a) Consider the linear system $$ \left\{\begin{array}{c}3 x+y=k \ k^{2} x+3 y=-9\end{array} ight. $$ For which values of $$ k $$ does the linear system has i- A unique solution. ii- Infinitely many solutions iii- No solution. b) Consider the linear system in part a. solve it at $$ k=2 $$.

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a) Consider the linear system $$ \left\{\begin{array}{c}3 x+y=k \ k^{2} x+3 y=-9\end{array}ight. $$ For which values of $$ k $$ does the linear system has i- A unique solution. ii- Infinitely many solutions iii- No solution. b) Consider the linear system in part a. solve it at $$ k=2 $$.

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**a)**  
We are given the system  
$$
\begin{cases}
3x + y = k,\[1mm]
k^2 x + 3y = -9.
\end{cases}
$$

1. **Find the determinant:**  
   Write the system in matrix form:
   $$
   \begin{pmatrix}
   3 & 1 \[1mm]
   k^2 & 3
   \end{pmatrix}
   \begin{pmatrix}
   x \[1mm]
   y
   \end{pmatrix}
   =
   \begin{pmatrix}
   k \[1mm]
   -9
   \end{pmatrix}.
   $$
   The determinant is given by:
   $$
   \det = 3 \cdot 3 - k^2 \cdot 1 = 9 - k^2.
   $$

2. **i- Unique solution:**  
   A unique solution exists if the determinant is nonzero:
   $$
   9 - k^2 \neq 0 \quad \Longrightarrow \quad k^2 \neq 9 \quad \Longrightarrow \quad k \neq 3 \text{ and } k \neq -3.
   $$

3. **ii- Infinitely many solutions:**  
   If the determinant is $$0$$ (so $$k = \pm3$$) the system might have infinitely many solutions if it is consistent.  
   - For $$k = -3$$:  
     The system becomes  
     $$
     \begin{cases}
     3x + y = -3,\[1mm]
     (-3)^2 x + 3y = 9x + 3y = -9.
     \end{cases}
     $$
     Solve the first equation for $$y$$:
     $$
     y = -3 - 3x.
     $$
     Substitute into the second equation:
     $$
     9x + 3(-3 - 3x) = 9x - 9 - 9x = -9.
     $$
     This is an identity ($$-9 = -9$$), so the system has infinitely many solutions when $$k=-3$$.

4. **iii- No solution:**  
   For $$k = 3$$:  
     The system becomes  
     $$
     \begin{cases}
     3x + y = 3,\[1mm]
     3^2 x + 3y = 9x + 3y = -9.
     \end{cases}
     $$
     Solve the first equation for $$y$$:
     $$
     y = 3 - 3x.
     $$
     Substitute into the second equation:
     $$
     9x + 3(3 - 3x) = 9x + 9 - 9x = 9.
     $$
     This yields $$9 = -9$$, which is a contradiction. Therefore, the system has no solution when $$k=3$$.

**b)**  
For $$k=2$$, the system becomes
$$
\begin{cases}
3x + y = 2,\[1mm]
2^2 x + 3y = 4x + 3y = -9.
\end{cases}
$$

1. Solve the first equation for $$y$$:
   $$
   y = 2 - 3x.
   $$

2. Substitute into the second equation:
   $$
   4x + 3(2 - 3x) = 4x + 6 - 9x = -5x + 6 = -9.
   $$

3. Solve for $$x$$:
   $$
   -5x = -9 - 6 \quad \Longrightarrow \quad -5x = -15 \quad \Longrightarrow \quad x = 3.
   $$

4. Substitute $$x=3$$ back into $$y = 2 - 3x$$:
   $$
   y = 2 - 3(3) = 2 - 9 = -7.
   $$

**Summary of Answers:**

- **i- Unique solution:** for $$k \neq 3$$ and $$k \neq -3$$.  
- **ii- Infinitely many solutions:** for $$k = -3$$.  
- **iii- No solution:** for $$k = 3$$.  
- For $$k=2$$, the solution is $$x = 3$$ and $$y = -7$$.
**a)** - **Unique Solution:** When $$ k \neq 3 $$ and $$ k \neq -3 $$. - **Infinitely Many Solutions:** When $$ k = -3 $$. - **No Solution:** When $$ k = 3 $$. **b)** For $$ k = 2 $$, the solution is $$ x = 3 $$ and $$ y = -7 $$.

a) Consider the linear system \( \left\{\begin{array}{c}3 x+y=k \\ k^{2} x+3 y=-9\end{array}\right. \) For which values of \( k \) does the linear system has i- A unique solution. ii- Infinitely many solutions iii- No solution. b) Consider the linear system in part a. solve it at \( k=2 \).

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