JEE MAIN - Mathematics (2007 - No. 13)
If $$D = \left| {\matrix{
1 & 1 & 1 \cr
1 & {1 + x} & 1 \cr
1 & 1 & {1 + y} \cr
} } \right|$$ for $$x \ne 0,y \ne 0,$$ then $$D$$ is :
divisible by $$x$$ but not $$y$$
divisible by $$y$$ but not $$x$$
divisible by neither $$x$$ nor $$y$$
divisible by both $$x$$ and $$y$$
Explanation
Given, $$D = \left| {\matrix{
1 & 1 & 1 \cr
1 & {1 + x} & 1 \cr
1 & 1 & {1 + y} \cr
} } \right|$$
Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$
and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$
$$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$$
Hence, $$D$$ is divisible by both $$x$$ and $$y$$
Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$
and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$
$$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$$
Hence, $$D$$ is divisible by both $$x$$ and $$y$$
Comments (0)
