First slide

Introduction to Determinants

Question

If $D_{k} = |\begin{array}{ccc} 1 & n & n \\ 2 k & n^{2} + n + 1 & n^{2} + n \\ 2 k - 1 & n^{2} & n^{2} + n + 1 \end{array}|$ and $\sum_{k = 1}^{n} D_{k} = 56$ , then n equals

Moderate

A

4
B

6
C

8
D

7

Solution

$\sum_{k = 1}^{n} D_{k} = 56$
$\Rightarrow |\begin{array}{ccc} \sum_{k = 1}^{n} 1 & n & n \\ \sum_{k = 1}^{n} 2 k & n^{2} + n + 1 & n^{2} + n \\ \sum_{k = 1}^{n} (2 k - 1) & n^{2} & n^{2} + n + 1 \end{array}| = 56$
or $|\begin{array}{ccc} n & n & n \\ n (n + 1) & n^{2} + n + 1 & n^{2} + n \\ n^{2} & n^{2} & n^{2} + n + 1 \end{array}| = 56$
Applying $C_{3} \to C_{3} - C_{1} and C_{2} \to C_{2} - C_{1}$ , we get
$|\begin{array}{ccc} n & 0 & 0 \\ n (n + 1) & 1 & 0 \\ n^{2} & 0 & n + 1 \end{array}| = 56$
or $n (n + 1) = 56 or n = 7$

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