JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 10)
If the following system of linear equations
2x + y + z = 5
x $$-$$ y + z = 3
x + y + az = b
has no solution, then :
2x + y + z = 5
x $$-$$ y + z = 3
x + y + az = b
has no solution, then :
$$a = - {1 \over 3},b \ne {7 \over 3}$$
$$a \ne {1 \over 3},b = {7 \over 3}$$
$$a \ne - {1 \over 3},b = {7 \over 3}$$
$$a = {1 \over 3},b \ne {7 \over 3}$$
Explanation
Here $$D = \left| {\matrix{
2 & 1 & 1 \cr
1 & { - 1} & 1 \cr
1 & 1 & a \cr
} } \right|\matrix{
{ = 2(a - 1) - 1(a - 1) + 1 + 1} \cr
{ = 1 - 3a} \cr
} $$
$${D_3} = \left| {\matrix{ 2 & 1 & 5 \cr 1 & { - 1} & 3 \cr 1 & 1 & b \cr } } \right|\matrix{ { = 2( - b - 3) - 1(b - 3) + 5(1 + 1)} \cr { = 7 - 3b} \cr } $$
for $$a = {1 \over 3},b \ne {7 \over 3}$$, system has no solutions.
$${D_3} = \left| {\matrix{ 2 & 1 & 5 \cr 1 & { - 1} & 3 \cr 1 & 1 & b \cr } } \right|\matrix{ { = 2( - b - 3) - 1(b - 3) + 5(1 + 1)} \cr { = 7 - 3b} \cr } $$
for $$a = {1 \over 3},b \ne {7 \over 3}$$, system has no solutions.
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