If the system of linear equations x + ky + 3z | vedclass

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(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 1

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$10$

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(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} 1 & k & 3 \ 3 & k & -2 \ 2 & 4 & -3 \end{vmatrix} = 0$Expanding along the first row:$1(-3k + 8) - k(-9 + 4) + 3(12 - 2k) = 0$$-3k + 8 + 5k + 36 - 6k = 0$$-4k + 44 = 0 \Rightarrow k = 11$Substituting $k = 11$ into the equations:$x + 11y + 3z = 0$ $(1)$$3x + 11y - 2z = 0$ $(2)$$2x + 4y - 3z = 0$ $(3)$Subtracting $(1)$ from $(2)$:$(3x - x) + (11y - 11y) + (-2z - 3z) = 0 \Rightarrow 2x - 5z = 0 \Rightarrow x = \frac{5}{2}z$Substituting $x = \frac{5}{2}z$ into $(1)$:$\frac{5}{2}z + 11y + 3z = 0 \Rightarrow 11y = -\frac{11}{2}z \Rightarrow y = -\frac{1}{2}z$Now,calculate $\frac{xz}{y^2}$:$\frac{xz}{y^2} = \frac{(\frac{5}{2}z)(z)}{(-\frac{1}{2}z)^2} = \frac{\frac{5}{2}z^2}{\frac{1}{4}z^2} = \frac{5}{2} 	imes 4 = 10$

If the system of linear equations $x + ky + 3z = 0$,$3x + ky - 2z = 0$,and $2x + 4y - 3z = 0$ has a non-zero solution $(x, y, z)$,then $\frac{xz}{y^2} = \dots$

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