Determine b1, b2, b3 for a Consistent Linear System

To solve the system and determine the values of $b_1$ , $b_2$ , and $b_3$ for consistency, we first express the system in matrix form. The system of equations is:

x + y - z &= b_1 \quad \text{(Equation 1)} \\ 2y + z &= b_2 \quad \text{(Equation 2)} \\ y - z &= b_3 \quad \text{(Equation 3)} \end{aligned}$$ We can write this as a matrix equation: $$\begin{pmatrix} 1 & 1 & -1 \\ 0 & 2 & 1 \\ 0 & 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$$ The matrix equation is $$A\mathbf{x} = \mathbf{b}$$, where: $$A = \begin{pmatrix} 1 & 1 & -1 \\ 0 & 2 & 1 \\ 0 & 1 & -1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$$ For the system to be consistent, the vector $$\mathbf{b}$$ must lie in the column space of the matrix $$A$$. We can find the conditions on $$b_1$$, $$b_2$$, and $$b_3$$ by performing row operations to reduce $$A$$ to its row echelon form (REF) and analyze the result. ### Step 1: Row Reduction of Matrix $$A$$ We start with the matrix $$A$$: $$\begin{pmatrix} 1 & 1 & -1 \\ 0 & 2 & 1 \\ 0 & 1 & -1 \end{pmatrix}$$ Let's perform row reduction: 1. **No changes to the first row:** $$\text{Row 1 (R1)} = \begin{pmatrix} 1 & 1 & -1 \end{pmatrix}$$ 2. **Row 2 (R2) stays as is:** $$\text{Row 2 (R2)} = \begin{pmatrix} 0 & 2 & 1 \end{pmatrix}$$ 3. **Row 3 (R3) stays as is:** $$\text{Row 3 (R3)} = \begin{pmatrix} 0 & 1 & -1 \end{pmatrix}$$ Next, divide R2 by 2: $$R2' = \begin{pmatrix} 0 & 1 & \frac{1}{2} \end{pmatrix}$$ Subtract R2' from R3: $$R3' = R3 - R2' = \begin{pmatrix} 0 & 1 & -1 \end{pmatrix} - \begin{pmatrix} 0 & 1 & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} 0 & 0 & -\frac{3}{2} \end{pmatrix}$$ The row echelon form of matrix $$A$$ is: $$\text{REF}(A) = \begin{pmatrix} 1 & 1 & -1 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & -\frac{3}{2} \end{pmatrix}$$ ### Step 2: Analyze the Row-Echelon Form The last row of the row echelon form gives the equation: $$-\frac{3}{2}z = b_3 - b_2$$ For the system to be consistent, the right-hand side of this equation must equal zero: $$b_3 - b_2 = 0 \quad \Rightarrow \quad b_3 = b_2$$ ### Conclusion The system is consistent if $$b_3 = b_2$$. So, $$b_1$$ can be any real number, and $$b_2$$ and $$b_3$$ must be equal for the system to be consistent. If $$b_3 = b_2$$, then the system has a solution for any $$b_1$$. ### Additional Questions: 1. How would the solution change if the third equation was different, e.g., $$2y - z = b_3$$? 2. What happens to the solution if the matrix $$A$$ has a determinant of zero? 3. How does the consistency of the system change if we alter the first equation to $$2x + y - z = b_1$$? 4. Can you provide an example where the system is inconsistent, and why does this occur? 5. How would you approach solving the system if one of the variables was eliminated beforehand? ### Tip: When working with linear systems, always check the determinant or row-echelon form to determine if a unique solution exists, if there are infinitely many solutions, or if the system is inconsistent.

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