1. The problem asks to determine the values of $k$ for which the system $$ kx + y + z = 1 $$ $$ x + ky + z = 1 $$ $$ x + y + kz = 1 $$ has a unique solution using the determinant. 2. Write the coefficient matrix $A$ of the system: $$ A = \begin{bmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{bmatrix} $$ 3. Calculate the determinant $\det(A)$: $$ \det(A) = k \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} + 1 \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} $$ Calculate each 2x2 determinant: $$ \begin{vmatrix} k & 1 \\ 1 & k \end{vmatrix} = k \cdot k - 1 \cdot 1 = k^2 - 1 $$ $$ \begin{vmatrix} 1 & 1 \\ 1 & k \end{vmatrix} = 1 \cdot k - 1 \cdot 1 = k - 1 $$ $$ \begin{vmatrix} 1 & k \\ 1 & 1 \end{vmatrix} = 1 \cdot 1 - k \cdot 1 = 1 - k $$ Substitute back: $$ \det(A) = k(k^2 - 1) - 1(k - 1) + 1(1 - k) $$ Simplify: $$ \det(A) = k^3 - k - k + 1 + 1 - k = k^3 - 3k + 2 $$ 4. For the system to have a unique solution, $\det(A) \neq 0$: $$ k^3 - 3k + 2 \neq 0 $$ 5. Find the roots of the cubic equation: $$ k^3 - 3k + 2 = 0 $$ Try possible rational roots $k = \pm1, \pm2$: - For $k=1$: $$ 1 - 3 + 2 = 0 $$ - For $k=2$: $$ 8 - 6 + 2 = 4 \neq 0 $$ - For $k=-1$: $$ -1 + 3 + 2 = 4 \neq 0 $$ - For $k=-2$: $$ -8 + 6 + 2 = 0 $$ Hence, roots are $k=1$ and $k=-2$. 6. Factor the cubic: $$ k^3 - 3k + 2 = (k-1)^2 (k+2) $$ 7. So $\det(A) = 0$ when $k=1$ or $k=-2$. 8. Therefore, the system has a unique solution for all values of $k$ except $k=1$ and $k=-2$. Final answer: $$ k \neq 1 \quad \text{and} \quad k \neq -2 $$