# Subspace of three equations

• MHB
• Johnathon1
In summary, S is a subspace of R4 when a = 1 and b = 0, as this is the only combination of values that satisfies all three equations and makes S closed under addition and scalar multiplication.

#### Johnathon1

S is the set of solutions for the set of three equations...

x + (1 - a)y-1 + 2z + b2w = 0
ax + y - 3z + (a - a2)|w| = a3 - a
x + (a - b)y + z + 2a2w = b

I worked out...

The first equation is a subset of R4 when a = 1, b is any real.

The second equation is a subset of R4 when a = 1 or a = 0.

The third equation is a subset of R4 when b = 0 and a is any real.

Now, I'm trying to work out the values of a and b that make S a subspace of R4.

Isn't S only a subspace for the values of a and b that are common to all three equations?

So in this case, S would only be a subspace when a = 1 and b = 0. This is because for a set to be a subspace, it must be closed under addition and scalar multiplication, meaning that if x and y are in the set, then ax + by must also be in the set.

In this case, if a = 1 and b = 0, all three equations will have the same coefficients and will be satisfied by the same set of solutions, making S a subspace of R4. However, if a and b have different values, the equations will have different coefficients and will not have the same set of solutions, making S not a subspace of R4.

Therefore, S is only a subspace of R4 when a = 1 and b = 0.