Show that the given matrix is nonsingular for every real val

Show that the given matrix is non-singular for every real value of t. A(t) (8e^t sin t -8e^t cos t e^t cos t e^t sin t) det A(t) = notequalto 0 Find A^-1(t) using Theorem II.2.

Solution

A(t) is a 2x2 matrix. Its determinant detA(t) = ( 8e^t sin t)*( e^t sin t) - (-8e^t cos t)(e^t cos t) = 8e^2t sin² t +  8e^2t cos² t =  8e^2t( sin² t + cos² t ) = 8 e^2t [ as ( sin² t + cos² t ) = 1]. Further, since 8 e^2t 0 (unless t = - ), hence detA(t) 0.

We have not been advised as to what the theorem II.2 states, but we know that if A = [ a b ] then [ c d ],

A^-1 = 1/(det A) [ d -b ] [ -c a ]

Here a =  8e^t sin t , b = -8e^t cos t , c = e^t cos t and d = e^t sin t.Therefore, A^-1(t) = 1/(8e^2t ) [  e^t sin t   8e^t cos t ] [ -e^t cos t   8e^t sin t ] =

(1/8) [ e^-tsin t 8e^-t cos t ] [ -e^-t cos t 8e^-t sin t ]