JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 16)
The following system of linear equations
7x + 6y – 2z = 0
3x + 4y + 2z = 0
x – 2y – 6z = 0, has
7x + 6y – 2z = 0
3x + 4y + 2z = 0
x – 2y – 6z = 0, has
no solution
infinitely many solutions, (x, y, z) satisfying
y = 2z
infinitely many solutions, (x, y, z) satisfying
x = 2z
only the trivial solution
Explanation
Given
7x + 6y – 2z = 0 .......(1)
3x + 4y + 2z = 0 ......(2)
x – 2y – 6z = 0 .......(3)
$$\Delta $$ = $$\left| {\matrix{ 7 & 6 & { - 2} \cr 3 & 4 & 2 \cr 1 & { - 2} & { - 6} \cr } } \right|$$
= 7(–24 + 4) – 6(–18 – 2) – 2(–6 – 4) = 0
$$ \therefore $$ $$\Delta $$ = 0
The system of equation has infinite non-trival solution.
Also adding equation (1) and 3$$ \times $$(3), we get
10x = 20z
$$ \Rightarrow $$ x = 2z
7x + 6y – 2z = 0 .......(1)
3x + 4y + 2z = 0 ......(2)
x – 2y – 6z = 0 .......(3)
$$\Delta $$ = $$\left| {\matrix{ 7 & 6 & { - 2} \cr 3 & 4 & 2 \cr 1 & { - 2} & { - 6} \cr } } \right|$$
= 7(–24 + 4) – 6(–18 – 2) – 2(–6 – 4) = 0
$$ \therefore $$ $$\Delta $$ = 0
The system of equation has infinite non-trival solution.
Also adding equation (1) and 3$$ \times $$(3), we get
10x = 20z
$$ \Rightarrow $$ x = 2z
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