The system of equations x+3by+bz=0, x+2ay+az= | vedclass

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(C) The given system of homogeneous linear equations is:$x+3by+bz=0$$x+2ay+az=0$$x+4cy+cz=0$This system has a non-zero solution if and only if the det

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non-zero solution, whenever $b(a+c)=2ac$

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(C) The given system of homogeneous linear equations is:$x+3by+bz=0$$x+2ay+az=0$$x+4cy+cz=0$This system has a non-zero solution if and only if the determinant of the coefficient matrix is zero:$\begin{vmatrix} 1 & 3b & b \ 1 & 2a & a \ 1 & 4c & c \end{vmatrix} = 0$Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$\begin{vmatrix} 1 & 3b & b \ 0 & 2a-3b & a-b \ 0 & 4c-3b & c-b \end{vmatrix} = 0$Expanding along the first column:$(2a-3b)(c-b) - (a-b)(4c-3b) = 0$$(2ac - 2ab - 3bc + 3b^2) - (4ac - 3ab - 4bc + 3b^2) = 0$$2ac - 2ab - 3bc + 3b^2 - 4ac + 3ab + 4bc - 3b^2 = 0$$-2ac + ab + bc = 0$$ab + bc = 2ac$$b(a+c) = 2ac$Thus, the system has a non-zero solution whenever $b(a+c) = 2ac$.

The system of equations $x+3by+bz=0$, $x+2ay+az=0$ and $x+4cy+cz=0$ has

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