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For the displayed system with constants k,p,q,r, find all k for which the solution is unique.

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

TOPIC 01

2021 JM02

A zero determinant does not distinguish no solution from infinitely many solutions without checking consistency.

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

01 / Standard#Your turn

For the displayed system with constants k,p,q,r, find all k for which the solution is unique.

(E):{kx+y−z=px+ky+z=q−x+y+kz=r(E):\begin{cases}kx+y-z=p\\x+ky+z=q\\-x+y+kz=r\end{cases}

Official paper · jm02-2021 · 5(b)(i) · PDF 7

Official original and suggested answers ↗ · Suggested answer PDF page 11

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use the coefficient determinant.
Hint 2
Factor the cubic determinant in k.
Worked solution
  1. Expand along the first row.

    det⁡A=k(k2−1)−(k+1)−(k+1)=k3−3k−2\det A=k(k^2-1)-(k+1)-(k+1)=k^3-3k-2
  2. A square system is uniquely solvable for any right side exactly when this determinant is nonzero.

    det⁡A=(k+1)2(k−2)≠0  ⟺  k≠−1,2\det A=(k+1)^2(k-2)\ne0\iff k\ne-1,2

All real k except −1 and 2.

Checks and common pitfalls: A zero determinant does not distinguish no solution from infinitely many solutions without checking consistency.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Expand along the first row.
    det⁡A=k(k2−1)−(k+1)−(k+1)=k3−3k−2\det A=k(k^2-1)-(k+1)-(k+1)=k^3-3k-2
  • A square system is uniquely solvable for any right side exactly when this determinant is nonzero.
    det⁡A=(k+1)2(k−2)≠0  ⟺  k≠−1,2\det A=(k+1)^2(k-2)\ne0\iff k\ne-1,2

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Curriculum and source notes ↗