KCET · Maths · Determinants
If \(A_n=\left[\begin{array}{cc}1-n & n \\ n & 1-n\end{array}\right]\), then \(\left|A_1\right|+\left|A_2\right|+\ldots\left|A_{2021}\right|=\)
- A -2021
- B \(-(2021)^2\)
- C \((2021)^2\)
- D 4042
Answer & Solution
Correct Answer
(B) \(-(2021)^2\)
Step-by-step Solution
Detailed explanation
Given, \(A_n=\left[\begin{array}{cc}1-n & n \\ n & 1-n\end{array}\right]\) and
\[
\begin{aligned}
& \left|A_n\right|=\left[\begin{array}{cc}
1-n & n \\
n & 1-n
\end{array}\right] \\
& =(1-n)^2-n^2=1+n^2-2 n-n^2 \\
& \left|A_n\right|=1-2 n \\
& \text { Now. }\left|A_1\right|+\left|A_2\right|+\left|A_3\right|+\ldots .+\left|A_{2021}\right| \\
& =(1-2)+(1-4)+(1-6)+\ldots+(1-4042) \\
& =(1+1+\ldots .+1)-(2+4+6+\ldots .+4042) \\
& =2021-\left[\frac{2021}{2}(2+4042)\right] \\
& =2021-\frac{2021}{2} \times 4044=2021-2021 \times 2022 \\
& =2021(1-2022)=-(2021)^2 \\
&
\end{aligned}
\]
\[
\begin{aligned}
& \left|A_n\right|=\left[\begin{array}{cc}
1-n & n \\
n & 1-n
\end{array}\right] \\
& =(1-n)^2-n^2=1+n^2-2 n-n^2 \\
& \left|A_n\right|=1-2 n \\
& \text { Now. }\left|A_1\right|+\left|A_2\right|+\left|A_3\right|+\ldots .+\left|A_{2021}\right| \\
& =(1-2)+(1-4)+(1-6)+\ldots+(1-4042) \\
& =(1+1+\ldots .+1)-(2+4+6+\ldots .+4042) \\
& =2021-\left[\frac{2021}{2}(2+4042)\right] \\
& =2021-\frac{2021}{2} \times 4044=2021-2021 \times 2022 \\
& =2021(1-2022)=-(2021)^2 \\
&
\end{aligned}
\]
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