Consider the system of equations : $$x + ay = 0$$, $$y + az = 0$$ and $$z + ax = 0$$. Then the set of all real values of 'a' for which the system has a unique solution is:

We are given the system of equations:

• $$ x + a y = 0 $$
• $$ y + a z = 0 $$
• $$ z + a x = 0 $$

This is a homogeneous system because all equations equal zero. For a homogeneous system to have a unique solution (which must be the trivial solution $$ x = 0 $$, $$ y = 0 $$, $$ z = 0 $$), the coefficient matrix must be invertible. This happens when its determinant is non-zero.

First, write the system in matrix form:

$$ \begin{pmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} $$

The coefficient matrix is:

$$ A = \begin{pmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \\ \end{pmatrix} $$

Now, compute the determinant of matrix $$ A $$. For a 3x3 matrix:

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

Substitute the elements:

• $$ a_{11} = 1 $$, $$ a_{12} = a $$, $$ a_{13} = 0 $$
• $$ a_{21} = 0 $$, $$ a_{22} = 1 $$, $$ a_{23} = a $$
• $$ a_{31} = a $$, $$ a_{32} = 0 $$, $$ a_{33} = 1 $$

So,

$$ \det(A) = 1 \cdot (1 \cdot 1 - a \cdot 0) - a \cdot (0 \cdot 1 - a \cdot a) + 0 \cdot (0 \cdot 0 - 1 \cdot a) $$

Simplify each part:

• First part: $$ 1 \cdot (1 - 0) = 1 \cdot 1 = 1 $$
• Second part: $$ -a \cdot (0 - a^2) = -a \cdot (-a^2) = a \cdot a^2 = a^3 $$
• Third part: $$ 0 \cdot (0 - a) = 0 \cdot (-a) = 0 $$

Therefore,

$$ \det(A) = 1 + a^3 + 0 = 1 + a^3 $$

For a unique solution, $$ \det(A) \neq 0 $$:

$$ 1 + a^3 \neq 0 $$

Solve:

$$ a^3 \neq -1 $$

$$ a \neq -1 $$

Now, verify the case when $$ a = -1 $$:

• Equation 1: $$ x + (-1)y = 0 $$ → $$ x - y = 0 $$ → $$ x = y $$
• Equation 2: $$ y + (-1)z = 0 $$ → $$ y - z = 0 $$ → $$ y = z $$
• Equation 3: $$ z + (-1)x = 0 $$ → $$ z - x = 0 $$ → $$ z = x $$

So, $$ x = y = z $$, meaning infinitely many solutions exist (not unique).

For any other real value of $$ a $$, such as $$ a = 0 $$ or $$ a = 1 $$, the system has only the trivial solution (as shown by substitution). Thus, the system has a unique solution for all real $$ a $$ except $$ a = -1 $$.

The set of all real values of $$ a $$ for which the system has a unique solution is $$ \mathbb{R} - \{-1\} $$.

Comparing with the options:

• A. $$ \mathbb{R} - \{1\} $$ → Incorrect, as $$ a = 1 $$ gives unique solution.
• B. $$ \mathbb{R} - \{-1\} $$ → Correct.
• C. $$ \{1, -1\} $$ → Incorrect, as it includes $$ a = -1 $$ and excludes other values.
• D. $$ \{1, 0, -1\} $$ → Incorrect, as it includes $$ a = -1 $$ and excludes other values.

Hence, the correct answer is Option B.