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3 and 4 .Determinants and Matrices
medium
If the system of linear equations $x+y+3 z=0$
$x+3 y+k^{2} z=0$
$3 x+y+3 z=0$
has a non-zero solution $(x, y, z)$ for some $k \in R ,$ then $x +\left(\frac{ y }{ z }\right)$ is equal to
A
$9$
B
$-3$
C
$-9$
D
$3$
(JEE MAIN-2020)
Solution
$x+y+3 z=0$
$x+3 y+k^{2} z=0$
$3 x+y+3 z=0$
$\left|\begin{array}{lll}1 & 1 & 3 \\ 1 & 3 & k^{2} \\ 3 & 1 & 3\end{array}\right|=0$
$\Rightarrow 9+3+3 k^{2}-27-k^{2}-3=0$
$\Rightarrow k ^{2}=9$
(i) $-$ (iii) $\Rightarrow-2 x =0 \Rightarrow x =0$
Now from (i) $\Rightarrow y +3 z =0$
$\Rightarrow \frac{y}{z}=-3$
$x+\frac{y}{z}=-3$
Standard 12
Mathematics
