Let $p, q, r$ be real numbers such that $p + q + r \neq 0$. The system of linear equations<br>$x + 2y - 3z = p$<br>$2x + 6y - 11z = q$<br>$x - 2y + 7z = r$<br>has at least one solution if :
Step-by-Step Solution
Key Concept: A system of linear equations has at least one solution if and only if the rank of the coefficient matrix equals the rank of the augmented matrix. This requires the augmented matrix rows to be linearly dependent, giving us a consistency condition.
<p><strong>Step 1:</strong> Write the coefficient matrix A and examine its determinant.</p><p>$$A = \begin{pmatrix} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7 \end{pmatrix}$$</p><p><strong>Step 2:</strong> Calculate det(A) by expanding along the first row.</p><p>$$\det(A) = 1(6 \cdot 7 - (-11)(-2)) - 2(2 \cdot 7 - (-11) \cdot 1) + (-3)(2(-2) - 6 \cdot 1)$$</p><p>$$= 1(42 - 22) - 2(14 + 11) - 3(-4 - 6)$$</p><p>$$= 1(20) - 2(25) - 3(-10)$$</p><p>$$= 20 - 50 + 30 = 0$$</p><p><strong>Step 3:</strong> Since det(A) = 0, the system is either inconsistent or has infinitely many solutions. For at least one solution to exist, the augmented matrix must have the same rank as A.</p><p><strong>Step 4:</strong> Find the linear dependence relation among rows. Row reduce or use the fact that if rows of the augmented matrix are dependent, then there exist constants $\lambda_1, \lambda_2, \lambda_3$ (not all zero) such that:</p><p>$$\lambda_1(R_1) + \lambda_2(R_2) + \lambda_3(R_3) = 0$$</p><p><strong>Step 5:</strong> Find which rows are linearly dependent in A. Testing: $R_2 - 2R_1 = (2, 6, -11) - 2(1, 2, -3) = (0, 2, -5)$. And $R_3 - R_1 = (0, -4, 10)$. Notice that $R_3 - R_1 = -2(R_2 - 2R_1)$, so: $R_3 - R_1 + 2(R_2 - 2R_1) = 0$, giving us $2R_2 + R_3 - 2R_1 = 0$.</p><p><strong>Step 6:</strong> Apply this same linear dependence to the constant terms:</p><p>$$2q + r - 2p = 0$$</p><p>$$5p - 2q - r = 0$$</p><p>∴ Answer: B</p>
Correct Answer: B