If A = [(cos a,sin a)(-sin a,cos a)], show that A^2 =

Question

If A = \( \begin{bmatrix} cos\,a& sin\,a \\[0.3em] -sin\,a & cos\,a \\[0.3em] \end{bmatrix}\) , show that A² = \( \begin{bmatrix} cos\,2a& sin\,2a \\[0.3em] -sin\,2a & cos\,2a \\[0.3em] \end{bmatrix}\)

A = [(cos a,sin a)(-sin a,cos a)]

Answer 1

Given : A = \( \begin{bmatrix} cos\,a& sin\,a \\[0.3em] -sin\,a & cos\,a \\[0.3em] \end{bmatrix}\),

To show : A² = \( \begin{bmatrix} cos\,2a& sin\,2a \\[0.3em] -sin\,2a & cos\,2a \\[0.3em] \end{bmatrix}\)

Formula used :

Where c_ij = a_i1b_1j + a_i2b_2j + a_i3b_3j + ……………… + a_inb_nj

If A is a matrix of order a x b and B is a matrix of order c x d ,then matrix AB exists and is of order a x d ,

if and only if b = c

We know that cos2α = cos²a - sin²a and sin 2a = 2sina cosa