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3 and 4 .Determinants and Matrices
hard
The following system of linear equations $7 x+6 y-2 z=0$ ; $3 x+4 y+2 z=0$ ; ${x}-2{y}-6{z}=0,$ has
A
infinitely many solutions, $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ satisfying $x=2 z$
B
no solution
C
only the trivial solution
D
infinitely many solutions, $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ satisfying $y=2 z$
(JEE MAIN-2020)
Solution
$7 \mathrm{x}+6 \mathrm{y}-2 \mathrm{z}=0\dots(1)$
$3 x+4 y+2 z=0\dots(2)$
$\mathrm{x}-2 \mathrm{y}-6 \mathrm{z}=0\dots(3)$
$\Delta=\left|\begin{array}{ccc}{7} & {6} & {-2} \\ {3} & {4} & {2} \\ {1} & {-2} & {-6}\end{array}\right|=0 \Rightarrow$ infinite solutions
Now $(1)+(2) \Rightarrow y=-x$ put in $(1),(2)$ and $(3)$
all will lead to $x=2 z$
Standard 12
Mathematics
