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Consider the system of equations
$ x-2 y+3 z=-1 $ ; $ -x+y-2 z=k $ ; $ x-3 y+4 z=1$
$STATEMENT -1$ : The system of equations has no solution for $\mathrm{k} \neq 3$. and
$STATEMENT - 2$ : The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & \mathrm{k} \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $\mathrm{k} \neq 3$.
Statement-$1$ is True, Statement -$2$ is True; Statement-$2$ is a correct explanation for Statement-$1$
Statement -$1$ is True, Statement -$2$ is True; Statement-$2$ is $NOT$ a correct explanation for Statement-$1$
Statement -$1$ is True, Statement -$2$ is False
Statement -$1$ is False, Statement - $2$ is True
Solution
$D=\left|\begin{array}{ccc}1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4\end{array}\right|=0$
and $D_1=\left|\begin{array}{ccc}-1 & -2 & 3 \\ k & 1 & -2 \\ 1 & -3 & 4\end{array}\right|=(3-k)=0$ if $k=3$
$D_2=\left|\begin{array}{ccc}1 & -1 & 3 \\ -1 & k & -2 \\ 1 & 1 & 4\end{array}\right|=(k-3)=0$, if $k=3$
$D_3=\left|\begin{array}{ccc}1 & -2 & -1 \\ -1 & 1 & k \\ 1 & -3 & 1\end{array}\right|=(k-3)=0$, if $k=3$
$\Rightarrow$ system of equations has no solution for $\mathrm{k} \neq 3$.