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 $k\ne 3$.
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

$D=\left|\begin{array}{ccc}1& -2& 3\\ -1& 1& -2\\ 1& -3& 4\end{array}\right|$
Here, $D=0$
For no solution, at least one of $D_1,D_2,D_3\ne 0$
$D_1=\left|\begin{array}{ccc}-1& -2& 3\\ k& 1& -2\\ 1& -3& 4\end{array}\right|$
$\Rightarrow k\ne 3$
$D_2=\left|\begin{array}{ccc}1& -1& 3\\ -1& k& -2\\ 1& 1& 4\end{array}\right|$
$\Rightarrow k\ne 3$
$D_3=\left|\begin{array}{ccc}1& -2& -1\\ -1& 1& k\\ 1& -3& 1\end{array}\right|$
$\left|\begin{array}{ccc}1& -2& -1\\ -1& 1& k\\ 1& -3& 1\end{array}\right|\ne 0$
$\Rightarrow k\ne 3$