3 and 4 .Determinants and Matrices
normal

If matrix  $A\, = \,\left[ {\begin{array}{*{20}{c}}
  1&{3k + \frac{1}{3}} \\ 
  0&1 
\end{array}} \right]$, then the value of  $\mathop \Pi \limits_{k = 1}^{36} \,\left[ {\begin{array}{*{20}{c}}
  1&{3k + \frac{1}{3}} \\ 
  0&1 
\end{array}} \right]$ is equal to :-

A

$\left[ {\begin{array}{*{20}{c}}
  1&{1998} \\ 
  0&1 
\end{array}} \right]$

B

$\left[ {\begin{array}{*{20}{c}}
  1&{2010} \\ 
  0&1 
\end{array}} \right]$

C

$\left[ {\begin{array}{*{20}{c}}
  1&{1005} \\ 
  0&1 
\end{array}} \right]$

D

$\left[ {\begin{array}{*{20}{c}}
  1&{999} \\ 
  0&1 
\end{array}} \right]$

Solution

$\prod\limits_{k = 1}^{36} {\left[ {\begin{array}{*{20}{c}}
1&{3k + \frac{1}{3}}\\
0&1
\end{array}} \right]} $

$ = \left[ {\begin{array}{*{20}{c}}
1&{3 + \frac{1}{3}}\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{6 + \frac{1}{3}}\\
0&1
\end{array}} \right]…….\left[ {\begin{array}{*{20}{c}}
1&{108 + \frac{1}{3}}\\
0&1
\end{array}} \right]$

$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
1&{\left( {3 + 6 + 9 + …. + 108} \right) + 12}\\
0&1
\end{array}} \right]$

$ = \left[ {\begin{array}{*{20}{c}}
1&{2010}\\
0&1
\end{array}} \right]$

Standard 12
Mathematics

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