Solution

1. Key Concept: Condition for Non-Zero Solution in Homogeneous Linear Equations

For a system of homogeneous linear equations of the form $AX = 0$ (where $A$ is the coefficient matrix, $X$ is the variable matrix, and $0$ is the zero matrix), a non-zero solution (also known as a non-trivial solution) exists if and only if the determinant of the coefficient matrix is zero, i.e., $\text{det}(A) = 0$. If $\text{det}(A) \neq 0$, then the only solution is the trivial solution $X=0$.

Given the system of equations:

1. $2x + 2ay + az = 0$
2. $2x + 3by + bz = 0$
3. $2x + 4cy + cz = 0$

The coefficient matrix $A$ for this system is: $A = \begin{pmatrix} 2 & 2a & a \\ 2 & 3b & b \\ 2 & 4c & c \end{pmatrix}$ Since the problem states that the system has a non-zero solution, we must have $\text{det}(A) = 0$.

2. Setting up the Determinant Equation

We set the determinant of the coefficient matrix to zero: $\left| \begin{matrix} 2 & 2a & a \\ 2 & 3b & b \\ 2 & 4c & c \end{matrix} \right| = 0$

3. Simplifying the Determinant using Row Operations

To simplify the calculation of the determinant, we can perform row operations to create zeros in a column or row. This makes expansion much easier. We observe that the first column has identical elements (all 2s). We can use this to create zeros in the first column below the first element.

• Perform the operation $R_2 \to R_2 - R_1$ (Subtract Row 1 from Row 2).
• Perform the operation $R_3 \to R_3 - R_1$ (Subtract Row 1 from Row 3).

These operations do not change the value of the determinant. $\left| \begin{matrix} 2 & 2a & a \\ 2-2 & 3b-2a & b-a \\ 2-2 & 4c-2a & c-a \end{matrix} \right| = 0$ This simplifies to: $\left| \begin{matrix} 2 & 2a & a \\ 0 & 3b-2a & b-a \\ 0 & 4c-2a & c-a \end{matrix} \right| = 0$

4. Expanding the Determinant

Now, we expand the determinant along the first column. Since the elements in the first column below the first row are zero, the expansion becomes very straightforward: $2 \cdot \left| \begin{matrix} 3b-2a & b-a \\ 4c-2a & c-a \end{matrix} \right| - 0 \cdot (\text{minor}) + 0 \cdot (\text{minor}) = 0$ $2 \cdot [ (3b-2a)(c-a) - (b-a)(4c-2a) ] = 0$ Since $2 \neq 0$, the expression inside the square brackets must be zero: $(3b-2a)(c-a) - (b-a)(4c-2a) = 0$

5. Algebraic Simplification

Now, we expand the products and simplify the algebraic expression: $(3bc - 3ab - 2ac + 2a^2) - (4bc - 2ab - 4ac + 2a^2) = 0$ Distribute the negative sign for the second parenthesis: $3bc - 3ab - 2ac + 2a^2 - 4bc + 2ab + 4ac - 2a^2 = 0$ Combine like terms: $(3bc - 4bc) + (-3ab + 2ab) + (-2ac + 4ac) + (2a^2 - 2a^2) = 0$ $-bc - ab + 2ac = 0$ Rearrange the terms to get positive coefficients where possible: $2ac = ab + bc$

6. Transforming to the Arithmetic Progression (A.P.) Condition

We are given that $a, b, c$ are non-zero distinct real numbers. This is crucial because it allows us to divide by $abc$ without dividing by zero. Divide the entire equation $2ac = ab + bc$ by $abc$: $\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$ Simplify each term: $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$ Rearranging the terms on the right side: $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$

7. Conclusion: Identifying the Arithmetic Progression

The condition $2/b = 1/a + 1/c$ is the defining property of an Arithmetic Progression (A.P.). If three numbers $x, y, z$ are in A.P., then the middle term $y$ is the arithmetic mean of the other two, i.e., $2y = x + z$. Comparing this with our result, where $x = 1/a$, $y = 1/b$, and $z = 1/c$, we can conclude that $1/a, 1/b, 1/c$ are in A.P.

The final answer is $\boxed{(A)}$.

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Tips and Common Mistakes to Avoid:

• Understanding Homogeneous Systems: Remember that a homogeneous system $AX=0$ always has the trivial solution ($X=0$). A non-zero solution exists only when $\text{det}(A)=0$.
• Determinant Calculation: Be careful with signs when expanding determinants. Row/column operations are powerful tools to simplify determinants, but apply them correctly. Remember that $R_i \to R_i + k R_j$ does not change the determinant value. Swapping rows/columns changes the sign. Multiplying a row/column by $k$ multiplies the determinant by $k$.
• Algebraic Manipulation: Pay close attention to parentheses and signs when expanding and simplifying algebraic expressions. A single sign error can lead to a completely different result.
• Division by Variables: Always check if the variables you are dividing by can be zero. In this problem, $a, b, c$ are given as non-zero, so division by $abc$ is valid. If they could be zero, this step would require careful case analysis.
• Recognizing Series Conditions: Be familiar with the conditions for Arithmetic Progression (A.P.), Geometric Progression (G.P.), and Harmonic Progression (H.P.).
  • A.P.: $2y = x + z$ (or $y-x = z-y$)
  • G.P.: $y^2 = xz$ (or $y/x = z/y$)
  • H.P.: $1/x, 1/y, 1/z$ are in A.P. (which means $2/y = 1/x + 1/z$)

Summary/Key Takeaway:

This problem beautifully connects the concept of non-trivial solutions for homogeneous linear systems (determinant equals zero) with algebraic manipulation leading to the definition of an Arithmetic Progression. It highlights the importance of determinant properties and careful algebraic simplification to arrive at the desired relationship between the coefficients. The condition $2ac = ab + bc$ is a common result that directly implies the reciprocals are in A.P. when all variables are non-zero.