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Set k=2 and suppose the system has more than one solution. Find the relation among p,q,r; then solve when p=5,q=1,r=−4.

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TOPIC 01

2021 JM02

State the free parameter domain and retain the compatibility condition; one particular solution is not the full answer.

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

01 / Standard#Your turn

Set k=2 and suppose the system has more than one solution. Find the relation among p,q,r; then solve when p=5,q=1,r=−4.

(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)(ii) · PDF 7

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

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
At k=2 the second coefficient row equals the sum of the first and third.
Hint 2
Take z as a free parameter after checking the right-side relation.
Worked solution
  1. The coefficient dependence forces the same relation on the right side.

    R2=R1+R3  ⟹  q=p+r  ⟺  p−q+r=0R_2=R_1+R_3\implies q=p+r\iff p-q+r=0
  2. The first two rows are independent, so this condition is also sufficient and leaves one free variable.

    ∣2112∣=3≠0\begin{vmatrix}2&1\\1&2\end{vmatrix}=3\ne0
  3. The specified right side is consistent; set z=t and solve the first two equations.

    5−1−4=0,2x+y−t=5,x+2y+t=15-1-4=0,\quad2x+y-t=5,\quad x+2y+t=1
  4. Substitution also satisfies the third equation.

    (x,y,z)=(3+t,−1−t,t),t∈R(x,y,z)=(3+t,-1-t,t),\quad t\in\mathbb R

p−q+r=0; for the specified values, (x,y,z)=(3+t,−1−t,t), t∈ℝ.

Checks and common pitfalls: State the free parameter domain and retain the compatibility condition; one particular solution is not the full answer.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The coefficient dependence forces the same relation on the right side.
    R2=R1+R3  ⟹  q=p+r  ⟺  p−q+r=0R_2=R_1+R_3\implies q=p+r\iff p-q+r=0
  • The first two rows are independent, so this condition is also sufficient and leaves one free variable.
    ∣2112∣=3≠0\begin{vmatrix}2&1\\1&2\end{vmatrix}=3\ne0
  • The specified right side is consistent; set z=t and solve the first two equations.
    5−1−4=0,2x+y−t=5,x+2y+t=15-1-4=0,\quad2x+y-t=5,\quad x+2y+t=1
  • Substitution also satisfies the third equation.
    (x,y,z)=(3+t,−1−t,t),t∈R(x,y,z)=(3+t,-1-t,t),\quad t\in\mathbb R

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