# Finding the linear combinations of two vectors

I am studying for my finals and I'm trying to answer the following question:

Consider the following two vectors in $\mathbb{R}^3$: $a=(1,2,3)$ and $b=(2,3,1)$. Decide whether it is possible to express the vector $c=(2,4,5)$ as a linear combination of $a$ and $b$.

I have used the following row operations to find what $C_1$,$C_2$ and $C_3$ are equal to: \begin{align*} \left[ \begin{array}{ccc|c} 1 & 2 &2 & 2\\ 2 & 3 & 4 & 4 \\ 3 & 1 & 5 & 5 \end{array}\right] &\overset{R_2 = 2R_1 -R_1}{\Longrightarrow} \left[ \begin{array}{ccc|c} 1 & 2 &2 &2 \\ 0 & 1 & 0 & 0 \\ 3 & 1 & 5 & 5 \end{array}\right] \overset{R_3=3R_1-R_3}{\Longrightarrow} \left[\begin{array}{ccc|c} 1 &2 & 2 & 2 \\ 0 &1 & 0 & 0 \\ 0 & 5 &1 & 1 \end{array} \right] \\ &\overset{R_3=5R_2+R_3}{\Longrightarrow} \left[\begin{array}{ccc|c} 1 & 2 & 2 & 2 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{array}\right] \end{align*}

Is the following argument correct?

By looking at the last row of the matrix, we see that it says $0 = 1$, which is impossible, and the system of equations therefore has no solutions. This means that we can not find the values for $C_1$ and $C_2$, and so $c$ can not be written as a linear combination of $a$ and $b$.

Thanks in advance!

• It seems you have an extra column in your matrices. There are only two constants to find. Omit the third column from all of your matrices, then you'll be ok. The equation $0=1$ would indeed tell you there is no solution. – David Mitra Apr 20 '12 at 13:42

## 5 Answers

What you did is actually incorrect, because you don't want to row-reduce
$$\left(\begin{array}{ccc|c}1 & 2 & 2 & 2\\ 2 & 3 & 4 & 4 \\ 3 & 1 & 5 & 5 \end{array}\right)$$ The problem with this is you're asking if there exist constants $C_1$, $C_2$, $C_3$ such that $C_1 a + C_2 b + C_3 c = c$, which is obvious (pick $C_3 = 1$, $C_1 = C_2 = 0$).

Instead, you want to row reduce the matrix $$\left(\begin{array}{cc|c}1 & 2 & 2\\ 2 & 3 & 4 \\ 3 & 1 & 5 \end{array}\right)$$ So in other words, you know that the vectors $a$ and $b$ are not linear multiples of each other, so you just want to check to see if you can write $c$ as a linear combination of $a$ and $b$, or you want to know if there exist $\alpha, \beta$ such that $\alpha a + \beta b = c$. (This would be the case of the first matrix)

Alternatively you could row reduce the matrix (this is most similar to what you have already done) $$\left(\begin{array}{ccc|c}1 & 2 & 2 & 0\\ 2 & 3 & 4 & 0 \\ 3 & 1 & 5 & 0 \end{array}\right)$$ This would be asking if there exist $C_1, C_2, C_3$ such that $C_1 a + C_2 b + C_3 c = 0$

It should be clear that these are equivalent methods. Once you have done the row-reduction correctly (just chop off the last column and replace it by a column of 0's and see what you find), then the answer should be fairly easy to see. I'll leave the rest to you.

By the way, if you find that the only possibility is $C_1 = C_2 = C_3 = 0$, this would mean that $c$ is not a linear combination of $a$ and $b$.

• Thanks, so i'm always meant to calculate the last column with the Zero vector? And because the last column is a zero, using the row operations C1 =C2=C3 would mean that C is not a linear combination of a and b? But say if C1=3, C2=4 and C3=1 then C would be a linear combination of a and b? And thanks for this Explanation. – Xabi Apr 20 '12 at 13:42
• @Fatz If $C_1 = 3$, $C_2 = 4$, and $C_3 = 1$, then this would say that $3a + 4b + c = 0$, or $c = -3a - 4b$, which would mean that $c$ would be a linear combination of $a$ and $b$. For your other question, it's convenient to do it with the zero vector as the last column, because writing $c$ as a linear combination of $a$ and $b$ is the same as writing the zero vector as a non-trivial combination of $a$, $b$, and $c$ (non-trivial meaning not all of $C_1$, $C_2$, and $C_3$ can be zero). – Nicholas Stull Apr 20 '12 at 13:52
• @Fatz, Also, if I've not made a mistake, you have done the row-reduction correctly, so you just need to replace the last column by a column of zeros, and you get that the only solution is $C_1 = C_2 = C_3 = 0$, whence $c$ is not a linear combination of $a$ and $b$. (It is very important that we have $C_1 = C_2 = C_3 = 0$, and not just $C_1 = C_2 = C_3$. For example, if $C_1 = C_2 = C_3 = 1$, then we could have $a = (1,0,1)$, $b = (0,1,0)$, and $c = (-1,-1,-1)$) – Nicholas Stull Apr 20 '12 at 13:59
• Thanks, i totally understand now! – Xabi Apr 20 '12 at 14:01
• Glad I could help. – Nicholas Stull Apr 20 '12 at 14:25

It is probably quicker to do this without matrices.

Suppose $\lambda a + \mu b = c$. Then you get three equations in two unknowns.

$\lambda + 2\mu = 2$

$2\lambda + 3\mu = 4$

$3\lambda + \mu = 5$

Solving the first two gives $\mu = 0, \lambda = 2$ but this solution doesn't work in the third equation so there is no solution to the equations, hence no such linear combination.

• Thanks for the answer, but i'm expected to solve it only using a matrix. – Xabi Apr 20 '12 at 13:33
• Well it is just exactly the same, matrices just make the working more concise to write down. If you MUST use matrices then just translate the equations in matrix form. – fretty Apr 20 '12 at 13:38
• Okie dokie, thanks – Xabi Apr 20 '12 at 13:43
• So in matrix form you will get a 3x2 matrix, unlike the 3x3 one you have in your working. – fretty Apr 20 '12 at 13:47

Note that your original linear system appears to be completely wrong for the purposes of solving the problem. What you're after is constants $c_1$ and $c_2$ such that $$c_1 \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \\ 5 \end{bmatrix},$$ or, equivalently,
$$\begin{bmatrix} 1 & 2 \\ 2 & 3 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \\ 5 \end{bmatrix}$$

$c=m\cdot a+n \cdot b$

Hence :

$(2,4,5)=m\cdot(1,2,3)+n\cdot(2,3,1)$

So we have following system of equations :

$\begin{cases} m+2n=2 \\ 2m+3n=4 \\ 3m+n=5 \end{cases}$

which has no solution ,therefore you cannot express $c$ as linear combination of $a$ and $b$ .

You can also do your example in following way,

Let $(2,4,5)=\alpha.(1,2,3)+\beta.(2,3,1)$

$(2,4,5)=(\alpha+2\beta,2\alpha+3\beta,3\alpha+\beta)$

$\therefore 2=\alpha+2\beta,4=2\alpha+3\beta,5=3\alpha+\beta$

Now, solving the first two equation, we get, $\alpha=2,\beta=0$ but it does not satisfied last equation therefore $c$ can not be expressed as a linear combination of $a$ and $b$.