2] ]D=\det[\,v,\ s\mathbf1,\ t\mathbf1-v^{[2]}\,]=-s\det[\,v,\mathbf1,v^{[2]}\,]=s\det[\,\mathbf1,v,v^{[2]}\,]
  • Evaluate the Vandermonde determinant and reorder factors.
    D=(a+b+c)(b−a)(c−a)(c−b)=(a−b)(b−c)(c−a)(a+b+c)D=(a+b+c)(b-a)(c-a)(c-b)=(a-b)(b-c)(c-a)(a+b+c)

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