Question

Consider the following system of equations: \[ x - 2y + 3z = -1 \] \[ x - 3y + 4z = 1 \] \[ -2x + 4y - 6z = k \] The value of \( k \) for which the system has infinitely many solutions is \dots\dots.

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( -1 \)

Hint:

Look for proportional rows directly. Notice that the expression on the left-hand side of the third equation, \(-2x + 4y - 6z\), is exactly \(-2\) times the left-hand side of the first equation, \(x - 2y + 3z\). For consistency, the right-hand side must follow the same scalar relationship: \(k = -2 \times (-1) = 2\).

Answer 1

The Correct Option is C

Concept: A system of linear equations can be represented in matrix form as \( AX = B \). According to the Rouché-Capelli theorem:
• The system has a unique solution if \(\text{rank}(A) = \text{rank}([A|B]) = \text{number of variables}\).
• The system has infinitely many solutions if \(\text{rank}(A) = \text{rank}([A|B]) < \text{number of variables}\).
• The system is inconsistent (no solution) if \(\text{rank}(A) \neq \text{rank}([A|B])\).

Step 1: Write down the Augmented Matrix \([A|B]\)
Construct the augmented matrix containing the coefficients and the right-hand side constraints: \[ [A|B] = \left[ \begin{array}{ccc|c} 1 & -2 & 3 & -1 \\ 1 & -3 & 4 & 1 \\ -2 & 4 & -6 & k \end{array} \right] \]

Step 2: Apply Elementary Row Operations to reach Row Echelon Form
To simplify the matrix, eliminate elements below the leading entry of the first column. Perform the operations: \[ R_2 \leftarrow R_2 - R_1 \] \[ R_3 \leftarrow R_3 + 2R_1 \] Let's calculate the new rows explicitly: Row 2 calculation: \[ [1 - 1, \,\, -3 - (-2), \,\, 4 - 3, \,\, 1 - (-1)] = [0, \,\, -1, \,\, 1, \,\, 2] \] Row 3 calculation: \[ [-2 + 2(1), \,\, 4 + 2(-2), \,\, -6 + 2(3), \,\, k + 2(-1)] = [0, \,\, 0, \,\, 0, \,\, k - 2] \] Updating our augmented matrix structure: \[ [A|B] \sim \left[ \begin{array}{ccc|c} 1 & -2 & 3 & -1 \\ 0 & -1 & 1 & 2 \\ 0 & 0 & 0 & k - 2 \end{array} \right] \]

Step 3: Analyze rank for infinite solution criterion
From the simplified matrix, the coefficient matrix row profile reveals: \[ \text{rank}(A) = 2 \] since the entire third row of matrix \(A\) contains only zeros. For the system to possess infinitely many solutions, the rank of the augmented matrix \([A|B]\) must also equal 2. This requires the last entry in the third row of the augmented matrix to be zero: \[ k - 2 = 0 \implies k = 2 \] If \( k \neq 2 \), the rank of \([A|B]\) would be 3, rendering the system completely inconsistent. Thus, \(k = 2\) is the unique value that yields an infinite solution space, corresponding to Option (C).