Question

let A =[0 √-1 -√-1 0] ( matrix) then prove that A^40= unit matrix​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{A=\left(\begin{array}{cc}0&\sqrt{-1}\\-\sqrt{-1}&0\end{array}\right)}$

$\underline{\textbf{To prove:}}$

$\mathsf{A^{40}=Unit\;matrix}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{A=\left(\begin{array}{cc}0&\sqrt{-1}\\-\sqrt{-1}&0\end{array}\right)}$

$\mathsf{A=\left(\begin{array}{cc}0&i\\-i&0\end{array}\right)}$

$\mathsf{A^2=\left(\begin{array}{cc}0&i\\-i&0\end{array}\right)\left(\begin{array}{cc}0&i\\-i&0\end{array}\right)}$

$\mathsf{A^2=\left(\begin{array}{cc}0-i^2&0+0\\0+0&-i^2+0\end{array}\right)}$

$\mathsf{A^2=\left(\begin{array}{cc}1&0\\0&1\end{array}\right)}$

$\implies\mathsf{A^2=I\;\;(unit\;matrix)}$

$\implies\mathsf{A^3=A^2{\times}A=I{\times}A=A}$

$\implies\mathsf{A^4=A^3{\times}A=A{\times}A=A^2=I}$

$\textsf{Proceeding like this, finally we get}$

$\mathsf{A^{40}=I\;\;(Unit\;matrix)}$

$\therefore\mathsf{A^{40}\;is\;a\;unit\;matrix}$

$\underline{\textbf{Find more:}}$

A square matrix A is said to be idempotent if A2 = A. Let A be an idempotent matrix.(a) Show that I - A is also idempotent.

(b) Show that if A is invertible, then A = I.

(c) Show that the only possible eigenvalues of A are 0 and 1. (Hint: Suppose x is an eigenvector with associated eigenvalue A and then multiply x on the left by A twice.)

https://brainly.in/question/38733698

Answer 2

Answer:

Step-by-step explanation:

1 0

0 1