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Find all p for which the system has a unique solution.

Read the idea, work independently, then explain what changed.

TOPIC 01

2022 JM02

The right-hand parameter q does not affect invertibility.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Find all p for which the system has a unique solution.

{x+y+pz=1px+y+z=px+py+z=q\begin{cases}x+y+pz=1\\px+y+z=p\\x+py+z=q\end{cases}

Official paper · jm02-2022 · 5(b)(i) · PDF 7

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

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Compute the coefficient determinant.
Hint 2
Factor the resulting cubic in p.
Worked solution
  1. Expand along the first row.

    Δ=∣11pp111p1∣=1−p−(p−1)+p(p2−1)=p3−3p+2\Delta=\begin{vmatrix}1&1&p\\p&1&1\\1&p&1\end{vmatrix}=1-p-(p-1)+p(p^2-1)=p^3-3p+2
  2. Require a nonzero determinant.

    Δ=(p−1)2(p+2)≠0  ⟺  p≠1,−2\Delta=(p-1)^2(p+2)\ne0\iff p\ne1,-2

p∈ℝ∖{1,−2}, for any q.

Checks and common pitfalls: The right-hand parameter q does not affect invertibility.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Expand along the first row.
    Δ=∣11pp111p1∣=1−p−(p−1)+p(p2−1)=p3−3p+2\Delta=\begin{vmatrix}1&1&p\\p&1&1\\1&p&1\end{vmatrix}=1-p-(p-1)+p(p^2-1)=p^3-3p+2
  • Require a nonzero determinant.
    Δ=(p−1)2(p+2)≠0  ⟺  p≠1,−2\Delta=(p-1)^2(p+2)\ne0\iff p\ne1,-2

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Curriculum and source notes ↗