Please help i dont know how to do this

Mathematics

1 answer:

maxonik [38]3 years ago

7 0

Hello once again!

When you see a question like this, you need to find the equation of the straight line.

The formular used is y = mx + c
Where
m = slope
c = constant

First find the slope, since it's a straight line, any 2 coordinates can be used.

$m = { \frac{y_1 - y_2}{x_1-x_2} } \\ m= { \frac{16 - (- 8)}{-2 - 2} } \\ m= { \frac{24}{-4} } \\ m = -6$

Now we need to substitude in the slope, and one of the coordinate you used to find the slope, to the formular to find the constant.

In this case i'm using the coordinate
(-2, 16)

y = mx + c
16 = -6(-2) + c
16 = 12 + c
c = 4

∴ The equation of the line is y = -6x + 4

The next step is to simply substitude in the x = 8 to the equation to find y.

y = -6(8) + 4
y = -48 + 4
y = -44

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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