Question:

Step 1: Write the given system of equations in matrix form. The given system of linear equations is: $x + ay = 0$ $y + az = 0$ $z + ax = 0$ We can rewrite these equations to explicitly show all three variables: $x + ay + 0z = 0$ $0x + y + az = 0$ $ax + 0y + z = 0$ The coefficient matrix, $A$, for this system is: $$A = \begin{pmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{pmatrix}$$ Step 2: State the condition for a unique solution of a homogeneous system. For a homogeneous system of linear equations (where all constant terms are zero, as in this case), the system always has the trivial solution $(x, y, z) = (0, 0, 0)$. For the system to have a unique solution (which must be the trivial solution), the determinant of the coefficient matrix must be non-zero. That is, for a unique solution, we must have $\det(A) \neq 0$. Step 3: Calculate the determinant of the coefficient matrix. We calculate the determinant of matrix $A$: $$\det(A) = \begin{vmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix}$$ Expanding along the first row: $$\det(A) = 1 \cdot \begin{vmatrix} 1 & a \\ 0 & 1 \end{vmatrix} - a \cdot \begin{vmatrix} 0 & a \\ a & 1 \end{vmatrix} + 0 \cdot \begin{vmatrix} 0 & 1 \\ a & 0 \end{vmatrix}$$ $$\det(A) = 1(1 \cdot 1 - a \cdot 0) - a(0 \cdot 1 - a \cdot a) + 0$$ $$\det(A) = 1(1 - 0) - a(0 - a^2)$$ $$\det(A) = 1 - a(-a^2)$$ $$\det(A) = 1 + a^3$$ Step 4: Apply the condition for a unique solution and solve for 'a'. For a unique solution, we must have $\det(A) \neq 0$. So, we set the calculated determinant not equal to zero: $1 + a^3 \neq 0$ $a^3 \neq -1$ Taking the cube root of both sides, for real values of $a$: $a \neq -1$ Step 5: Conclude the set of all real values of 'a'. The system of equations has a unique solution for all real values of $a$ except for $a = -1$. Therefore, the set of all real values of 'a' for which the system has a unique solution is $R - \{-1\}$. The final answer is $\boxed{\text{R - \{-1\}}}$.