what is the basis of a subspace or R3 defined by the equation

2x1+3x2+x3=0

I know the basis is vectors <-3,2,0> and <-1,0,2>
but how do you get that?

The equation:

2x1+3x2+x3=0
Defines a plane.
Any two different (not linearly dependent) vectors in that plane form a basis.
Pick any old values for x and y then solve for z
like 1,1 then -5
and 1,-1 then 1
so I would say
(1,1,-5) and (1,-1,1)
the ones you gave are
-3,2 then you get -6+6+x3=0 so x3= 0 sure enough (-3,2,0)
If you chose -1,0 then you would get x3 = 2 so (-1,0,2)
both answers and an infinite number of other answers, are fine.

Well, let me illuminate the path to that basis for you.

To find the basis of a subspace, we need to focus on the solutions of the equation 2x1 + 3x2 + x3 = 0. The variables x1, x2, and x3 are the coefficients that define our vectors.

To make things simpler, let's solve the equation using the method of elimination.

First, we can freely choose a value for x1 and x2, and solve for x3. Let's set x1 = 1 and x2 = 0:

2(1) + 3(0) + x3 = 0
2 + x3 = 0
x3 = -2

So, we have one solution x = <1, 0, -2>.

Now, let's freely choose a different value for x1 and x2, and solve for x3 once again. Let's set x1 = 0 and x2 = 1:

2(0) + 3(1) + x3 = 0
3 + x3 = 0
x3 = -3

This gives us another solution x = <0, 1, -3>.

Therefore, the basis of the subspace defined by the equation 2x1 + 3x2 + x3 = 0 is the set of vectors <-3, 2, 0> and <-1, 0, 2>. Voilà!

Now, aren't you glad you embarked on this enlightening journey with me?

To find the basis of a subspace defined by an equation in R3, you need to consider the coefficients of the variables in the equation. In this case, the equation is 2x1 + 3x2 + x3 = 0.

We can rewrite this equation as:

x1 = (-3/2)x2 - (1/2)x3.

Now, we can consider x2 and x3 as free variables. This means that we can set any values for x2 and x3 and solve for x1.

Let's assign a value to x2, say x2 = 2. By substituting this value into the equation, we can solve for x1 and x3:

x1 = (-3/2)(2) - (1/2)x3
x1 = -3 - (1/2)x3

We can rewrite x1 and x3 as a vector:

<-3, -1/2, 0> = -3<1, 0, 0> - (1/2)<0, 1, 0> + 0<0, 0, 1>

Next, let's assign a value to x3, say x3 = 2. By substituting this value into the equation, we can solve for x1 and x2:

x1 = (-3/2)x2 - (1/2)(2)
x1 = (-3/2)x2 - 1

We can rewrite x1 and x2 as a vector:

<-1, -3/2, 2> = -1<1, 0, 0> - (3/2)<0, 1, 0> + 2<0, 0, 1>

So, the vectors -3<1, 0, 0> - (1/2)<0, 1, 0> + 0<0, 0, 1> and -1<1, 0, 0> - (3/2)<0, 1, 0> + 2<0, 0, 1> form a basis for the subspace defined by the equation 2x1 + 3x2 + x3 = 0.

To find the basis of a subspace defined by an equation, you need to find a set of vectors that satisfy the given equation and span the subspace.

In this case, the equation that defines the subspace is 2x1 + 3x2 + x3 = 0.

To find the basis, we need to solve this equation. We can rewrite it as a system of linear equations:

2x1 + 3x2 + x3 = 0

By assigning a parameter, we can express this equation in terms of free variables:

x1 = t
x2 = s
x3 = -2t + s

Now, we can express the solutions to this system as a linear combination of vectors:

<-3, 2, 0> = -3<1, 0, 0> + 2<0, 1, 0>
<-1, 0, 2> = -<1, 0, 0> + 2<0, 1, 0> + 2<0, 0, 1>

Therefore, the basis of the subspace is {<-3, 2, 0>, <-1, 0, 2>}. These vectors satisfy the equation 2x1 + 3x2 + x3 = 0 and span the subspace.