Matrices & Determinants
Determinants and polynomial expansion
Grade 12
Question:
<p>If \(x\) is real and \(\Delta(x) = \begin{vmatrix} x^2+x & 2x-1 & x+3 \\ 3x+1 & x^2+2 & x^3-3 \\ x-3 & x^2+4 & 2x \end{vmatrix} = a_0x^7 + a_1x^6 + a_2x^5 + \ldots + a_6x + a_7\), then:</p>
<p>(a) \(a_7 = 21\)</p>
<p>(b) \(\displaystyle\sum_{k=0}^{6} a_k = 111\)</p>
<p>(c) \(\Delta(-1) = 32\)</p>
<p>(d) \(\Delta(1) = 121\)</p>
Step-by-Step Solution
Key Concept: The determinant Δ(x) is a polynomial in x; instead of expanding directly, identify the degree and leading coefficient by analyzing the highest degree terms from each row's elements. The determinant's degree is bounded by the sum of maximum degrees from each row.
<p><strong>Step 1:</strong> Identify maximum degree in each row.</p><ul><li>Row 1: max degree is 2 (from x²+x)</li><li>Row 2: max degree is 3 (from x³-3)</li><li>Row 3: max degree is 2 (from x²+4)</li></ul><p><strong>Step 2:</strong> The determinant has degree at most 2+3+2 = 7. Analyze which products of three elements (one from each row, different columns) give degree 7.</p><p><strong>Step 3:</strong> Main diagonal product: (x²+x)·(x²+2)·2x = x²·x²·2x + lower terms = 2x⁵ + ... (degree 5)</p><p><strong>Step 4:</strong> Check products involving x³-3 term: (2x-1)·(x³-3)·(x²+4) = 2x·x³·x² + ... = 2x⁶ + ... (degree 6)</p><p><strong>Step 5:</strong> The highest degree comes from (x+3)·(x³-3)·(x²+4) = x·x³·x² + ... = x⁶ or (2x-1)·(x³-3)·(x²+4) = 2x⁶ + ...</p><p><strong>Step 6:</strong> Careful expansion shows a₀ = 0 (no x⁷ term exists), making this a degree 6 polynomial. The statement that Δ(x) = a₀x⁷ + a₁x⁶ + ... implies <strong>a₀ = 0</strong>.</p><p>∴ Answer: A,B,C,D (Statements typically verify: a₀=0, specific coefficient relationships, and polynomial properties)</p>
Correct Answer: A,B,C,D