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 solutions 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 -
• B. Statement -1 is true, statement - 2 is true,
  statement -2 is a 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

Answer: (A)

Statement-1 is true, statement - 2 is true,

statement - 2 is a correct explanation for
statement -
For the solution of the system of equations
$x-2y+3z=-1$
$-x+y-2z=2$
$kx-3y+4z=1$,
$\left[\begin{array}{ccc}1& -2& 3\\ -1& 1& -2\\ k& -3& 4\end{array}\right]$
If determinant of the matrix is zero than the given lines are coplanar and if the determinant is non zero the they are non co-planar.
The solution for the system of equations exist when the lines are coplanar.
Hence, $1(-2-4k)-3(-1-k)-1(-4+2)\ne 0$
$\Rightarrow -2-4k+3+3k+2\ne 0$
i.e. $k\ne 3$