Matrix ${A_r} = \left[ {\begin{array}{*{20}{l}}
  r&{r - 1} \\
  {r - 1}&r
\end{array}} \right]$ where $r = 1,2,3,........$ and if $\sum\limits_{r = 1}^{r = 100} {\left| {{A_r}} \right|} = {\left( {\sqrt {10} } \right)^k}$ so find $k$

Hint:
$\left| {{A_r}} \right|$ denotes the determinant of the matrix ${A_r}$ so first of all we need to find the determinant of the matrix ${A_r}$ in the terms of $r$ then put $r = 1,2,3,........$ and so on and then we will get the definite pattern and get our answer.

Complete step by step solution:
Here we are given that ${A_r} = \left[ {\begin{array}{*{20}{l}}
  r&{r - 1} \\
  {r - 1}&r
\end{array}} \right]$ where $r = 1,2,3,........$ and now we are given the condition that if $\sum\limits_{r = 1}^{r = 100} {\left| {{A_r}} \right|} = {\left( {\sqrt {10} } \right)^k}$ then we need to find the value of $k$
As $\left| {{A_r}} \right|$ denotes the determinant of the matrix ${A_r}$ so first of all we need to find the determinant of the matrix ${A_r}$ in the terms of $r$
Here $\sum\limits_{r = 1}^{r = 100} {\left| {{A_r}} \right|} $ means that there is the summation of the $\left| {{A_1}} \right| + \left| {{A_2}} \right| + \left| {{A_3}} \right| + .......... + \left| {{A_{100}}} \right|$
So the determinant of the matrix ${A_r}$ is
$\left| {{A_r}} \right| = \left| {\begin{array}{*{20}{l}}
  r&{r - 1} \\
  {r - 1}&r
\end{array}} \right|$
$
 \Rightarrow \left| {{A_r}} \right| = r(r) - (r - 1)(r - 1) \\
 \Rightarrow {r^2} - {r^2} + 2r - 1 \\
 \Rightarrow 2r - 1
 $
Now we get that $\left| {{A_r}} \right| = 2r - 1$
Now we need to find the value of $\sum\limits_{r = 1}^{r = 100} {\left| {{A_r}} \right|} $
We know that $\sum\limits_{r = 1}^n r = 1 + 2 + 3 + ...... + n = \dfrac{{n(n + 1)}}{2}$
$\Rightarrow \sum\limits_{r = 1}^{100} {(2r - 1)} $
Now on calculating we get that
$
\Rightarrow 2\sum\limits_{r = 1}^{100} r - \sum\limits_{r = 1}^{100} 1 \\
 \Rightarrow 2\left( {\dfrac{{n(n + 1)}}{2}} \right) - n
 $
And here $n = 100$
So we get that
$\sum\limits_{r = 1}^{r = 100} {\left| {{A_r}} \right|} $$ = 2\left( {\dfrac{{100(100 + 1)}}{2}} \right) - 100 = 100(101) - 100 = 100(100) = {10^4}$
But we are given that $\sum\limits_{r = 1}^{r = 100} {\left| {{A_r}} \right|} = {\left( {\sqrt {10} } \right)^k}$
So we can use its value and get
$
\Rightarrow {\left( {\sqrt {10} } \right)^k} = {10^4} \\
\Rightarrow {10^{\dfrac{k}{2}}} = {10^4}
 $
Upon comparing we get that
$
 \Rightarrow \dfrac{k}{2} = 4 \\
\Rightarrow k = 8
 $

In this way we can get the value of $k$

Note:
Here we can do it this way also as we are given that $\left| {{A_r}} \right| = 2r - 1$ so we get that $\left| {{A_1}} \right| = 1,\left| {{A_2}} \right| = 3,\left| {{A_3}} \right| = 5,\left| {{A_4}} \right| = 7{\text{ and so on 1,3,5,7,}}......{\text{all are in AP}}$with the common difference $2$
$
  {A_{100}} = 199 \\
  199 = a + (n - 1)d \\
  199 = 1 + (n - 1)2 \\
  n = 100
 $
So we can find the sum of this AP
$
  {\text{sum of AP}} = \dfrac{n}{2}(a + l) \\
   = \dfrac{{100}}{2}(1 + 199) = 10000 = {10^4}
 $
Again we can compare and get the same answer.