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.