Suppose Ax = b always has at least one solution no matter what b is. Why AT y = 0 has only the trivial solution y = 0?

For the first equation we have:

A*x = b

solving for x, we get:

x = b/A

If it always has a solution, then we can not have A = 0, because that causes an undefined operation.

so for example, if we have A = 1 and b = 2

x = b/A = 2/1

For the other case,

A*y = 0

dividing both sides by A

y = 0/A = 0

y = 0

Here we have only one possible solution, the trivial one, y = 0.

And the dependence on A disappears (because the quotient between zero and a number different than zero is always zero)

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.