Question:

If $A = \begin{bmatrix}1&1\\ 1&1\end{bmatrix}$ then $A^{100}$ :

Updated On: Jun 4, 2023
  • $2^{100}A$
  • $2^{99}A$
  • $2^{101}A$
  • None of the above
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The Correct Option is B

Solution and Explanation

Let $ A = \begin{bmatrix}1&1\\ 1&1\end{bmatrix}$
$A^2 = \begin{bmatrix}1&1\\ 1&1\end{bmatrix} \begin{bmatrix}1&1\\ 1&1\end{bmatrix} $
$= 2 \begin{bmatrix}1&1\\ 1&1\end{bmatrix} = 2A$
$A^3 = 2^2 \begin{bmatrix}1&1\\ 1&1\end{bmatrix}$,
$A^4 = 2^3 \begin{bmatrix}1&1\\ 1&1\end{bmatrix}$
$A^3 = 2^2 A, \, \, \, A^4 = 2^3 A$
$\therefore \, A^n = 2^{n - 1} \begin{bmatrix}1&1\\ 1&1\end{bmatrix}$
$\Rightarrow \, A^{100} = 2^{100 - 1} A $
$ \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.