# Linear combination of a matrix

• Derill03

#### Derill03

im given four vectors as a 1x4 matrices:

[1,4,2,8]^t = v1
[2,5,3,9]^t = v2
[11,14,12,18]^t = v3
[4,3,2,1]^t = v4

and I am asked to express the vector v= [7,9,6,8]^t in two ways as a linear combination of {v1,v2,v3,v4}?

I know the vectors are linearly independent cause none of the vectors are multiples of each other, but i can't seem to find a c1,c2,c3,c4 that will work, am i missing something?

Your first step would be to write out a formula for what is required. So, using your notation, you need to find c1, c2, c3, and c4 such that
c1 v1 + c2 v2 + c3 v3 + c4 v4 = v
Plug in the vectors into the equation and then write out the equation coordinate wise (so you will have four equations in total).

I tried to solve it this way:

c1 + 2c2 + 11c3 + 4c4 = 7
4c1 + 5c2 + 14c3 + 3c4 = 9
2c1 + 3c2 + 12c3 + 2c4 = 6
8c1 + 9c2 + 18c3 + c4 = 8

but I am havin trouble finding a solution to this equation, any suggestions? I used the equation solver on my ti-89 and it comes back "false", is there really no solution?

Wait, your vector's aren't linearly independent! Check the determinant and you'll see that they are linearly dependant. In fact, the question asks for two ways of writing a vector using a certain basis. If you have a linearly independent set of vectors, there exists a unique (or none, if the set doesn't span the space your vector is in) coordinate that defines an arbitrary vector using that set of vectors. If it is asking for two, then obviously you are dealing with a linearly dependent set.

Remember, being a linear combination of vectors doesn't mean one vector is simply a multiple of another, but that it can be expressed as a combination of the other vectors given; that is, v= av1 + bv2 + cv3 where a,b,c, don't equal 0 and you can pick each v out of those 4 that you have.

Last edited:
ow can this be expressed as a linear combination if the simultaneous equations have no solution? Bc i would need to solve for c1,c2,c3,c4 in order to write a linear combination right?'

I know how to find linear combinations by solving for the constants, but i am just not sure if this set of vectors can be wrote as a linear combo? Can these vectors be wrote as a linear combo?

Derill03 said:
ow can this be expressed as a linear combination if the simultaneous equations have no solution?
If the vectors v1, v2, v3, and v4 in your first post are linearly dependent, they can't possibly span R4, and this means that some vectors in R4 are not linear combinations of your four vectors.
Derill03 said:
Bc i would need to solve for c1,c2,c3,c4 in order to write a linear combination right?'
Right.
Derill03 said:
I know how to find linear combinations by solving for the constants, but i am just not sure if this set of vectors can be wrote as a linear combo? Can these vectors be wrote as a linear combo?
If (7, 9, 6, 8) is a linear combination of v1, v2, v3, and v4, then your system of equations will produce at least one solution. If not, then the system of equations has no solutions (is inconsistent).

Your system of equations is a good start.
c1 + 2c2 + 11c3 + 4c4 = 7
4c1 + 5c2 + 14c3 + 3c4 = 9
2c1 + 3c2 + 12c3 + 2c4 = 6
8c1 + 9c2 + 18c3 + c4 = 8

It would be good practice to solve this system manually, not relying on your calculator. You can do this by using an augmented matrix:

[1 2 11 4 | 7]
[4 5 14 3 | 9]
[2 3 12 2 | 6]
[8 9 13 1 | 8]

If there really is no solution to the system of equations, you'll end up with a row that looks like this:
[0 0 0 0 | 2]

The 2 on the right side is just for example. In an inconsistent system, you'll end up with some nonzero number there and zeros on the other entries in that row.

I do get an inconsistency for the system of equations so does this mean there is no linear combination for the vectors?

Yes.

How can i tell if any of the vectors can be removed from the set without changing the span?

If your set of vectors is linearly dependent, then at least one of them is a linear combination of the others.

You check for linear dependence or linear independence in the same way--by finding solutions for the constants in c1*v1 + c2*v2 + c3*v3 + c4*v4 = 0.

If the vectors are linearly independent, there will be only one solution for the constants: c1 = c2 = c3 = c4 = 0.

If the vectors are linearly dependent, there will be an infinite number of solutions for the constants, where at least one of the constants is nonzero. Pick a vector whose coefficient is nonzero and solve for it. That vector is a linear combination of the other vectors, and can be removed without changing the span of the set of vectors.