Let \(S ={ (\begin{matrix} -1 & 0 \\ a & b \end{matrix}), a,b, ∈(1,2,3,.....100)}\) and
let \(T_n = {A ∈ S : A^{n(n + 1)} = I}. \)
Then the number of elements in \(\bigcap_{n=1}^{100}\) \(T_n \) is
Let \(S ={ (\begin{matrix} -1 & 0 \\ a & b \end{matrix}), a,b, ∈(1,2,3,.....100)}\) and
let \(T_n = {A ∈ S : A^{n(n + 1)} = I}. \)
Then the number of elements in \(\bigcap_{n=1}^{100}\) \(T_n \) is
Correct Answer: 100
Solution and Explanation
The correct answer is 100
\(S ={ (\begin{matrix} -1 & 0 \\ a & b \end{matrix}), a,b, ∈(1,2,3,.....100)}\)
\(∴ A =\) \((\begin{matrix} -1 & 0 \\ a & b \end{matrix})\)
then even powers of A as
\(A(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})\)
if b = 1 and a ∈ {1,….., 100}
Here, n(n + 1) is always even.
∴ \(T_1, T_2, T_3\), …, \(T_n\) are all I for b = 1 and each value of a.
\(∴\) \(\bigcap_{n=1}^{100}\) \(T_n = 100\)
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Concepts Used:
Cartesian Products of Sets
Cartesian products of sets here are explained with the help of an example. Consider A and B to be the 2 sets such that A is a set of 3 colors of tables and B is a set of 3 colors of chairs objects, i.e.,
A = {red, blue, purple}
B = {brown, green, yellow},
Now let us find the number of pairs of colored objects that we can make from a set of tables and chairs in various combinations. They can be grouped as given below:
(red, brown), (red, green), (red, yellow), (blue, brown), (blue, green), (blue, yellow), (purple, brown), (purple, green), (purple, yellow)
There are 9 such pairs in the Cartesian product since 3 elements are there in each of the defined sets A and B. The above-ordered pairs shows the definition for the Cartesian product of sets given. This product is resembled by “A × B”.