Question:

If A=[1111]A = \begin{bmatrix}1&1\\ 1&1\end{bmatrix} then A100A^{100} :

Updated On: Jan 30, 2025
  • 2100A2^{100}A
  • 299A2^{99}A
  • 2101A2^{101}A
  • None of the above
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The Correct Option is B

Solution and Explanation

Let A=[1111] A = \begin{bmatrix}1&1\\ 1&1\end{bmatrix}
A2=[1111][1111]A^2 = \begin{bmatrix}1&1\\ 1&1\end{bmatrix} \begin{bmatrix}1&1\\ 1&1\end{bmatrix}
=2[1111]=2A= 2 \begin{bmatrix}1&1\\ 1&1\end{bmatrix} = 2A
A3=22[1111]A^3 = 2^2 \begin{bmatrix}1&1\\ 1&1\end{bmatrix},
A4=23[1111]A^4 = 2^3 \begin{bmatrix}1&1\\ 1&1\end{bmatrix}
A3=22A,   A4=23AA^3 = 2^2 A, \, \, \, A^4 = 2^3 A
An=2n1[1111]\therefore \, A^n = 2^{n - 1} \begin{bmatrix}1&1\\ 1&1\end{bmatrix}
A100=21001A\Rightarrow \, A^{100} = 2^{100 - 1} A
A100=299A \therefore \, A^{100} = 2^{99} A
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Concepts Used:

Invertible matrices

A matrix for which matrix inversion operation exists, given that it satisfies the requisite conditions is known as an invertible matrix. Any given square matrix A of order n × n is called invertible if and only if there exists, another n × n square matrix B such that, AB = BA = In, where In  is an identity matrix of order n × n.

For example,

It can be observed that the determinant of the following matrices is non-zero.