Consider the system of equations
$x-2y+3z = -1$
$-x+y-2z = k$
$x-3y+4z = 1$
Statement-1: The system of equations has no solution for $k ≠ 3$
Statement-2: The determinant $\begin{vmatrix}1&3&-1\\-1&-2&k\\1&4&1\end{vmatrix}≠0$ for $k ≠ 3$
Answer & explanation
Correct answer: option 1
We have,
$D=\begin{vmatrix}1&-2&3\\-1&1&-2\\1&-3&4\end{vmatrix}$
$⇒D=\begin{vmatrix}1&-2&3\\0&-1&0\\0&-1&1\end{vmatrix}=0$ [Applying $R_2 → R_2-R_1,R_3→R_3-R_1$]
and, $D_2=\begin{vmatrix}1&-1&3\\-1&k&-2\\1&1&4\end{vmatrix}=-\begin{vmatrix}1&3&-1\\-1&-2&k\\1&4&1\end{vmatrix}$ [Applying $C_2↔C_3$]
$⇒D_2=-\begin{vmatrix}1&3&-1\\0&1&k-1\\0&1&2\end{vmatrix}=k-3$ [Applying $R_2 → R_2-R_1, R_3→R_3-R_1$]
Clearly, $D_2≠0$ for $k ≠3$. However, D = 0 for all k. Hence, the system is inconsistent for $k ≠ 3$ and the determinant is not zero for $k ≠ 3$.
Hence, both the statement are true and statement-2 is a correct explanation for statement-1.