Find all real matrices #A# , such that #A²= I(2)# (#A# is a matrix of second order)?

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Answer 1 · 2016-07-03T15:25:26

Solutions:

#((1,0),(0,1))#, #((-1,0),(0,-1))#, #((1,0),(c,-1))#, #((-1,0),(c,1)),((a, b),((1-a^2)/b,-a))#

Suppose #A=((a, b),(c, d))#

Then:

#A^2 = ((a, b),(c, d))((a, b),(c, d)) = ((a^2+bc, b(a+d)),(c(a+d), d^2+bc))#

So if we want #A^2=((1,0),(0,1))# then we get the following equations:

#{ (a^2+bc = 1), (b(a+d) = 0), (c(a+d) = 0), (d^2+bc = 1) :}#

From the second equation, we have #b=0# and/or #a+d = 0#

#color(white)()#
Case #bb(b=0)#

#a=+-1#, #d=+-1#

If additionally #a=-d# then #c# can have any value. Otherwise #c=0#.

So the case #b=0# results in solutions:

#((1,0),(0,1))#, #((-1,0),(0,-1))#, #((1,0),(c,-1))#, #((-1,0),(c,1))#

#color(white)()#
Case #bb(a+d=0)#

#bc = 1 - a^2#

This results in solutions:

#((a, b),((1-a^2)/b,-a))#

#color(white)()#
All of these solutions work.

Answer 2 · 2016-07-03T15:32:45

See below

Given

#A = ((a_{11},a_{12}),(a_{21},a_{22}))#

we need all #A# such that

#A cdot A = I(2)#

where

#I(2) = ((1,0),(0,1))#

Solving the system of resulting equations

#{(a_(11)^2 + a_(12) a_(21) = 1), (a_(11) a_(12) + a_(12) a_(22) = 0), (a_(11) a_(21) + a_(21) a_(22) = 0), (a_(12) a_(21) + a_(22)^2 = 1) :}#

we have

#A_1 = ((lambda_1,lambda_2),((1-lambda_1^2)/lambda_2,-lambda_1))#
#A_2=((-1,0),(lambda_1,1))#
#A_3=-A_2#
#A_4 = I(2)#
#A_5 = -I(2)#

For #lambda_1 in RR# and #(lambda_2 ne 0) in RR#