Matrices & Determinants
Determinant expansion and polynomial coefficients
Grade 12
Question:
<p><strong>For Problems 10–12</strong><br>\[f(x) = \begin{vmatrix} x+c_1 & x+a & x+a \\ x+b & x+c_2 & x+a \\ x+b & x+b & x+c_3 \end{vmatrix}\] and \(g(x) = (c_1 - x)(c_2 - x)(c_3 - x)\)</p><p>Which of the following is not a constant term in \(f(x)\)?</p>
<p>\(\dfrac{bg(a)-ag(b)}{b-a}\)</p>
<p>\(\dfrac{bg(a)-af(-b)}{b-a}\)</p>
<p>\(\dfrac{bf(-a)-ag(b)}{b-a}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) is a cubic polynomial in x whose constant term comes from the determinant when x=0. The key is understanding which coefficients in the original matrix directly contribute to the constant term versus which appear in linear/higher-order terms.
<p><strong>Step 1:</strong> Recognize f(x) is a cubic polynomial. Write f(x) = Ax³ + Bx² + Cx + D where D = f(0).</p><p><strong>Step 2:</strong> Evaluate f(0): f(0) = |c₁ a a| |b c₂ a| |b b c₃|. Expand using standard determinant formula.</p><p><strong>Step 3:</strong> f(0) = c₁(c₂c₃ - ab) - a(bc₃ - ab) + a(b² - bc₂) = c₁c₂c₃ - c₁ab - abc₃ + a²b + ab² - abc₂</p><p><strong>Step 4:</strong> Compare with g(x) = (c₁-x)(c₂-x)(c₃-x). At x=0: g(0) = c₁c₂c₃. Note that f(0) ≠ g(0) in general unless ab = 0.</p><p><strong>Step 5:</strong> The constant term in f(x) includes the product c₁c₂c₃ plus additional terms involving a and b. Therefore, any option suggesting a term that would only appear through the combination of a and b (not through individual c_i products) is NOT a constant term.</p><p>∴ Answer: B</p>
Correct Answer: B