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

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(B) For the system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero: $\left|\begin{array}{lll}1

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are in $H.P.$

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(B) For the system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero: $\left|\begin{array}{lll}1 & 2a & a \ 1 & 3b & b \ 1 & 4c & c\end{array}ight| = 0$ Applying row operations $R_{2} ightarrow R_{2} - R_{1}$ and $R_{3} ightarrow R_{3} - R_{1}$: $\left|\begin{array}{ccc}1 & 2a & a \ 0 & 3b - 2a & b - a \ 0 & 4c - 2a & c - a\end{array}ight| = 0$ Expanding along the first column: $(3b - 2a)(c - a) - (4c - 2a)(b - a) = 0$ Expanding the terms: $(3bc - 3ab - 2ac + 2a^{2}) - (4bc - 4ab - 2ac + 2a^{2}) = 0$ Simplifying the expression: $3bc - 3ab - 2ac + 2a^{2} - 4bc + 4ab + 2ac - 2a^{2} = 0$ $-bc + ab = 0$ $ab = bc$ Wait,let us re-evaluate the expansion: $3bc - 3ab - 2ac + 2a^{2} - (4bc - 4ab - 2ac + 2a^{2}) = 0$ $3bc - 3ab - 2ac + 2a^{2} - 4bc + 4ab + 2ac - 2a^{2} = 0$ $-bc + ab = 0 \Rightarrow ab = bc$. Actually,let's re-calculate: $(3b-2a)(c-a) = 3bc - 3ab - 2ac + 2a^2$ $(4c-2a)(b-a) = 4bc - 4ac - 2ab + 2a^2$ Subtracting: $(3bc - 3ab - 2ac + 2a^2) - (4bc - 4ac - 2ab + 2a^2) = 0$ $-bc - ab + 2ac = 0$ $2ac = ab + bc$ Dividing by $abc$: $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$ This implies that $a, b, c$ are in $H.P.$

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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