If `A=[[1,tanx],[-tanx,1]]`
, show that `A^T\ A^(-1)=[[cos2x,-sin2x],[sin2x,cos2x]]`
.
Text Solution
Verified by Experts
Given,`A=[[1,tanx],[-tanx,1]]`
`|A|=1+tan^2x !=0`
So A is invertiable. Also,
`adj(A)=[[1,tanx],[-tanx,1]]^T`
=`[[1,-tanx],[tanx,1]]`
Thus,`A^(-1)=adj(A)/|A|`
=`1/(1+tan^2x)[[1,-tanx],[tanx,1]]`
=`[[1/(1+tan^2x),-tanx/(1+tan^2x)],[tanx/(1+tan^2x),1/(1+tan^2x)]]`
...
Topper's Solved these Questions
ADJOINTS AND INVERSE OF MATRIX
RD SHARMA|Exercise QUESTION|1 Videos
ALGEBRA OF MATRICES
RD SHARMA|Exercise Solved Examples And Exercises|410 Videos
Similar Questions
Explore conceptually related problems
If A=[[1,tan x-tan x,1]], show that A^(T)A^(-1)=[[cos2x,-sin2xsin2x,cos2x]]
If A=[(1,tanx),(-tanx, 1)], " show that " A' A^(-1)=[(cos 2x,-sin 2x),(sin 2x, cos 2x)] .
(d)/(dx)[(1+cos2x+sin2x)/(1+sin2x-cos2x)]=
D((1-cos2x)/(sin2x))=
int(1)/(cos2x+sin2x)dx=
int(sin2x-cos2x)/(sin2x*cos2x)dx=?
If y = {(1-tanx)/(1+tanx)} , show that (dy)/(dx) = (-2)/((1+sin2x)) .
If y=(1-tanx)/(1+tanx) , prove that (dy)/(dx)=(-2)/(1+sin2x) .
