## 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* = **I**_{2 × 2} ,

*A* = *A*^{T} , 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) *a*^{2} + *b*^{2} = 1 ;

Bearing in mind that b = c , compare the NW entries in AA = I_{2×2} , to
see that a^{2} + b^{2} = a^{2} + bc = 1 .

(c) det(*A*) = – 1 ;

On the one hand, det(AA) = det(I_{2×2}) = 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

But from AA = I_{2×2} 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 ).

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