Algebra 443The determinant of the system isu^2 +uvtrA+v^2 detA, and an easy algebraic computation
shows that this is equal to det(uI 2 +vA), which is nonzero by hypothesis. Hence the
system can be solved, and its solution determines the desired inverse.
259.Rewriting the matrix equation as
X^2 (X− 3 I 2 )=
(
− 2 − 2
− 2 − 2
)
and taking determinants, we obtain that either detX=0ordet(X− 3 I 2 )=0. In the
first case, the Cayley–Hamilton equation implies thatX^2 =(trX)X, and the equation
takes the form
[(trX)^2 −3trX]X=(
− 2 − 2
− 2 − 2
)
.
Taking the trace of both sides, we find that the trace ofXsatisfies the cubic equation
t^3 − 3 t^2 + 4 =0, with real rootst=2 andt=−1. In the case trX=2, the matrix
equation is
− 2 X=
(
− 2 − 2
− 2 − 2
)
with the solution
X=(
11
11
)
.
When trX=−1, the matrix equation is
4 X=
(
− 2 − 2
− 2 − 2
)
with the solution
X=(
−^12 −^12
−^12 −^12
)
.
Let us now study the case det(X− 3 I 2 )=0. One of the two eigenvalues ofXis 3. To
determine the other eigenvalue, add 4I 2 to the equation from the statement. We obtain
X^3 − 3 X^2 + 4 I 2 =(X− 2 I 2 )(X+I 2 )=
(
− 2 − 2
− 2 − 2
)
.
Taking determinants we find that either det(X− 2 I 2 )=0ordet(X+I 2 )=0. So the
second eigenvalue ofXis either 2 or−1. In the first case, the Cayley–Hamilton equation
forXis