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>.
I think the answer is (3, 6)
Step-by-step explanation:
Your work is correct, but you need to round. x and y have 2 significant figures, so round your final answer to 2 significant figures:
2.4×10⁶
For this case we have that by definition, if we draw the diagonal of a square two rectangular triangles are formed. If the diagonal measures "x", then the Pythagorean theorem is fulfilled:
Where:
l: It's the side of the square.
We know that the area of a square is given by:
So, the area is:
Answer:
Answer:
4 and 1/4 i think
Step-by-step explanation:
