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pishuonlain [190]
2 years ago
12

SOMEONE HELP ME PLEASE thank you!

Mathematics
2 answers:
faltersainse [42]2 years ago
7 0

Answer: See below

Step-by-step explanation:

Same steps as the last question I answered

5) a ➤ x=-3

b ➤ y=7

6) x=-1

7) y=-4

Alexxx [7]2 years ago
5 0

Answer:

5. X=-3 Y=0

6.-3/11

7.None

Step-by-step explanation:

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15x2=30

80-30=50

50/2=25

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3 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 <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 A=J^{-1}AJ. 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 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.

<em>Transitive</em>: 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

A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP).

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A=R^{-1}CR. Hence, A↔C.

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