The idea of a linear mix of vectors is really important come the examine of linear algebra. We can use direct combinations to recognize spanning sets, the column an are of a matrix, and a large number of various other topics. Among the most useful an abilities when working with straight combinations is determining when one vector is a linear mix of a given collection of vectors.

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Suppose the we have a vector (vecv) and we want to understand the answer come the question “is (vecv) a linear combination of the vectors (veca_1), (veca_2), and (veca_3)?”. Utilizing the meaning of a linear mix of vectors, this question can be restated to the following:

Are there scalars (x_1), (x_2), and (x_3) together that:(vecv = x_1veca_1 +x_2veca_2+ x_3veca_3)?

If the vectors space in (R^n) for some (n), then this is a inquiry that deserve to be answered making use of the identical augmented matrix:

(left< eginarrayc veca_1 & veca_2 & veca_3 & vecv \ endarray ight>)

If this procession represents a regular system the equations, climate we have the right to say that (vecv) is a linear mix of the various other vectors.

### Example

Determine if the vector (eginbmatrix 5 \ 3 \ 0 \ endbmatrix) is a linear mix of the vectors:(eginbmatrix 2 \ 0 \ 1 \ endbmatrix), (eginbmatrix 1 \ 4 \ 3 \ endbmatrix), (eginbmatrix 8 \ 1 \ 1 \ endbmatrix), and also (eginbmatrix -4 \ 6 \ 1 \ endbmatrix)

### Solution

Remember the this means we desire to discover constants (x_1), (x_2), (x_3), and (x_4) together that:

(eginbmatrix 5 \ 3 \ 0 \ endbmatrix = x_1eginbmatrix 2 \ 0 \ 1 \ endbmatrix + x_2eginbmatrix 1 \ 4 \ 3 \ endbmatrix + x_3eginbmatrix 8 \ 1 \ 1 \ endbmatrix + x_4eginbmatrix -4 \ 6 \ 1 \ endbmatrix)

This vector equation is indistinguishable to one augmented matrix. Setup this matrix up and row reducing, we uncover that:

(left< eginarrayc 2 & 1 & 8 & -4 & 5 \0 & 4 & 1 & 6 & 3 \1 & 3 & 1 & 1 & 0 \endarray ight>)

Is indistinguishable to:

(left< eginarraycccc 1 & 0 & 0 & -dfrac10329 & -dfrac7429 \0 & 1 & 0 & dfrac4229 & dfrac1329 \0 & 0 & 1 & dfrac629 & dfrac3529 \endarray ight>)

While it isn’t pretty, this procession does no contain a heat such as (eginbmatrix 0 & 0 & 0 & 0 & c \ endbmatrix) whereby (c eq 0) which would suggest the underlying system is inconsistent. Therefore the underlying system is consistent (has a solution) which way the vector equation is likewise consistent.

So, we deserve to say that (eginbmatrix 5 \ 3 \ 0 \ endbmatrix) is a linear combination of the various other vectors.

## The step-by-step process

In general, if you desire to determine if a vector (vecu) is a linear combination of vectors (vecv_1), (vecv_2), … , (vecv_p) (for any whole number (p > 2)) girlfriend will carry out the following.

### Step 1

Set increase the augmented matrix

(left< eginarraycccc vecv_1 & vecv_2 & cdots & vecv_p & vecu \ endarray ight>)

and row mitigate it.

### Step 2

Use the reduced kind of the matrix to identify if the augmented matrix represents a constant system that equations. If so, climate (vecu) is a linear mix of the others. Otherwise, the is not.

In the second step, the is important to remember the a mechanism of equations is constant if over there is one systems OR plenty of solutions. The number of solutions is not crucial – only that there IS at least one solution. That means there is at least one way to compose the provided vector as a linear combination of the others.

## Writing a Vector as a Linear combination of various other Vectors

Sometimes you can be asked to create a vector together a linear combination of various other vectors. This needs the same job-related as over with one more step. You have to use a equipment to the vector equation to compose out just how the vectors are linked to make the new vector.

Let’s start with an easier case than the one us did before and then come ago to it due to the fact that it is a little complicated.

### Example

Write the vector (vecv = eginbmatrix 2 \ 4 \ 2 \ endbmatrix) as a linear mix of the vectors: (eginbmatrix 2 \ 0 \ 1 \ endbmatrix), (eginbmatrix 0 \ 1 \ 0 \ endbmatrix), and (eginbmatrix -2 \ 0 \ 0 \ endbmatrix)

### Solution

Step 1We collection up our augmented matrix and also row mitigate it.

(left< eginarrayc 2 & 0 & -2 & 2 \0 & 1 & 0 & 4 \1 & 0 & 0 & 2 \endarray ight>)

is identical to

(left< eginarrayc 1 & 0 & 0 & 2 \0 & 1 & 0 & 4 \0 & 0 & 1 & 1 \endarray ight>)

Step 2We determine if the procession represents a continuous system that equations.

Based on the lessened matrix, the underlying mechanism is consistent. Again, this is since there space no rows of all zeros in the coefficient component of the matrix and a single nonzero worth in the augment. (you could likewise use the number of pivots to make the argument.)

Unlike before, us don’t only want to verify the we have actually a straight combination. We desire to display the linear combination itself. This means that we require an really solution. In this case, there is only one:

(x_1 = 2), (x_2 = 4), (x_3 = 1)

Using these values, we have the right to write (vecv) as:

(vecv = eginbmatrix 2 \ 4 \ 2 \ endbmatrix = (2)eginbmatrix 2 \ 0 \ 1 \ endbmatrix + (4)eginbmatrix 0 \ 1 \ 0 \ endbmatrix + (1)eginbmatrix -2 \ 0 \ 0 \ endbmatrix)

Now let’s go ago to our first example (the one v the crazy fractions) but change the accuse a bit.

