do the problem again now assume b1,b2,b3 to be 0 in augmented matrix

If we assume \( b_1 = 0 \), \( b_2 = 0 \), and \( b_3 = 0 \), the system of equations becomes:

\[
\begin{aligned}
x + y - z &= 0 \quad \text{(Equation 1)} \\
2y + z &= 0 \quad \text{(Equation 2)} \\
y - z &= 0 \quad \text{(Equation 3)}
\end{aligned}
\]

We can write this system in augmented matrix form:

\[
\left[\begin{array}{ccc|c}
1 & 1 & -1 & 0 \\
0 & 2 & 1 & 0 \\
0 & 1 & -1 & 0
\end{array}\right]
\]

### Step 1: Perform Row Reduction to Row Echelon Form

Starting with the augmented matrix:

\[
\left[\begin{array}{ccc|c}
1 & 1 & -1 & 0 \\
0 & 2 & 1 & 0 \\
0 & 1 & -1 & 0
\end{array}\right]
\]

1. **No changes to the first row:**
   \[
   R1 = \begin{pmatrix} 1 & 1 & -1 & 0 \end{pmatrix}
   \]

2. **Divide the second row (R2) by 2:**
   \[
   R2' = \frac{1}{2} \times \begin{pmatrix} 0 & 2 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & \frac{1}{2} & 0 \end{pmatrix}
   \]

3. **Subtract the new R2 from R3:**
   \[
   R3' = R3 - R2' = \begin{pmatrix} 0 & 1 & -1 & 0 \end{pmatrix} - \begin{pmatrix} 0 & 1 & \frac{1}{2} & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -\frac{3}{2} & 0 \end{pmatrix}
   \]

The resulting row echelon form of the augmented matrix is:

\[
\left[\begin{array}{ccc|c}
1 & 1 & -1 & 0 \\
0 & 1 & \frac{1}{2} & 0 \\
0 & 0 & -\frac{3}{2} & 0
\end{array}\right]
\]

### Step 2: Interpret the Solution

The last row of the reduced matrix gives:

\[
-\frac{3}{2}z = 0 \quad \Rightarrow \quad z = 0
\]

Substituting \( z = 0 \) into the second row:

\[
y + \frac{1}{2}(0) = 0 \quad \Rightarrow \quad y = 0
\]

Substituting \( y = 0 \) and \( z = 0 \) into the first row:

\[
x + 0 - 0 = 0 \quad \Rightarrow \quad x = 0
\]

### Conclusion

The solution to the system is:

\[
x = 0, \quad y = 0, \quad z = 0
\]

Thus, the solution to the system is the trivial solution where all variables are zero.

### Additional Questions:
1. What would be the solution if the augmented matrix had non-zero entries for \( b_1 \), \( b_2 \), or \( b_3 \)?
2. How does the rank of the matrix relate to the existence of solutions?
3. Can this system have infinitely many solutions under any circumstances?
4. What happens to the solution if we introduce a non-homogeneous term (e.g., \( b_1 = 1 \))?
5. How would the solution change if we added another equation to the system?

### Tip:
When solving homogeneous systems of linear equations (where the right-hand side is all zeros), the system always has at least the trivial solution (all variables equal zero). Checking for non-trivial solutions involves examining the rank of the matrix.

Explore how to solve a homogeneous system of linear equations using an augmented matrix. This problem involves setting specific variables to zero in the augmented matrix and performing row reduction to determine the solution. Suitable for advanced high school or college-level students studying linear algebra.

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