JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 16)
The following system of linear equations
2x + 3y + 2z = 9
3x + 2y + 2z = 9
x $$-$$ y + 4z = 8
2x + 3y + 2z = 9
3x + 2y + 2z = 9
x $$-$$ y + 4z = 8
does not have any solution
has a solution ($$\alpha$$, $$\beta$$, $$\gamma$$) satisfying $$\alpha$$ + $$\beta$$2 + $$\gamma$$3 = 12
has a unique solution
has infinitely many solutions
Explanation
$$\Delta = \left| {\matrix{
2 & 3 & 2 \cr
3 & 2 & 2 \cr
1 & { - 1} & 4 \cr
} } \right| = - 20 \ne 0$$ $$ \therefore $$ unique solution
$${\Delta _x} = \left| {\matrix{ 9 & 3 & 2 \cr 9 & 2 & 2 \cr 8 & { - 1} & 4 \cr } } \right| = 0$$
$${\Delta _y} = \left| {\matrix{ 2 & 9 & 2 \cr 3 & 9 & 2 \cr 1 & 8 & 4 \cr } } \right| = - 20$$
$${\Delta _z} = \left| {\matrix{ 2 & 3 & 9 \cr 3 & 2 & 9 \cr 1 & { - 1} & 8 \cr } } \right| = - 40$$
$$ \therefore $$ $$x = {{{\Delta _x}} \over \Delta } = 0$$
$$y = {{{\Delta _y}} \over \Delta } = 1$$
$$z = {{{\Delta _z}} \over \Delta } = 2$$
Unique solution : (0, 1, 2)
$${\Delta _x} = \left| {\matrix{ 9 & 3 & 2 \cr 9 & 2 & 2 \cr 8 & { - 1} & 4 \cr } } \right| = 0$$
$${\Delta _y} = \left| {\matrix{ 2 & 9 & 2 \cr 3 & 9 & 2 \cr 1 & 8 & 4 \cr } } \right| = - 20$$
$${\Delta _z} = \left| {\matrix{ 2 & 3 & 9 \cr 3 & 2 & 9 \cr 1 & { - 1} & 8 \cr } } \right| = - 40$$
$$ \therefore $$ $$x = {{{\Delta _x}} \over \Delta } = 0$$
$$y = {{{\Delta _y}} \over \Delta } = 1$$
$$z = {{{\Delta _z}} \over \Delta } = 2$$
Unique solution : (0, 1, 2)
Comments (0)
