If $A = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$ and $B = \sum\limits_{r = 1}^{2021} {{A^r}} $ then the value of $|B|$ is
a) $2021$
b) ${2021^2}$
c) $ - 2021$
d) $0$

Hint: Given, $A = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$ and $B = \sum\limits_{r = 1}^{2021} {{A^r}} $. Firstly, we will find the determinant of $A$ using method of finding determinant of any \[2 \times 2\] square matrix or a square matrix of order \[2 \times 2\], then we will find value of B by given formula $B = \sum\limits_{r = 1}^{2021} {{A^r}} $. Lastly, we will find the determinant of $B$.

Formula Used:
 Formula used: $|kA| = {k^n}|A|$

Complete step by step Solution:
 In mathematics, the determinant is a scalar value. It allows for characterizing some properties of the matrix. In particular, a matrix is invertible if and only if the determinant is nonzero and the linear map represented by the matrix is an isomorphism. The determinant of a product of matrices is the product of their determinants. The determinant of a matrix $A$ is denoted $\det A$, $\det (A)$or $|A|$.
 For any \[2 \times 2\] square matrix or a square matrix of order \[2 \times 2\], we can use the determinant formula to calculate its determinant:
Let $X = \left[ {\begin{array}{*{20}{c}}
  x&y \\
  z&w
\end{array}} \right]$
$|X| = xw - yz$
Given, $A = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
Similarly, we will find the determinant of $A$
$|A| = 1 - 0$
$|A| = 1$
And $B = \sum\limits_{r = 1}^{2021} {{A^r}} $
$B = A + A + A + .......$(2021 times)
$B = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] + ............$(2021 times)
We know that
$I = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
So, we can use $A = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = I$
We know that the determinant of every identity matrix is $1$
This also implies $A = I$
$B = I + I + I + .........$(2021 times)
We know $X = y + y + y + .............$(n times)
$X = ny$
So, $B = 2021I$
$|B| = |2021A|$
We know that if $A$ is a square matrix of order $n \times n$ and $k$is any scalar,
Then the determinant of $|kA| = {k^n}|A|$
Similarly, here the order of $A$ is $2 \times 2$
$ \Rightarrow |B| = {2021^2}|A|$
As we calculate above the determinant of $A$ is $1$, using this
$|B| = {2021^2} \times 1$
$|B| = {2021^2}$

Hence, the correct option is (b).

Note: Students solve questions step by step without skipping any step to avoid mistakes. They can make mistakes while finding the value B if they do not calculate it properly this can change the answer and cannot get the correct solution. Also, we should be careful while applying the formula to get the correct answer.