Matrices & Determinants
Determinant Properties
Grade 12
Question:
<p><strong>Question 82:</strong> If $\phi(r) = \begin{vmatrix} r & r-1 \\ r-3 & r-4 \end{vmatrix}$, then $\sum_{r=1}^{n} \phi(r) = ?$</p><p><strong>Statement-1 (Assertion):</strong> $\sum_{r=1}^{n} \phi(r) = \frac{n(n+1)}{2}$</p><p><strong>Statement-2 (Reason):</strong> If $\phi(r) = \begin{vmatrix} f_1(r) & f_2(r) \\ f_3(r) & f_4(r) \end{vmatrix}$, then $\sum_{r=1}^{n} \phi(r) = \begin{vmatrix} \sum_{r=1}^{n} f_1(r) & \sum_{r=1}^{n} f_2(r) \\ \sum_{r=1}^{n} f_3(r) & \sum_{r=1}^{n} f_4(r) \end{vmatrix}$</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: Expansion of determinants and use of the property that sum of determinants equals determinant of sums.
<p>First, compute $\phi(r) = \begin{vmatrix} r & r-1 \\ r-3 & r-4 \end{vmatrix} = r(r-4) - (r-1)(r-3) = r^2 - 4r - (r^2 - 4r + 3) = -3$</p><p>Thus $\sum_{r=1}^{n} \phi(r) = -3n$. Statement-1 needs verification of the actual sum.</p><p>Statement-2 is true: the sum of determinants equals the determinant of sums (property of determinants).</p><p>The answer is (a).</p>
Correct Answer: a