JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 3)
Consider the following system of equations :
x + 2y $$-$$ 3z = a
2x + 6y $$-$$ 11z = b
x $$-$$ 2y + 7z = c,
where a, b and c are real constants. Then the system of equations :
x + 2y $$-$$ 3z = a
2x + 6y $$-$$ 11z = b
x $$-$$ 2y + 7z = c,
where a, b and c are real constants. Then the system of equations :
has no solution for all a, b and c
has a unique solution when 5a = 2b + c
has infinite number of solutions when 5a = 2b + c
has a unique solution for all a, b and c
Explanation
$$D = \left| {\matrix{
1 & 2 & { - 3} \cr
2 & 6 & { - 11} \cr
1 & { - 2} & 7 \cr
} } \right|$$
= 20 $$-$$ 2(25) $$-$$3($$-$$10)
= 20 $$-$$ 50 + 30 = 0
$${D_1} = \left| {\matrix{ a & 2 & { - 3} \cr b & 6 & { - 11} \cr c & { - 2} & 7 \cr } } \right|$$
= 20a $$-$$ 2(7b + 11c) $$-$$3($$-$$2b $$-$$ 6c)
= 20a $$-$$ 14b $$-$$ 22c + 6b +18c
= 20a $$-$$ 8b $$-$$ 4c
= 4(5a $$-$$ 2b $$-$$ c)
$${D_2} = \left| {\matrix{ 1 & a & { - 3} \cr 2 & b & { - 11} \cr 1 & c & 7 \cr } } \right|$$
= 7b + 11c $$-$$ a(25) $$-$$3(2c $$-$$ b)
= 7b + 11c $$-$$ 25a $$-$$ 6c + 3b
= $$-$$25a + 10b + 5c
= $$-$$5(5a $$-$$ 2b $$-$$ c)
$${D_3} = \left| {\matrix{ 1 & 2 & a \cr 2 & 6 & b \cr 1 & { - 2} & c \cr } } \right|$$
= 6c + 2b $$-$$ 2(2c $$-$$ b) $$-$$ 10a
= $$-$$10a + 4b + 2c
= $$-$$2(5a $$-$$ 2b $$-$$ c)
for infinite solution
$$D = {D_1} = {D_2} = {D_3} = 0$$
$$ \Rightarrow $$ 5a = 2b + c
= 20 $$-$$ 2(25) $$-$$3($$-$$10)
= 20 $$-$$ 50 + 30 = 0
$${D_1} = \left| {\matrix{ a & 2 & { - 3} \cr b & 6 & { - 11} \cr c & { - 2} & 7 \cr } } \right|$$
= 20a $$-$$ 2(7b + 11c) $$-$$3($$-$$2b $$-$$ 6c)
= 20a $$-$$ 14b $$-$$ 22c + 6b +18c
= 20a $$-$$ 8b $$-$$ 4c
= 4(5a $$-$$ 2b $$-$$ c)
$${D_2} = \left| {\matrix{ 1 & a & { - 3} \cr 2 & b & { - 11} \cr 1 & c & 7 \cr } } \right|$$
= 7b + 11c $$-$$ a(25) $$-$$3(2c $$-$$ b)
= 7b + 11c $$-$$ 25a $$-$$ 6c + 3b
= $$-$$25a + 10b + 5c
= $$-$$5(5a $$-$$ 2b $$-$$ c)
$${D_3} = \left| {\matrix{ 1 & 2 & a \cr 2 & 6 & b \cr 1 & { - 2} & c \cr } } \right|$$
= 6c + 2b $$-$$ 2(2c $$-$$ b) $$-$$ 10a
= $$-$$10a + 4b + 2c
= $$-$$2(5a $$-$$ 2b $$-$$ c)
for infinite solution
$$D = {D_1} = {D_2} = {D_3} = 0$$
$$ \Rightarrow $$ 5a = 2b + c
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