JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 13)
An ordered pair ($$\alpha $$, $$\beta $$) for which the system of linear equations
(1 + $$\alpha $$) x + $$\beta $$y + z = 2
$$\alpha $$x + (1 + $$\beta $$)y + z = 3
$$\alpha $$x + $$\beta $$y + 2z = 2
has a unique solution, is :
(1 + $$\alpha $$) x + $$\beta $$y + z = 2
$$\alpha $$x + (1 + $$\beta $$)y + z = 3
$$\alpha $$x + $$\beta $$y + 2z = 2
has a unique solution, is :
(–3, 1)
(1, –3)
(–4, 2)
(2, 4)
Explanation
For unique solution
$$\Delta $$ $$ \ne $$ 0 $$ \Rightarrow $$ $$\left| {\matrix{ {1 + \alpha } & \beta & 1 \cr \alpha & {1 + \beta } & 1 \cr \alpha & \beta & 2 \cr } } \right| \ne 0$$
$$\left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr \alpha & \beta & 2 \cr } } \right| \ne 0 \Rightarrow \alpha + \beta \ne - 2$$
$$\Delta $$ $$ \ne $$ 0 $$ \Rightarrow $$ $$\left| {\matrix{ {1 + \alpha } & \beta & 1 \cr \alpha & {1 + \beta } & 1 \cr \alpha & \beta & 2 \cr } } \right| \ne 0$$
$$\left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr \alpha & \beta & 2 \cr } } \right| \ne 0 \Rightarrow \alpha + \beta \ne - 2$$
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