If the system of equations
$(k - 1) x + (3 k + 1) y + 2 k z = 0 (k - 1) x + (4 k - 2) y + (k + 3) z = 0 2 x + (3 k + 1) y + 3 (k - 1) z = 0$
has a non - zero solution, then prove that $k = 0, 3$

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Solution

For non-trivial solution $△ = 0$
$∣ ∣ ∣ ∣ \begin{matrix} k - 1 & 3 k + 1 & 2 k k - 1 & 4 k - 2 & k + 3 2 & 3 k + 1 & 3 k - 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Apply $R_{2} - R_{1}, R_{3} - R_{1}$
$△ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} k - 1 & 3 k + 12 k 0 & k - 3 & - (k - 3) - (k - 3) & 0 & k - 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
= $(k - 3)^{2} ∣ ∣ ∣ ∣ \begin{matrix} k - 1 & 3 k + 1 & 2 k 0 & 1 & - 1 - 1 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣$
Apply $C_{2} + C_{3}$
$(k - 3)^{2} ∣ ∣ ∣ ∣ \begin{matrix} k - 1 & 5 k + 1 & 2 k 0 & 0 & - 1 - 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
Expanding with $2 n d$ row.
$(k - 3)^{2} 6 k = 0 ∴ k = 0 o r 3$ .

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