Question

If the system of equations \[ x + y + z = 1,\quad x + 2y + 4z = K,\quad x + 4y + 10z = K^2 \] is consistent, then \(K =\)

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

Hint:

For consistency of three equations in three unknowns, the determinant of the coefficient matrix may be zero; then check the augmented matrix rank.

Answer 1

The Correct Option is C

Concept: For a system \(A\mathbf{x} = \mathbf{b}\) to be consistent, the rank of coefficient matrix \(A\) must equal the rank of augmented matrix \([A|\mathbf{b}]\).

Step 1: Write the augmented matrix. \[ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 1 & 2 & 4 & K 1 & 4 & 10 & K^2 \end{array}\right] \]

Step 2: Row reduce. \(R_2 \leftarrow R_2 - R_1\): \[ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 0 & 1 & 3 & K-1 1 & 4 & 10 & K^2 \end{array}\right] \] \(R_3 \leftarrow R_3 - R_1\): \[ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 0 & 1 & 3 & K-1 0 & 3 & 9 & K^2-1 \end{array}\right] \] \(R_3 \leftarrow R_3 - 3R_2\): \[ \left[\begin{array}{ccc|c} 1 & 1 & 1 & 1 0 & 1 & 3 & K-1 0 & 0 & 0 & K^2 - 1 - 3(K-1) \end{array}\right] \]

Step 3: Consistency condition. Last row: \(0\cdot x + 0\cdot y + 0\cdot z = K^2 - 1 - 3K + 3\) \[ 0 = K^2 - 3K + 2 \] \[ K^2 - 3K + 2 = 0 \quad \Rightarrow \quad (K-1)(K-2) = 0 \] \[ K = 1 \quad \text{or} \quad K = 2 \]

Step 4: Check if system has a solution. For \(K=1\) or \(K=2\), the rank of \(A\) = rank of augmented matrix = 2 < 3, so infinitely many solutions. System is consistent.