### Example

Write the vector (vecv = eginbmatrix 5 \ 3 \ 0 \ endbmatrix) together a linear mix of the vectors:(eginbmatrix 2 \ 0 \ 1 \ endbmatrix), (eginbmatrix 1 \ 4 \ 3 \ endbmatrix), (eginbmatrix 8 \ 1 \ 1 \ endbmatrix), and (eginbmatrix -4 \ 6 \ 1 \ endbmatrix)

When we did action 1, we had actually the following work. This showed that the indistinguishable vector equation was consistent and verified that (vecv) was a linear mix of the other vectors.

(left< eginarraycccc 2 & 1 & 8 & -4 & 5 \0 & 4 & 1 & 6 & 3 \1 & 3 & 1 & 1 & 0 \endarray ight>)

Is equivalent to:

(left< eginarraycccc 1 & 0 & 0 & -dfrac10329 & -dfrac7429 \0 & 1 & 0 & dfrac4229 & dfrac1329 \0 & 0 & 1 & dfrac629 & dfrac3529 \endarray ight>)

What if we wanted to compose out the straight combination. This is different from the previous instance in the there are infinitely plenty of solutions to the vector equation.

Looking much more closely in ~ this augmented matrix, we can see that there is one free variable (x_4). If we compose out the equations, us have:

(x_1 – left(dfrac10329 ight)x_4 = -dfrac7429)

(x_2 + left(dfrac4229 ight)x_4 = dfrac1329)

(x_3 + left(dfrac629 ight)x_4 = dfrac3529)

Since (x_4) is a cost-free variable, we can let that have any value and find a equipment to this device of equations. A really “nice” value would it is in zero. If (x_4 = 0), then:

(x_1 – dfrac10329(0) = -dfrac7429)

(x_2 + dfrac4229(0) = dfrac1329)

(x_3 + dfrac629(0) = dfrac3529)

Using this solution, we can write (vecv) as a linear combination of the other vectors.

(vecv = eginbmatrix 5 \ 3 \ 0 \ endbmatrix = left(-dfrac7229 ight)eginbmatrix 2 \ 0 \ 1 \ endbmatrix + left(dfrac1329 ight)eginbmatrix 1 \ 4 \ 3 \ endbmatrix + left(dfrac3529 ight)eginbmatrix 8 \ 1 \ 1 \ endbmatrix + (0)eginbmatrix -4 \ 6 \ 1 \ endbmatrix)

This would certainly be one solution, but since (x_4) is free, there space infinitely many. For each feasible value that (x_4), friend have another correct way to create (vecv) together a linear mix of the other vectors. For example, if (x_4 = 1):

(eginalignx_1 &= -dfrac7429 + dfrac10329 \ &= dfrac2929 \ &= 1endalign)

(eginalignx_2 &= dfrac1329 – dfrac4229\ &= -dfrac2929 \ &= -1endalign)

(eginalignx_3 &= dfrac3529 – dfrac629\ &= dfrac2929 \ &= 1endalign)

Using this, we can also write:

(vecv = eginbmatrix 5 \ 3 \ 0 \ endbmatrix = (1)eginbmatrix 2 \ 0 \ 1 \ endbmatrix + (-1)eginbmatrix 1 \ 4 \ 3 \ endbmatrix + (1)eginbmatrix 8 \ 1 \ 1 \ endbmatrix + (1)eginbmatrix -4 \ 6 \ 1 \ endbmatrix)

How pretty is that? (note: normally, we wouldn’t create out the 1 in the equation showing the linear combination. I left it there so you might see whereby each number native the solution ended up).

Again, a difficulty like this has actually infinitely numerous answers. All you need to do is choose a value for the complimentary variables and also you will have one specific solution you have the right to use in writing the straight combination.

## When the Vector is no a Linear mix of the Others

It is precious seeing one instance where a vector is no a linear mix of some provided vectors. Once this happens, we will finish up v an augmented matrix indicating an inconsistent device of equations.

### Example

Determine if the vector (eginbmatrix 1 \ 2 \ 1 \ endbmatrix) is a linear combination of the vectors:(eginbmatrix 1 \ 1 \ 0 \ endbmatrix), (eginbmatrix 0 \ 1 \ -1 \ endbmatrix), and also (eginbmatrix 1\ 2 \ -1 \ endbmatrix).

### Solution

Step 1We set up our augmented matrix and also row mitigate it.

(left< eginarrayccc 1 & 0 & 1 & 1 \1 & 1 & 2 & 2 \0 & -1 & -1 & 1 \endarray ight>)

is equivalent to:

(left< eginarrayc 1 & 0 & 1 & 0 \0 & 1 & 1 & 0 \0 & 0 & 0 & 1 \endarray ight>)

Step 2We recognize if the matrix represents a continuous system that equations.

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Given the type of the last row, this matrix represents an *inconsistent* system of equations. That means there is no means to compose this vector as a linear mix of the various other vectors. That’s that – nothing rather to say! This will certainly be ours conclusion any time heat reduction results in a row through zeros and also a nonzero value on the augment.

## Study overview – straight combinations and also span

Need an ext practice with direct combinations and also span? This 40-page study guide will help! It has explanations, examples, exercise problems, and also full step-by-step solutions.