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Step 1: Write Down the System of Equations
The given system of linear equations is:
1) $2x + 2ay + az = 0$
2) $2x + 3by + bz = 0$
3) $2x + 4cy + cz = 0$
where $a,\, b,\, c \in \mathbb{R}$ and are non-zero distinct constants.
Step 2: Condition for a Non-zero Solution
For the system to have a non-zero solution (i.e., a non-trivial solution), the determinant of the coefficient matrix must be zero:
$ \begin{vmatrix}
2 & 2a & a \\
2 & 3b & b \\
2 & 4c & c
\end{vmatrix} = 0.
$
Step 3: Simplify the Determinant
We can factor out common factors or perform row operations. One approach is to subtract the first row from the second and third rows to make the determinant simpler. For instance, replace the first row with half of itself (just for easier numbers) and then perform suitable row operations:
$ \begin{vmatrix}
2 & 2a & a \\
2 & 3b & b \\
2 & 4c & c
\end{vmatrix}
= 2 \cdot
\begin{vmatrix}
1 & a & \frac{a}{2} \\
2 & 3b & b \\
2 & 4c & c
\end{vmatrix}.
$
(Note: There are various possible ways to simplify; we will follow the simplification shown in the given solution reference since it’s straightforward.)
In the provided reference solution, it shows a row operation scaling directly to get:
$
\begin{vmatrix}
1 & 2a & a \\
0 & 3b - 2a & b - a \\
0 & 4c - 2a & c - a
\end{vmatrix} = 0.
$
Step 4: Expand and Further Simplify
By expanding along the first column (which now has two zeros), the determinant becomes:
$ (3b - 2a)(c - a) - (b - a)(4c - 2a) = 0.
$
Expanding both products and then simplifying leads to:
$ 3bc - 3ba - 2ac + 2a^2 - (4bc - 4ac - 2ab + 2a^2) = 0.
$
Simplifying this expression eventually yields the key relation:
$2ac = bc + ab.
Step 5: Conclude the Relation in Terms of 1/a, 1/b, and 1/c
Rewriting $2ac = bc + ab$ as:
$
2ac = a(b + c),
$
we can divide through by $abc$ (since $a, b, c \neq 0$) to obtain:
$
\frac{2}{b} = \frac{1}{a} + \frac{1}{c}.
$
This is precisely the condition that the three numbers $ \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ form an arithmetic progression (A.P), because in an A.P., the middle term is the average of the other two, i.e.
$
\frac{1}{b} = \frac{\frac{1}{a} + \frac{1}{c}}{2}.
$
Final Answer
Therefore, $ \frac{1}{a}, \frac{1}{b}, \frac{1}{c} $ are in A.P.
