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

Options

A. A.P
B. G.P
C. H.P
D. None of these

Answer

To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) such that the system of linear equations has a non-zero solution. This condition is satisfied when the determinant of the coefficients of the system is equal to zero.

The given equations are:
1. \( x + 2ay + az = 0 \)
2. \( x + 3by + bz = 0 \)
3. \( x + 4cy + cz = 0 \)

We can represent these equations in matrix form as follows:

\[ \begin{bmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{bmatrix} \]

We need to find the determinant of this matrix and set it equal to zero.

### Step 1: Calculate the Determinant

The determinant of a 3x3 matrix

\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

is given by the formula:

\[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Applying this to our matrix:

\[ \text{det} = 1 \cdot (3b \cdot c - b \cdot 4c) - 2a \cdot (1 \cdot c - b \cdot 1) + a \cdot (1 \cdot 4c - 3b \cdot 1) \]

### Step 2: Simplify the Determinant Expression

Calculating each term:

1. First term: \( 1 \cdot (3bc - 4bc) = -bc \)
2. Second term: \( -2a \cdot (c - b) = -2a(c - b) \)
3. Third term: \( a \cdot (4c - 3b) = a(4c - 3b) \)

Putting it all together, we have:

\[ \text{det} = -bc - 2a(c - b) + a(4c - 3b) \]

### Step 3: Set the Determinant to Zero

Setting the determinant equal to zero for non-zero solutions:

\[ -bc - 2a(c - b) + a(4c - 3b) = 0 \]

### Step 4: Rearranging the Equation

Rearranging the equation gives:

\[ -bc - 2ac + 2ab + 4ac - 3ab = 0 \]

Combining like terms:

\[ (-bc + 2ab - 3ab + 4ac - 2ac) = 0 \]

This simplifies to:

\[ -bc - ab + 2ac = 0 \]

### Step 5: Factor the Equation

Rearranging gives:

\[ 2ac = ab + bc \]

### Step 6: Divide by abc (assuming \( a, b, c \neq 0 \))

Dividing through by \( abc \):

\[ \frac{2}{b} = \frac{1}{c} + \frac{1}{a} \]

This indicates that \( a, b, c \) are in harmonic progression.

### Final Result

Thus, the condition for \( a, b, c \) to have a non-zero solution is that they are in harmonic progression.

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