Question

Let \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 3 & 1 \\ -2 & -3 & -3 \end{bmatrix}, \quad b = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}. \] For \( Ax = b \) to be solvable, which one of the following options is the correct condition on \( b_1, b_2, \) and \( b_3 \)?

• A. \( b_1 + b_2 + b_3 = 1 \)
• B. \( 3b_1 + b_2 + 2b_3 = 0 \)
• C. \( b_1 + 3b_2 + b_3 = 2 \)
• D. \( b_1 + b_2 + b_3 = 2 \)

Hint:

When solving systems of linear equations, converting the augmented matrix to echelon form helps identify the conditions for consistency.

Answer 1

The Correct Option is B

Understanding the Problem: 
We are given the system \( Ax = b \), where \( A \) is a \( 3 \times 3 \) matrix and \( b \) is a column vector. 
To determine if the system is consistent, we must analyze the augmented matrix \( [A : b] \) and ensure that it does not lead to an inconsistent row (such as \( 0 = c \) where \( c \neq 0 \)). 
Step 1: Constructing the Augmented Matrix 
The given system can be represented as: \[ [A : b] = \begin{bmatrix} 1 & 1 & 1 & b_1 \\ 1 & 3 & 1 & b_2 \\ -2 & -3 & -3 & b_3 \end{bmatrix}. \] We will now perform row operations to convert it into row echelon form. 
Step 2: Row Reduction 
- Subtract the first row from the second row: \[ R_2 \to R_2 - R_1 \] \[ \begin{bmatrix} 1 & 1 & 1 & b_1 \\ 0 & 2 & 0 & b_2 - b_1 \\ -2 & -3 & -3 & b_3 \end{bmatrix}. \] - Add 2 times the first row to the third row: \[ R_3 \to R_3 + 2R_1 \] \[ \begin{bmatrix} 1 & 1 & 1 & b_1 \\ 0 & 2 & 0 & b_2 - b_1 \\ 0 & -1 & -1 & b_3 + 2b_1 \end{bmatrix}. \] - Add \( \frac{1}{2} \) of the second row to the third row: \[ R_3 \to R_3 + \frac{1}{2} R_2 \] \[ \begin{bmatrix} 1 & 1 & 1 & b_1 \\ 0 & 2 & 0 & b_2 - b_1 \\ 0 & 0 & -1 & b_3 + 2b_1 + \frac{b_2 - b_1}{2} \end{bmatrix}. \] The last row should not lead to an inconsistency (such as \( 0 = c \neq 0 \)). 
This condition holds when: \[ b_3 + 2b_1 + \frac{b_2 - b_1}{2} = 0. \] Multiplying through by 2 for simplification: \[ 2b_3 + 4b_1 + b_2 - b_1 = 0. \] \[ 3b_1 + b_2 + 2b_3 = 0. \] Final Conclusion: 
The system remains consistent when the condition \[ 3b_1 + b_2 + 2b_3 = 0 \] is satisfied. 
Correct Answer: Option (B) ✅