Is it possible for a linear system to have a unique solution if it has more equations than variables? if yes, give an example. if no, justify why it is impossible.

Yes

Step-by-step explanation:

Yes it is possible, if the equations are equivalent. For example in one variable, we have x=2 and if we add the equation 2x=4 both considered on the Real numbers, then the system has two equations and one variable, more equations than variables and yet the solution is unique x=2.

Yes, it's possible.

Step-by-step explanation:

We know that a linear system has a unique solution if it has the same number of equations than variables (but remember that you musn't have a linear combination of equations).

So it is possible to create a system with more equations than variables, with a unique solution.

Example:

1)  X + Y + Z = 3

2) 2X + 2Y + 2Z = 6

3) X + Y = 2

4) X + Z = 2

This is a 4 equation and 3 variable system.  Pay attention to equation number 1) and 2). They are a linear combination (number 2 is the number 1 multiplied by 2). So, for this system you can basically ignore one of them.

If you do that, you will have 3 equations and 3 variables, with a unique solution. In this case solution is X = 1, Y = 1, Z = 1.

Now, let's see if equation number 2 satisfies:

2x1 + 2x1 + 2x1 = 6

                     6 = 6

With this simple example we proved that you can have a linear system with a unique solution if it has more equations than variables.

the data of the detroit pistons in numerical order from least to greatest is 0.30, 0.37, 0.55, 0.62, 0.69, 0.72, 0.73, 0.75, 0.77, 0.78, 0.80, 0.81, 0.81, 0.83, 0.84, 0.89

the data of the miami heat in numerical order from least to greatest is 0.50, 0.50, 0.56, 0.57, 0.67, 0.72, 0.72, 0.72, 0.75, 0.75, 0.76, 0.81, 0.82, 0.90, 1.00, 1.00

the detroit pistons median : 1.52

the miami heat median: 1.47

the iqr of this data is 0.05