https://file.aiquickdraw.com/mathbot/file/1726748375625t0e78ksl.jpg

I see the image you uploaded contains three linear algebra problems. I'll guide you through each of them in detail.

### Problem 1: Matrix Multiplication for Production Cost

You are given the cost per portion for rice, meat, and salad, as well as the portions used in different dish types \( P_1, P_2, P_3 \).

- **Cost per portion**:
  \[
  \text{Rice} = 2, \text{Meat} = 4, \text{Salad} = 3
  \]
  
- **Portions in dish types**:
  \[
  P_1 = [2, 1, 2], \quad P_2 = [1, 2, 2], \quad P_3 = [2, 1, 1]
  \]
  
You need to multiply the matrices to find the total cost for each dish type.

The matrix multiplication is:

\[
\begin{bmatrix}
2 & 1 & 2 \\
1 & 2 & 2 \\
2 & 1 & 1
\end{bmatrix}
\times
\begin{bmatrix}
2 \\
4 \\
3
\end{bmatrix}
=
\begin{bmatrix}
14 \\
16 \\
11
\end{bmatrix}
\]

So, the correct answer is \( \boxed{a} \).

---

### Problem 2: Solving the Determinant-Based Equation

You are asked to solve the equation involving a determinant:

\[
\left|
\begin{matrix}
1 & 0 & x \\
x & 3 & 1 \\
2 & 1 & x
\end{matrix}
\right| = x^3 - 13x + 5
\]

The solution involves:
1. **Expanding the determinant** using the Sarrus rule (repeating the first two rows and calculating the diagonals).
2. You get the quadratic equation \( x^2 - 8x + 12 = 0 \).
3. Solving it yields \( x = 2 \) or \( x = 6 \).

Thus, the solution set is \( \boxed{S = \{2, 6\}} \).

---

### Problem 3: Solving the Matrix Equation \( AX + B = C \)

You are given matrices:

\[
A = \begin{bmatrix} 1 & -2 \\ 4 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} -2 \\ 6 \end{bmatrix}, \quad C = \begin{bmatrix} -5 \\ 5 \end{bmatrix}
\]

You need to find \( X = \begin{bmatrix} x \\ y \end{bmatrix} \) such that \( AX + B = C \).

1. Start by calculating \( AX \):
   \[
   AX = \begin{bmatrix} 1 & -2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x - 2y \\ 4x + 3y \end{bmatrix}
   \]
   
2. Set up the equation \( AX + B = C \), and substitute values:
   \[
   \begin{bmatrix} x - 2y \\ 4x + 3y \end{bmatrix} + \begin{bmatrix} -2 \\ 6 \end{bmatrix} = \begin{bmatrix} -5 \\ 5 \end{bmatrix}
   \]
   
   This results in the system of equations:
   \[
   x - 2y - 2 = -5 \quad \text{(Equation 1)}
   \]
   \[
   4x + 3y + 6 = 5 \quad \text{(Equation 2)}
   \]

3. Solving this system gives \( x = 1 \) and \( y = 2 \).

Therefore, \( X = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \).

---

Let me know if you'd like further details on any part, or have any questions! Here are 5 follow-up questions to deepen your understanding:

1. How would the result change in Problem 1 if the costs per portion were different?
2. Could you explain in more detail how Sarrus' rule is applied in Problem 2?
3. How would you solve the system of equations for different matrices \( A, B, C \) in Problem 3?
4. Can you generalize the matrix multiplication process for larger matrices?
5. What are some real-life applications of matrix equations like \( AX + B = C \)?

**Tip**: Always double-check matrix dimensions when multiplying to ensure compatibility.

Three linear algebra problems covering matrix multiplication, determinants, and solving systems of linear equations.

This page covers three linear algebra problems involving matrix multiplication, solving determinants using Sarrus' rule, and finding solutions to the system of equations. These problems are suitable for university-level students studying linear algebra.

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