Question

For a matrix $A = \left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)$ , the value of $\prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} $ is equal to
A.$\left( {\begin{array}{*{20}{c}}
  1&{100} \\
  0&1
\end{array}} \right)$
B.$\left( {\begin{array}{*{20}{c}}
  1&{4950} \\
  0&1
\end{array}} \right)$
C.$\left( {\begin{array}{*{20}{c}}
  1&{5050} \\
  0&1
\end{array}} \right)$
D.$\left( {\begin{array}{*{20}{c}}
  1&{2500} \\
  0&1
\end{array}} \right)$

Answer 1

Hint: The symbol $\prod\limits_{r = 1}^{50} {} $ represents the product of matrices obtained on putting the values r=1,2,3,4….50 in matrix A. Multiplying all the 50 matrices will be tough work so simplifying the product of matrices and by observing a pattern in the results, we can get to the correct answer.

Complete step-by-step answer:
$
  \prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} = \left( {\begin{array}{*{20}{c}}
  1&{2(1) - 1} \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&{2(2) - 1} \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&{2(3) - 1} \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&{2(4) - 1} \\
  0&1
\end{array}} \right)......\left( {\begin{array}{*{20}{c}}
  1&{2(50) - 1} \\
  0&1
\end{array}} \right) \\
  \prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} = \left( {\begin{array}{*{20}{c}}
  1&1 \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&3 \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&5 \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&7 \\
  0&1
\end{array}} \right)......\left( {\begin{array}{*{20}{c}}
  1&{99} \\
  0&1
\end{array}} \right) \\
 $
On multiplying the first two matrices we get,
$
  \prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} = \left( {\begin{array}{*{20}{c}}
  {1 \times 1 + 1 \times 0}&{1 \times 3 + 1 \times 1} \\
  {0 \times 1 + 1 \times 0}&{0 \times 3 + 1 \times 1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&5 \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&7 \\
  0&1
\end{array}} \right)......\left( {\begin{array}{*{20}{c}}
  1&{99} \\
  0&1
\end{array}} \right) \\
  \prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} = \left( {\begin{array}{*{20}{c}}
  1&4 \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&5 \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&7 \\
  0&1
\end{array}} \right)......\left( {\begin{array}{*{20}{c}}
  1&{99} \\
  0&1
\end{array}} \right) \\
 $
Now multiplying the matrix obtained by the multiplication of 1st and 2nd matrix with the 3rd matrix,
$
  \prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} = \left( {\begin{array}{*{20}{c}}
  {1 \times 1 + 4 \times 0}&{1 \times 5 + 4 \times 1} \\
  {0 \times 1 + 1 \times 0}&{0 \times 5 + 1 \times 1}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&7 \\
  0&1
\end{array}} \right)......\left( {\begin{array}{*{20}{c}}
  1&{99} \\
  0&1
\end{array}} \right) \\
  \prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} = \left( {\begin{array}{*{20}{c}}
  1&9 \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&7 \\
  0&1
\end{array}} \right)......\left( {\begin{array}{*{20}{c}}
  1&{99} \\
  0&1
\end{array}} \right) \\
 $
Now multiplying the obtained matrix with the 4th matrix,
$
  \prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} = \left( {\begin{array}{*{20}{c}}
  {1 \times 1 + 9 \times 0}&{1 \times 7 + 9 \times 1} \\
  {0 \times 1 + 1 \times 0}&{0 \times 7 + 1 \times 1}
\end{array}} \right)......\left( {\begin{array}{*{20}{c}}
  1&{99} \\
  0&1
\end{array}} \right) \\
  \prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} = \left( {\begin{array}{*{20}{c}}
  1&{16} \\
  0&1
\end{array}} \right)......\left( {\begin{array}{*{20}{c}}
  1&{99} \\
  0&1
\end{array}} \right) \\
 $
Now on observing the result of the multiplication of matrices, we see a common trend as we go on.
On the multiplication of the first two matrices, we got $\left( {\begin{array}{*{20}{c}}
  1&4 \\
  0&1
\end{array}} \right)$ which can also be written as $\left( {\begin{array}{*{20}{c}}
  1&{{{(2)}^2}} \\
  0&1
\end{array}} \right)$
On the multiplication of the first three matrices, we got $\left( {\begin{array}{*{20}{c}}
  1&9 \\
  0&1
\end{array}} \right)$ which can also be written as $\left( {\begin{array}{*{20}{c}}
  1&{{{(3)}^2}} \\
  0&1
\end{array}} \right)$
On the multiplication of the first four matrices, we got $\left( {\begin{array}{*{20}{c}}
  1&{16} \\
  0&1
\end{array}} \right)$ which can also be written as $\left( {\begin{array}{*{20}{c}}
  1&{{{(4)}^2}} \\
  0&1
\end{array}} \right)$
Thus we can say that, on multiplication of first n matrices, we get $\left( {\begin{array}{*{20}{c}}
  1&{{n^2}} \\
  0&1
\end{array}} \right)$
So, on multiplying the 50 matrices, we get $\left( {\begin{array}{*{20}{c}}
  1&{{{(50)}^2}} \\
  0&1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1&{2500} \\
  0&1
\end{array}} \right)$
Therefore, $\prod\limits_{r = 1}^{50} {\left( {\begin{array}{*{20}{c}}
  1&{2r - 1} \\
  0&1
\end{array}} \right)} = \left( {\begin{array}{*{20}{c}}
  1&{2500} \\
  0&1
\end{array}} \right)$
So, the correct answer is “Option D”.

Additional information:
$\sum\limits_{r = a}^b {} $ Represents the sum of a given quantity from $a$ to $b$ .
For example, $\sum\limits_{x = 1}^5 x = 1 + 2 + 3 + 4 + 5$ whereas $\prod\limits_{x = 1}^5 x = 1 \times 2 \times 3 \times 4 \times 5$ .

Note: The product of two matrices $\left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right)$ and $\left( {\begin{array}{*{20}{c}}
  e&f \\
  g&h
\end{array}} \right)$ is equal to
$\left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  e&f \\
  g&h
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {a \times e + b \times g}&{a \times f + b \times h} \\
  {c \times e + d \times g}&{c \times f + d \times h}
\end{array}} \right)$
Multiplication of matrices is not commutative, that is AB≠BA so matrices should be multiplied carefully in the given order to get the correct answer.