JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 3)
Consider the system of linear equations
$$-$$x + y + 2z = 0
3x $$-$$ ay + 5z = 1
2x $$-$$ 2y $$-$$ az = 7
Let S1 be the set of all a$$\in$$R for which the system is inconsistent and S2 be the set of all a$$\in$$R for which the system has infinitely many solutions. If n(S1) and n(S2) denote the number of elements in S1 and S2 respectively, then
$$-$$x + y + 2z = 0
3x $$-$$ ay + 5z = 1
2x $$-$$ 2y $$-$$ az = 7
Let S1 be the set of all a$$\in$$R for which the system is inconsistent and S2 be the set of all a$$\in$$R for which the system has infinitely many solutions. If n(S1) and n(S2) denote the number of elements in S1 and S2 respectively, then
n(S1) = 2, n(S2) = 2
n(S1) = 1, n(S2) = 0
n(S1) = 2, n(S2) = 0
n(S1) = 0, n(S2) = 2
Explanation
$$\Delta = \left| {\matrix{
{ - 1} & 1 & 2 \cr
3 & { - a} & 5 \cr
2 & { - 2} & { - a} \cr
} } \right|$$
$$ = - 1({a^2} + 10) - 1( - 3a - 10) + 2( - 6 + 2a)$$
$$ = - {a^2} - 10 + 3a + 10 - 12 + 4a$$
$$\Delta = - {a^2} + 7a - 12$$
$$\Delta = - [{a^2} - 7a + 12]$$
$$\Delta = - [(a - 3)(a - 4)]$$
$${\Delta _1} = \left| {\matrix{ 0 & 1 & 2 \cr 1 & { - a} & 5 \cr 7 & { - 2} & { - a} \cr } } \right|$$
$$ = a + 35 - 4 + 14a$$
= $$15a + 31$$
Now, $${\Delta _1} = 15a + 31$$
For inconsistent $$\Delta$$ = 0 $$\therefore$$ a = 3, a = 4 and for a = 3 and 4, $$\Delta$$1 $$\ne$$ 0
n(S1) = 2
For infinite solution : $$\Delta$$ = 0 and $$\Delta$$1 = $$\Delta$$2 = $$\Delta$$3 = 0
Not possible
$$\therefore$$ n(S2) = 0
$$ = - 1({a^2} + 10) - 1( - 3a - 10) + 2( - 6 + 2a)$$
$$ = - {a^2} - 10 + 3a + 10 - 12 + 4a$$
$$\Delta = - {a^2} + 7a - 12$$
$$\Delta = - [{a^2} - 7a + 12]$$
$$\Delta = - [(a - 3)(a - 4)]$$
$${\Delta _1} = \left| {\matrix{ 0 & 1 & 2 \cr 1 & { - a} & 5 \cr 7 & { - 2} & { - a} \cr } } \right|$$
$$ = a + 35 - 4 + 14a$$
= $$15a + 31$$
Now, $${\Delta _1} = 15a + 31$$
For inconsistent $$\Delta$$ = 0 $$\therefore$$ a = 3, a = 4 and for a = 3 and 4, $$\Delta$$1 $$\ne$$ 0
n(S1) = 2
For infinite solution : $$\Delta$$ = 0 and $$\Delta$$1 = $$\Delta$$2 = $$\Delta$$3 = 0
Not possible
$$\therefore$$ n(S2) = 0
Comments (0)
