JEE MAIN - Mathematics (2011 - No. 13)
The number of values of $$k$$ for which the linear equations
$$4x + ky + 2z = 0,kx + 4y + z = 0$$ and $$2x+2y+z=0$$ possess a non-zero solution is :
$$4x + ky + 2z = 0,kx + 4y + z = 0$$ and $$2x+2y+z=0$$ possess a non-zero solution is :
$$2$$
$$1$$
zero
$$3$$
Explanation
$$\Delta = 0 \Rightarrow \left| {\matrix{
4 & k & 2 \cr
k & 4 & 1 \cr
2 & 2 & 1 \cr
} } \right| = 0$$
$$ \Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) + $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0$$
$$ \Rightarrow 8 - {k^2} + 2k + 4k - 16 = 0$$
$$ \Rightarrow {k^2} - 6k + 8 = 0$$
$$ \Rightarrow \left( {k - 4} \right)\left( {k - 2} \right) = 0,k = 4,2$$
$$ \Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) + $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0$$
$$ \Rightarrow 8 - {k^2} + 2k + 4k - 16 = 0$$
$$ \Rightarrow {k^2} - 6k + 8 = 0$$
$$ \Rightarrow \left( {k - 4} \right)\left( {k - 2} \right) = 0,k = 4,2$$
Comments (0)
