Matrices & Determinants
Determinants
nta_pyq_2025_jan
Grade 12
Question:
The system of equations $x+y+z=6,\ x+2y+5z=9,\ x+5y+\lambda z=\mu$ has no solution if:
$\lambda=15,\,\mu\ne 17$
$\lambda\ne 17,\,\mu\ne 18$
$\lambda=17,\,\mu\ne 18$
$\lambda=17,\,\mu=18$
Step-by-Step Solution
Key Concept: For no solution: $D=0$ AND at least one $D_{i}\ne 0$. Compute $D$ first to fix $\lambda$, then $D_{z}$ to constrain $\mu.$
$D=\begin{vmatrix}1&1&1\\1&2&5\\1&5&\lambda\end{vmatrix}=1(2\lambda-25)-(\lambda-5)+(5-2)=\lambda-17.$
$D=0\Leftrightarrow \lambda=17.$
$D_{z}=\begin{vmatrix}1&1&6\\1&2&9\\1&5&\mu\end{vmatrix}=(2\mu-45)-(\mu-9)+6(3)=\mu-18.$
For no solution: $\lambda=17$ and $\mu\ne 18.$
Correct Answer: 3