<p>The system of equations<br>\(-2x + y + z = a\)<br>\(x - 2y + z = b\)<br>\(x + y - 2z = c\)<br>has:</p>
<p>no solution if \(a + b + c \neq 0\)</p>
<p>unique solution if \(a + b + c = 0\)</p>
<p>infinite number of solutions if \(a + b + c = 0\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: The coefficient matrix has a special structure (cyclic form with -2 on diagonal and 1s elsewhere); check if its determinant is zero to determine consistency conditions. The sum of all three equations reveals a critical constraint on a, b, c.
<p><strong>Step 1:</strong> Find the determinant of coefficient matrix A:</p><p>A = [[-2, 1, 1], [1, -2, 1], [1, 1, -2]]</p><p>det(A) = -2(4-1) - 1(-2-1) + 1(1+2) = -2(3) + 3 + 3 = 0</p><p><strong>Step 2:</strong> Since det(A) = 0, the system either has no solution or infinitely many solutions. Add all three equations:</p><p>(-2x + y + z) + (x - 2y + z) + (x + y - 2z) = a + b + c</p><p>0 = a + b + c</p><p><strong>Step 3:</strong> This is the compatibility condition. If a + b + c = 0, the system is consistent (infinitely many solutions exist since rank = 2). If a + b + c ≠ 0, the system is inconsistent (no solution).</p><p><strong>Step 4:</strong> The rank of A is 2 (since det = 0 but not all minors are zero), so when consistent, we have 3 - 2 = 1 free variable.</p><p>∴ Answer: System has infinitely many solutions if and only if a + b + c = 0; otherwise no solution exists. (Options A and C typically state these outcomes)</p>
Correct Answer: A,C