Matrices & Determinants
Determinant Properties
Grade 12
Question:
<p><strong>Question 85:</strong> <strong>Statement-1:</strong> If $f(x) = \begin{vmatrix} (1-x)^{11} & (1-x)^{12} & (1-x)^{13} \\ (1-x)^{21} & (1-x)^{22} & (1-x)^{23} \\ (1-x)^{31} & (1-x)^{32} & (1-x)^{33} \end{vmatrix}$, then the coefficient of $x$ in $f(x) = 0$.</p><p><strong>Statement-2:</strong> If $P(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots + a_n x^n$, then $a_1 = P'(0)$, where dash denotes the differential coefficient.</p>
<p>(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1</p>
<p>(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1</p>
<p>(c) Statement-1 is true, Statement-2 is false</p>
<p>(d) Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: Recognize that the determinant has identical rows after factoring, making it zero. Use derivatives to extract coefficients.
<p>For Statement-1: Factor out $(1-x)^{11}$ from row 1, $(1-x)^{21}$ from row 2, and $(1-x)^{31}$ from row 3:</p><p>$f(x) = (1-x)^{11}(1-x)^{21}(1-x)^{31} \begin{vmatrix} 1 & (1-x) & (1-x)^2 \\ 1 & (1-x) & (1-x)^2 \\ 1 & (1-x) & (1-x)^2 \end{vmatrix} = 0$ (all rows identical)</p><p>So Statement-1 is true: the coefficient of $x$ is 0.</p><p>For Statement-2: If $P(x) = a_0 + a_1 x + a_2 x^2 + \cdots$, then $P'(x) = a_1 + 2a_2 x + 3a_3 x^2 + \cdots$, so $P'(0) = a_1$. Statement-2 is true.</p><p>Statement-2 does not explain Statement-1 (different contexts).</p><p>The answer is (b).</p>
Correct Answer: b