
Exam 3 Sol
... a) (A + B)~x = A~x + B~x = λ1 ~x + λ2 ~x = (λ1 + λ2 )~x, so the corresponding eigenvalue is λ1 + λ2 . b) λ = 2 will have at most 3 and as few as one associated LI eigenvector, λ = 3 can have at most two and as few as one associated LI eigenvector and λ = 0 has a single LI eigenvector so C has at lea ...
... a) (A + B)~x = A~x + B~x = λ1 ~x + λ2 ~x = (λ1 + λ2 )~x, so the corresponding eigenvalue is λ1 + λ2 . b) λ = 2 will have at most 3 and as few as one associated LI eigenvector, λ = 3 can have at most two and as few as one associated LI eigenvector and λ = 0 has a single LI eigenvector so C has at lea ...
Solutions
... must be exactly one card numbered 1/2 and exactly one numbered 3/2. Then there must be two pairs of cards with sum 3; this can be accomplished by either 1/2 + 5/2 or 3/2 + 3/2. However, we already know there is only one card numbered 3/2, so there must be two ways to draw 1/2 + 5/2. Only one card is ...
... must be exactly one card numbered 1/2 and exactly one numbered 3/2. Then there must be two pairs of cards with sum 3; this can be accomplished by either 1/2 + 5/2 or 3/2 + 3/2. However, we already know there is only one card numbered 3/2, so there must be two ways to draw 1/2 + 5/2. Only one card is ...