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The image describes a matrix equation solution process using **back substitution**. Back substitution is typically applied after forward elimination in Gaussian elimination, starting with the last equation and working upwards. Let me break down the contents of the image:

1. **The Matrix System**: 
   The system of equations is represented by the augmented matrix:
   \[
   \begin{bmatrix}
   1 & 0.5 & 0.5 & 5 \\
   0 & 1 & 3 & 6 \\
   0 & 0 & -2 & -10
   \end{bmatrix}
   \]
   This represents a triangular system where you can solve the equations from bottom to top.

2. **Back Substitution Formulas**:
   The image provides the formulas for solving the unknowns \(x_2\), \(x_1\), and \(x_0\) step by step:
   
   - \( x_2 = \frac{b_2}{a_{22}} \)
   - \( x_1 = \frac{b_1 - a_{12}x_2}{a_{11}} \)
   - \( x_0 = \frac{b_0 - a_{01}x_1 - a_{02}x_2}{a_{00}} \)

   These represent how, after solving for \(x_2\), you substitute its value into the equation for \(x_1\), and then substitute both values into the equation for \(x_0\).

### Steps to Solve:

- From the third row: 
  \[
  -2x_2 = -10 \implies x_2 = 5
  \]
- From the second row:
  \[
  x_1 + 3x_2 = 6 \implies x_1 + 3(5) = 6 \implies x_1 = 6 - 15 = -9
  \]
- From the first row:
  \[
  x_0 + 0.5x_1 + 0.5x_2 = 5 \implies x_0 + 0.5(-9) + 0.5(5) = 5
  \]
  Simplifying this:
  \[
  x_0 - 4.5 + 2.5 = 5 \implies x_0 - 2 = 5 \implies x_0 = 7
  \]

Thus, the solution is:
\[
x_0 = 7, \quad x_1 = -9, \quad x_2 = 5
\]

Would you like further clarification on any of the steps?

### 5 Relative Questions:
1. What is the significance of back substitution in solving linear systems?
2. How does Gaussian elimination relate to the back substitution method?
3. Can this method be applied to non-triangular matrices?
4. What modifications would be needed if there were more variables in the system?
5. How does the accuracy of the back substitution method compare to other solution methods?

### Tip:
When performing back substitution, always start from the equation corresponding to the last row (the simplest one) to avoid unnecessary complexity in solving.

How do I perform back substitution on the given matrix to solve for the variables x0, x1, and x2?

This page provides a detailed step-by-step solution using back substitution for solving a system of linear equations represented by a triangular matrix. Key concepts include Gaussian elimination, back substitution formulas, and solving for x0, x1, and x2. Perfect for students studying linear algebra at the college level.

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