Quiz 3, 4/13/2006, Math 221, s. 02: Solutions

Suppose the 2 × 2 matrix A = [ a b] is invertible and satisfies:
c d
A A = I2 × 2 ,
A = AT , and  
det(A) < 0 .  
Prove:
(a) b = c ; 
Since A and its transpose are the same, the NE entries must be equal, whence b = c .
(b) a2 + b2 = 1 ; 
Bearing in mind that b = c , compare the NW entries in AA = I2×2 , to 
see that a2 + b2 = a2 + bc = 1 .
(c) det(A) = – 1 ; 
On the one hand, det(AA) = det(I2×2) = 1 . On the other hand, det(AA) = 
(det(A))2 . It follows that det(A) = ±1 . But the +1 option is unavailable, 
given that det(A) < 0 . So we must have det(A) = –1 .
(d) d = – a . 
It is easy to compute the inverse of the matrix A (now that we know det(A) = –1 ): 
using the method of minors, that inverse is just
[ – d     b] .
  c     – a
But from AA = I2×2 we know that the inverse of A is just A itself. 
Comparing NW (or SE) corners of A and A-1 , we obtain a = –d (or d = –a ).
[Please do not consult outside sources (books, notes, web, people) while working out this quiz.]
[Please do record the clock times at which you begin and end work on this quiz.]
[(The most likely expectation is a total elapsed time of not more than 15 minutes.)]
Created: 16 Apr 2006. First posted: 18 Apr 2006 Math 221 Home   Contents © 2006 by Fred E.J. Linton