Consider the system of equations $x-2 y+3 z=-1, x-3 y+4 z=1$ and $-x+y-2 z=k$ Statement 1 The system of equations has no solution for $k \neq 3$. Statement 2 The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & k \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $k \neq 3$.

• Correct answer
  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

Solution

$$ \text { The given system of equations can be expressed as } $$
When $k \neq 3$, the given system of equations has no solution. $\Rightarrow$ Statement 1 is true.Clearly, Statement 2 is also true as it is rearrangement of rows and columns of
Hence, option (a) is correct.

Hence, the correct answer is option: 'A'.