Matrices & Determinants
Consistency of system of equations
Grade 12
Question:
<p>If \(n = 1\) and the system of equations \(x + y - 1 = 0\), \(2x - y - c = 0\), and \(bx + 3by - c = 0\) is consistent, then the possible real values of \(b\) are</p>
<p>\(b \in \left(-3, \dfrac{3}{4}\right)\)</p>
<p>\(b \in \left(-\dfrac{3}{2}, 4\right)\)</p>
<p>\(b \in \left(-\dfrac{3}{4}, 3\right)\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: For a system of 3 equations in 2 unknowns to be consistent, the three lines must be concurrent (meet at a point). Use the condition that the coefficient matrix and augmented matrix have the same rank, which requires the determinant of any 3×3 system to equal zero.
<p><strong>Step 1:</strong> Given n = 1, we have three equations: x + y - 1 = 0, 2x - y - c = 0, and bx + 3by - c = 0.</p><p><strong>Step 2:</strong> For consistency, solve the first two equations. From x + y = 1 and 2x - y = c: Adding gives 3x = 1 + c, so x = (1+c)/3. Then y = 1 - x = (2-c)/3.</p><p><strong>Step 3:</strong> For the system to be consistent, the third equation must be satisfied by this point: bx + 3by - c = 0. Substitute the values of x and y:</p><p>b·(1+c)/3 + 3b·(2-c)/3 - c = 0</p><p>b(1+c)/3 + b(2-c) - c = 0</p><p>b(1+c)/3 + 3b(2-c)/3 - c = 0</p><p>[b(1+c) + 3b(2-c)]/3 = c</p><p>[b + bc + 6b - 3bc]/3 = c</p><p>[7b - 2bc]/3 = c</p><p>7b - 2bc = 3c</p><p>7b = 3c + 2bc = c(3 + 2b)</p><p><strong>Step 4:</strong> If b ≠ -3/2, then c = 7b/(3 + 2b). For this to hold for all consistent systems, or we can also have the third equation be a linear combination of the first two (dependent). Checking when the third equation is dependent: bx + 3by - c = k(x + y - 1) + m(2x - y - c) for some k, m. This gives b = k + 2m, 3b = k - m, and -c = -k - mc. Solving: from first two equations, k = 7b/3 and m = b/3. These relationships must hold consistently.</p><p>∴ Answer: C</p>
Correct Answer: C