`x+ ay = 0, y + az=0, z+ ax=0` The value of a for which the system of equations has infinitely many solutions is

Options

A. a = 1
B. a = 0
C. a = -1
D. no value

Answer

To find the value of \( a \) for which the system of equations

\[ x + ay = 0, \quad y + az = 0, \quad z + ax = 0 \]

has infinitely many solutions, we can express this system in matrix form and then find the determinant of the coefficient matrix. If the determinant is zero, the system has either no solutions or infinitely many solutions.

### Step 1: Write the system in matrix form

The given equations can be rewritten in the form:

\[ \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix

The coefficient matrix is

\[ A = \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{bmatrix} \]

We need to calculate the determinant \( \text{det}(A) \).

Using the formula for the determinant of a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

Substituting the values from matrix \( A \):

\[ \text{det}(A) = 1 \cdot (1 \cdot 1 - a \cdot 0) - a \cdot (0 \cdot 1 - a \cdot a) + 0 \cdot (0 \cdot 0 - 1 \cdot a) \]

This simplifies to:

\[ \text{det}(A) = 1 \cdot (1) - a \cdot (-a^2) + 0 \]

\[ = 1 + a^3 \]

### Step 3: Set the determinant to zero for infinitely many solutions

To find the value of \( a \) for which the system has infinitely many solutions, we set the determinant equal to zero:

\[ 1 + a^3 = 0 \]

### Step 4: Solve for \( a \)

Rearranging gives:

\[ a^3 = -1 \]

Taking the cube root of both sides, we find:

\[ a = -1 \]

### Conclusion

Thus, the value of \( a \) for which the system of equations has infinitely many solutions is

\[ \boxed{-1} \]