Question

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 $k\ne 3$.
and
STATEMENT -2 : The determinant $\left|\begin{array}{ccc}1& 3& -1\\ -1& -2& k\\ 1& 4& 1\end{array}\right|\ne 0$, for $k\ne 3$

• A. Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1
• B. Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement -1 is True, Statement -2 is False
• D. Statement -1 is False, Statement -2 is True

Answer 1

The correct option is A Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1

$D=\left|\begin{array}{ccc}1& -2& 3\\ -1& 1& -2\\ 1& -3& 4\end{array}\right|=0$
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 $k\ne 3.$
i.e Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1
Hence, option A.