Hi Eliza,

For (a) and (b):

A linear equation in 3 variables x,y,z is called linear if it has the form ax + by + cz = d, where a,b,c,d are constants and not all of them can be zero at the same time.

Can you now do parts (a) and (b)?

For (c): consider the system

x + y + z = 1

x + y + z = 2

This system has fewer equations than variables. Does it have a solution? What does that mean for the answer to part (c)?

For (d): There are several ways to go about answering this one, it would depend on what you've covered. Have you done looking at systems as matrix equations? Assuming not, look at the system

x + y = 1

It has 2 variables but only one equation. Does it have a unique solution? Can you expand on this?

For (e): By inspection we can see that (2, 3, 1) = 2*(1,1,1) + 1*(0,1,-1). So that should help you answer right there.

In general v is a linear combination of v1 and v2 if we can express it as v = a*v1 + b*v2, where a and b are constants.

To solve problems like this in general, you can set up a system. For example, with this problem,

Set (2,3,1) = a*(1,1,1) + b*(0,1,-1)

=> (2,3,1) = (a, a + b, a - b)

By equating coordinates, you get the system:

a = 2,

a+b = 3,

a - b = 1

For which you can solve for a and b. If no such a and b exist, then that means you canNOT write (2,3,1) as a linear combination of (1,1,1) and (0,1,-1). But earlier, I showed you a solution exists.

Hope that helps,

Jhevon

Eliza T.

(a) would be true because it's in the form whereas (b) would be false because it isn't. (c) is false because if I write in this form: [ 1 1 1 | 1 ] [ 1 1 1 | 2 ] Then subtract row 1 from row 2: [ 1 1 1 | 1 ] [ 0 0 0 | 1 ] But 0 can't equal one, therefore false. (d) would be true according to your example, because there infinitely many possibilities, as x can be 3 and y can be -2, x can also be 0 and y can also be 1 and so on. (e) would be true Thanks for the help so far!03/02/20