Question:

For a system of linear equations to have a non-zero solution, the determinant of its coefficient matrix must be zero. The given system of equations is: $x + ky + 3z = 0$ $3x + ky - 2z = 0$ $2x + 4y - 3z = 0$ The coefficient matrix is: $$ A = \begin{pmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{pmatrix} $$ Setting the determinant of $A$ to zero: $$ \det(A) = \begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{vmatrix} = 0 $$ Expanding the determinant along the first row: $$ 1(k(-3) - (-2)(4)) - k(3(-3) - (-2)(2)) + 3(3(4) - k(2)) = 0 $$ $$ 1(-3k + 8) - k(-9 + 4) + 3(12 - 2k) = 0 $$ $$ -3k + 8 - k(-5) + 36 - 6k = 0 $$ $$ -3k + 8 + 5k + 36 - 6k = 0 $$ $$ -4k + 44 = 0 $$ $$ -4k = -44 $$ $$ k = 11 $$ Substitute $k=11$ back into the first two equations of the system: $$ x + 11y + 3z = 0 \quad (1) $$ $$ 3x + 11y - 2z = 0 \quad (2) $$ Subtract equation (1) from equation (2): $$ (3x + 11y - 2z) - (x + 11y + 3z) = 0 $$ $$ 2x - 5z = 0 $$ $$ 2x = 5z $$ $$ x = \frac{5z}{2} $$ Substitute $x = \frac{5z}{2}$ into equation (1): $$ \frac{5z}{2} + 11y + 3z = 0 $$ $$ 11y = -3z - \frac{5z}{2} $$ $$ 11y = -\frac{6z}{2} - \frac{5z}{2} $$ $$ 11y = -\frac{11z}{2} $$ $$ y = -\frac{z}{2} $$ Now, calculate the expression $\frac{xz}{y^2}$: $$ \frac{xz}{y^2} = \frac{\left(\frac{5z}{2}\right)z}{\left(-\frac{z}{2}\right)^2} $$ $$ = \frac{\frac{5z^2}{2}}{\frac{z^2}{4}} $$ $$ = \frac{5z^2}{2} \cdot \frac{4}{z^2} $$ $$ = \frac{5}{2} \cdot 4 $$ $$ = 10 $$