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Find the parameter values giving a unique solution.

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TOPIC 01

2025 JM02

A zero determinant needs a separate consistency analysis, not an automatic no-solution conclusion.

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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 the parameter values giving a unique solution.

kx+2y−z=p,ky+z=q,kx+3y=6kx+2y-z=p,\quad ky+z=q,\quad kx+3y=6

Official paper · jm02-2025 · 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
Use the determinant of the coefficient matrix.
Hint 2
Subtract the first row from the third.
Worked solution
  1. The row operation preserves the determinant.

    ∣k2−10k1k30∣=∣k2−10k1011∣=k(k−1)\begin{vmatrix}k&2&-1\\0&k&1\\k&3&0\end{vmatrix}=\begin{vmatrix}k&2&-1\\0&k&1\\0&1&1\end{vmatrix}=k(k-1)
  2. A nonsingular coefficient matrix gives exactly one solution for every p,q.

    k≠0,1k\ne0,1

Unique solution exactly when k≠0 and k≠1.

Checks and common pitfalls: A zero determinant needs a separate consistency analysis, not an automatic no-solution conclusion.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The row operation preserves the determinant.
    ∣k2−10k1k30∣=∣k2−10k1011∣=k(k−1)\begin{vmatrix}k&2&-1\\0&k&1\\k&3&0\end{vmatrix}=\begin{vmatrix}k&2&-1\\0&k&1\\0&1&1\end{vmatrix}=k(k-1)
  • A nonsingular coefficient matrix gives exactly one solution for every p,q.
    k≠0,1k\ne0,1

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Curriculum and source notes ↗