If the system of equations 3x - 2y + z = 0,la | vedclass

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(D) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be equal to zero.Given

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$29$

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(D) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be equal to zero.Given the system:$3x - 2y + z = 0$$\lambda x - 14y + 15z = 0$$x + 2y + 3z = 0$The determinant of the coefficient matrix is:$\Delta = \begin{vmatrix} 3 & -2 & 1 \ \lambda & -14 & 15 \ 1 & 2 & 3 \end{vmatrix} = 0$Expanding along the first row:$3((-14)(3) - (15)(2)) - (-2)((\lambda)(3) - (15)(1)) + 1((\lambda)(2) - (-14)(1)) = 0$$3(-42 - 30) + 2(3\lambda - 15) + (2\lambda + 14) = 0$$3(-72) + 6\lambda - 30 + 2\lambda + 14 = 0$$-216 + 8\lambda - 16 = 0$$8\lambda - 232 = 0$$8\lambda = 232$$\lambda = \frac{232}{8} = 29$Thus,the value of $\lambda$ is $29$.

If the system of equations $3x - 2y + z = 0$,$\lambda x - 14y + 15z = 0$,and $x + 2y + 3z = 0$ has a non-trivial solution,then $\lambda = $

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