JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 23)
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 $$k\ne3$$.
and
Statement - 2 : The determinant $$\left| {\matrix{ 1 & 3 & { - 1} \cr { - 1} & { - 2} & k \cr 1 & 4 & 1 \cr } } \right| \ne 0$$, for $$k \ne 3$$.
Explanation
The given equations are
$$x - 2y + 3z = - 1$$
$$ - x + y - 2z = k$$
$$x - 3y + 4z = 1$$
$$D = \left| {\matrix{ 1 & { - 2} & 3 \cr { - 1} & 1 & { - 2} \cr 1 & { - 3} & 4 \cr } } \right| = 0$$
$$D = \left| {\matrix{ 1 & { - 1} & 3 \cr { - 1} & k & { - 2} \cr 1 & 1 & 4 \cr } } \right| = k - 3 \ne 0$$
If $$k \ne 3$$, the system has no solutions.
Hence, Statement - 1 is True Statement - 2 is correct explanation for Statement - 1.
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