If the system of linear equations x+y+3z=0,x+ | vedclass

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(B) The given system of linear equations is:$x+y+3z=0$  $(i)$$x+3y+k^{2}z=0$  (ii)$3x+y+3z=0$  (iii)For a non-zero solution to exist,the determinant o

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(B) The given system of linear equations is:$x+y+3z=0$  $(i)$$x+3y+k^{2}z=0$  (ii)$3x+y+3z=0$  (iii)For a non-zero solution to exist,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 1 & 1 & 3 \ 1 & 3 & k^{2} \ 3 & 1 & 3 \end{vmatrix} = 0$Expanding the determinant:$1(9 - k^{2}) - 1(3 - 3k^{2}) + 3(1 - 9) = 0$$9 - k^{2} - 3 + 3k^{2} - 24 = 0$$2k^{2} - 18 = 0$$2k^{2} = 18 \Rightarrow k^{2} = 9$Now,substitute $k^{2} = 9$ into the equations:$(i)$ $x+y+3z=0$(iii) $3x+y+3z=0$Subtracting $(i)$ from (iii): $(3x+y+3z) - (x+y+3z) = 0 \Rightarrow 2x = 0 \Rightarrow x = 0$Substituting $x=0$ into $(i)$: $0 + y + 3z = 0 \Rightarrow y = -3z$Therefore,$\frac{y}{z} = -3$Finally,$x + \frac{y}{z} = 0 + (-3) = -3$

If the system of linear equations $x+y+3z=0$,$x+3y+k^{2}z=0$,and $3x+y+3z=0$ has a non-zero solution $(x, y, z)$ for some $k \in R$,then $x + (y/z)$ is equal to

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