Suppose A = {2,5,7,9,13, 25, 26).(a) Find n(A).n(A) =(b) T/F: 7. A

Mathematics

1 answer:

charle [14.2K]2 years ago

6 0

Answer:

a) n (A) = 7

Step-by-step explanation:

Here, set A = {2,5,7,9,13, 25, 26)

Noe, for any set A,

n (A) is defined as he total number of elements in the set A.

or, n(A) = CARNALITY OF SET A

Here,the number of elements in set A = 7

⇒ n (A) = 7

You might be interested in

NEED HELP ASAP! Triangle PQR has vertices at the following coordinates: P(0, 1), Q(3, 2), and R(5, -4). Determine wether or not

morpeh [17]

Orchards minced minister porch jochen

4 0

2 years ago

Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

Sophie [7]

Answer:

Recall that a relation is an equivalence relation if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get