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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Determinants

NDA 2023 · Mathematics · Determinants

If A_k=[ array cc k-1 & k k-2 & k+1 array ] , then what is (A_1)+...

If A_k=\left[k−1kk−2k+1\begin{array}{cc}k-1 & k \\ k-2 & k+1\end{array}\right], then what is det⁡⁡(A1)+det⁡⁡(A2)+det⁡⁡(A3)+…+det⁡⁡(A100)equalto?\operatorname\det\left(A_1\right)+\operatorname\det \left(A_2\right)+\operatorname\det \left(A_3\right)+\ldots+\operatorname\det\left(A_{100}\right) equal to?

Options

  1. A100
  2. B1000
  3. C9900
  4. D10000

Correct answer

D. 10000

Step-by-step solution

Since, A_K=\left[K−1KK−2K+1\begin{array}{cc}K-1 & K \\ K-2 & K+1\end{array}\right]⇒∣AK∣=(K−1)(K+1)−K( K−2)=K2−1−K2+2 K=2 K−1\begin{aligned} \Rightarrow\left|A_K\right| & =(\mathrm{K}-1)(\mathrm{K}+1)-\mathrm{K}(\mathrm{~K}-2) \\ & =\mathrm{K}^2-1-\mathrm{K}^2+2 \mathrm{~K}=2 \mathrm{~K}-1 \end{aligned}Now, $\left|\mathrm{A}_1\right|+\left|\mathrm{A}_2\right|+\left|\mathrm{A}_3\right|+\ldots . .+\left|\mathrm{A}_{100}\right|\begin{aligned} & =(2 \times 1-1)+(2 \times 2-1)+(2 \times 3-1)+\ldots . \\ & =1+(2 \times 100-1) \\ & =1+3+5+\ldots .+199 \\ & =\frac{100}{2}(1+199)=100 \times 100=10000 \end{aligned}$$$
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