Find the values of k giving a unique solution.

01 / Standard#Your turn

\begin{cases}x+y+z=1\\kx+5y+z=3\\x+ky-z=1\end{cases}

Official paper · jm02-2024 · 5(b)(i) · PDF 7

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Form the coefficient determinant.

Hint 2

A square linear system is uniquely solvable when its determinant is nonzero.

Worked solution

Expand the determinant.
$\begin{vmatrix}1&1&1\\k&5&1\\1&k&-1\end{vmatrix}=(-5-k)+(k+1)+(k^2-5)=k^2-9$
Exclude its two zeros.
$k\in\mathbb R\setminus\{-3,3\}$

k≠±3.

Checks and common pitfalls: A zero determinant does not automatically imply no solution; it only rules out uniqueness.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Expand the determinant.
$\begin{vmatrix}1&1&1\\k&5&1\\1&k&-1\end{vmatrix}=(-5-k)+(k+1)+(k^2-5)=k^2-9$
Exclude its two zeros.
$k\in\mathbb R\setminus\{-3,3\}$

Think first. Reveal a hint when the class is ready.

2024 JM02