1. **Problem statement:** Given the linear system with parameter $k \in \mathbb{R}$: $$\begin{cases} x_1 + k x_2 + 2 x_3 = 2 k \\ k x_1 + x_2 + (k + 2) x_3 = 2 k^2 - 2 k - 1 \\ k x_1 + k^2 x_2 + (k - 1) x_3 = 2 k^2 - 2 k - 2 \end{cases}$$ Find the values of $k$ for which the system has (a) a unique solution, (b) infinitely many solutions, and (c) no solutions. 2. **Represent the system as an augmented matrix:** $$\left[\begin{array}{ccc|c} 1 & k & 2 & 2 k \\ k & 1 & k+2 & 2 k^2 - 2 k - 1 \\ k & k^2 & k-1 & 2 k^2 - 2 k - 2 \end{array}\right]$$ 3. **Perform row operations to simplify:** - Use $R_1$ as pivot row. - Replace $R_2$ by $R_2 - k R_1$: $$R_2 = \left[k, 1, k+2, 2 k^2 - 2 k - 1\right] - k \times \left[1, k, 2, 2 k\right] = \left[0, 1 - k^2, k + 2 - 2 k, 2 k^2 - 2 k - 1 - 2 k^2\right]$$ Simplify: $$1 - k^2, \quad k + 2 - 2 k = 2 - k, \quad 2 k^2 - 2 k - 1 - 2 k^2 = -2 k - 1$$ So, $$R_2 = \left[0, 1 - k^2, 2 - k, -2 k - 1\right]$$ - Replace $R_3$ by $R_3 - k R_1$: $$R_3 = \left[k, k^2, k - 1, 2 k^2 - 2 k - 2\right] - k \times \left[1, k, 2, 2 k\right] = \left[0, k^2 - k^2, k - 1 - 2 k, 2 k^2 - 2 k - 2 - 2 k^2\right]$$ Simplify: $$0, 0, k - 1 - 2 k = -k - 1, 2 k^2 - 2 k - 2 - 2 k^2 = -2 k - 2$$ So, $$R_3 = \left[0, 0, -k - 1, -2 k - 2\right]$$ 4. **Matrix after these operations:** $$\left[\begin{array}{ccc|c} 1 & k & 2 & 2 k \\ 0 & 1 - k^2 & 2 - k & -2 k - 1 \\ 0 & 0 & -k - 1 & -2 k - 2 \end{array}\right]$$ 5. **Analyze the last row for solutions:** - The last row corresponds to: $$(-k - 1) x_3 = -2 k - 2$$ - If $-k - 1 \neq 0$, then $$x_3 = \frac{-2 k - 2}{-k - 1} = \frac{2(k + 1)}{k + 1}$$ - Cancel $k + 1$ (assuming $k \neq -1$): $$x_3 = \cancel{\frac{2 \cancel{(k + 1)}}{\cancel{k + 1}}} = 2$$ 6. **Case 1: $k \neq -1$** - The last row gives a unique $x_3$. - The second row has pivot $1 - k^2$. - If $1 - k^2 \neq 0$, i.e., $k \neq \pm 1$, then second row pivot is nonzero. - So for $k \neq \pm 1$, the system has 3 pivots and thus a unique solution. 7. **Case 2: $k = 1$** - Substitute $k=1$ into $R_2$: $$1 - 1^2 = 0, \quad 2 - 1 = 1, \quad -2(1) - 1 = -3$$ - So $R_2$ becomes: $$[0, 0, 1, -3]$$ - $R_3$ becomes: $$[0, 0, -1 - 1, -2(1) - 2] = [0, 0, -2, -4]$$ - Multiply $R_3$ by $-\frac{1}{2}$: $$[0, 0, 1, 2]$$ - Now $R_2$ and $R_3$ contradict: $$x_3 = -3 \quad \text{and} \quad x_3 = 2$$ - Contradiction means **no solution** for $k=1$. 8. **Case 3: $k = -1$** - Substitute $k=-1$ into $R_3$: $$-(-1) - 1 = 1 - 1 = 0, \quad -2(-1) - 2 = 2 - 2 = 0$$ - So last row is: $$[0, 0, 0, 0]$$ - Substitute $k=-1$ into $R_2$: $$1 - (-1)^2 = 1 - 1 = 0, \quad 2 - (-1) = 3, \quad -2(-1) - 1 = 2 - 1 = 1$$ - So $R_2$ is: $$[0, 0, 3, 1]$$ - This corresponds to: $$3 x_3 = 1 \implies x_3 = \frac{1}{3}$$ - Since $R_3$ is zero row, no contradiction. - $R_1$ and $R_2$ give pivots in $x_1$ and $x_3$, but $x_2$ is free (since $R_2$ has zero coefficient for $x_2$). - So infinitely many solutions for $k = -1$. 9. **Summary:** - Unique solution if $k \neq \pm 1$. - No solution if $k = 1$. - Infinitely many solutions if $k = -1$. **Final answers:** - (a) Unique solution for $k \in \mathbb{R} \setminus \{-1, 1\}$. - (b) Infinitely many solutions for $k = -1$. - (c) No solutions for $k = 1$.