Call a polynomial $ P(x_{1}, \ldots, x_{k})$ [i]good[/i] if there exist $ 2\times 2$ real matrices $ A_{1}, \ldots, A_{k}$ such that 
$ P(x_{1}, \ldots, x_{k}) = \det \left(\sum_{i=1}^{k}x_{i}A_{i}\right).$ 

Find all values of $ k$ for which all homogeneous polynomials with $ k$ variables of degree 2 are good. (A polynomial is homogeneous if each term has the same total degree.)
1. **Understanding the Problem:**
   We need to determine for which values of \( k \) all homogeneous polynomials of degree 2 in \( k \) variables can be expressed as the determinant of a linear combination of \( 2 \times 2 \) real matrices. Specifically, we need to find \( k \) such that for any homogeneous polynomial \( P(x_1, \ldots, x_k) \) of degree 2, there exist \( 2 \times 2 \) matrices \( A_1, \ldots, A_k \) such that:
   \[
   P(x_1, \ldots, x_k) = \det \left( \sum_{i=1}^{k} x_i A_i \right).
   \]

2. **Case \( k = 1 \) and \( k = 2 \):**
   - For \( k = 1 \), any homogeneous polynomial of degree 2 is of the form \( P(x_1) = c x_1^2 \). We can choose \( A_1 = \begin{pmatrix} \sqrt{c} & 0 \\ 0 & \sqrt{c} \end{pmatrix} \), and then:
     \[
     \det(x_1 A_1) = \det \left( x_1 \begin{pmatrix} \sqrt{c} & 0 \\ 0 & \sqrt{c} \end{pmatrix} \right) = x_1^2 c = P(x_1).
     \]
   - For \( k = 2 \), consider any homogeneous polynomial of degree 2, \( P(x_1, x_2) = a x_1^2 + b x_2^2 + c x_1 x_2 \). We can choose:
     \[
     A_1 = \begin{pmatrix} \sqrt{a} & 0 \\ 0 & \sqrt{a} \end{pmatrix}, \quad A_2 = \begin{pmatrix} 0 & \sqrt{b} \\ \sqrt{b} & 0 \end{pmatrix}, \quad A_3 = \begin{pmatrix} 0 & \frac{c}{2} \\ \frac{c}{2} & 0 \end{pmatrix}.
     \]
     Then:
     \[
     \det(x_1 A_1 + x_2 A_2 + x_3 A_3) = a x_1^2 + b x_2^2 + c x_1 x_2 = P(x_1, x_2).
     \]

3. **Case \( k = 3 \):**
   - We need to show that not all homogeneous polynomials of degree 2 in 3 variables can be expressed in the desired form. Consider the polynomial \( P(x_1, x_2, x_3) = x_1^2 + x_2^2 + x_3^2 \).
   - Assume there exist \( 2 \times 2 \) matrices \( A, B, C \) such that:
     \[
     \det(x_1 A + x_2 B + x_3 C) = x_1^2 + x_2^2 + x_3^2.
     \]
   - This implies that \( A, B, C \) must satisfy certain properties. Specifically, since the determinant of a \( 2 \times 2 \) matrix is a quadratic form, the matrices must be such that their combination yields the polynomial \( x_1^2 + x_2^2 + x_3^2 \).
   - By Cayley-Hamilton theorem, for any \( 2 \times 2 \) matrix \( M \), \( M^2 - (\text{tr}(M))M + \det(M)I = 0 \). For \( A, B, C \) to satisfy the polynomial, they must have specific traces and determinants.
   - If we assume \( A = I \) (identity matrix), then:
     \[
     \det(x_1 I + x_2 B + x_3 C) = x_1^2 + x_2^2 + x_3^2.
     \]
   - This implies \( B^2 = -I \) and \( C^2 = -I \), which means \( B \) and \( C \) are similar to the imaginary unit \( i \) in the complex numbers, leading to a contradiction since the group generated by such matrices is non-abelian and finite, which cannot be the case for real matrices.

4. **Conclusion:**
   - Therefore, \( k = 3 \) does not work, and the only values of \( k \) for which all homogeneous polynomials of degree 2 are good are \( k \leq 2 \).

The final answer is \( \boxed{ k \leq 2 } \).