If the system of linear equations x + 2ay + a | vedclass

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

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Are in $H$.$P$.

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(C) For the system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix} 1 & 2a & a \ 1 & 3b & b \ 1 & 4c & c \end{vmatrix} = 0$Applying column operation $C_2 	o C_2 - 2C_3$:$\begin{vmatrix} 1 & 0 & a \ 1 & b & b \ 1 & 2c & c \end{vmatrix} = 0$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$:$\begin{vmatrix} 1 & 0 & a \ 0 & b & b-a \ 0 & 2c-b & c-b \end{vmatrix} = 0$Expanding along the first column:$1 \cdot [b(c - b) - (b - a)(2c - b)] = 0$$bc - b^2 - (2bc - b^2 - 2ac + ab) = 0$$bc - b^2 - 2bc + b^2 + 2ac - ab = 0$$2ac - ab - bc = 0$$2ac = ab + bc$Dividing both sides by $abc$:$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$This is the condition for $a, b, c$ to be in Harmonic Progression ($H$.$P$.).

If the system of linear equations $x + 2ay + az = 0$,$x + 3by + bz = 0$,and $x + 4cy + cz = 0$ has a non-zero solution,then $a, b, c$:

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