Q.The system $3x-y+4z=3$, $x+2y-3z=-2$ and $6x+5y+\lambda z=-3$ has at least one solution when
A) $\lambda=-5$
B) $\lambda=5$
C) $\lambda=3$
D) $\lambda=-13$

$D=\begin{vmatrix}3&-1&4\\1&2&-3\\6&5&\lambda\end{vmatrix}=3(2\lambda+15)+1(\lambda+18)+4(5-12)=6\lambda+45+\lambda+18-28=7\lambda+35$.

For $\lambda\neq-5$, $D\neq0$ and there is always exactly one solution. At $\lambda=-5$, $D=0$; checking consistency by eliminating x between the equations gives $-7y+13z=9$ from (eq1$-3\times$eq2) and $7y-13z=-9$ from (eq3$-2\times$eq2) — these are the same equati …

Compute D as a linear function of lambda; the interesting boundary case is where D= …

• Assuming D=0 automatically means "no solution" instead of checking whether it's actually a consistent (dependent) system …