Answer:
Recall that a relation is an <em>equivalence relation</em> 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.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that
. Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that
. In this equality we can perform a right multiplication by
and obtain
. Then, in the obtained equality we perform a left multiplication by P and get
. If we write
and
we have
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have
and from B↔C we have
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and
. So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.
Answer
300
Step-by-step explanation:
multiply by each other
Answer:
3=-8y+5
Step-by-step explanation:
Gather your like terms, - 3y and - 5y which in turn in - 8y. You cant plus by 5 as it is not a like term. IF we were finding y we would rearrange and bring +5 to the other side and so on...
Answer:
4y + 9
Step-by-step explanation:
2y+
+9
2y + 2y +9
4y + 9
EF=9 cm. Both triangles are the same.