Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Determine if the following set of vectors is linearly independent:


I've done the following system of equations, and I think I did it right... It's been such a long time since I did this sort of thing...

Assume the following: \begin{equation*} a\left[\begin{array}{r}2\\2\\0\end{array}\right]+b\left[\begin{array}{r}1\\-1\\1\end{array}\right]+c\left[\begin{array}{r}4\\2\\-2\end{array}\right]=\left[\begin{array}{r}0\\0\\0\end{array}\right] \end{equation*} Determine if $a=b=c=0$: \begin{align} 2a+b+4c&=0&&(1)\\ 2a-b+2c&=0&&(2)\\ b-2c&=0&&(3) \end{align} Subtract (2) from (1): \begin{align} b+c&=0&&(4)\\ b-2c&=0&&(5) \end{align} Substitute (5) into (4): \begin{equation} c=0 \end{equation}

So now what do I do with this fact? I'm tempted to say that only $c=0$, and $a$ and $b$ can be something else... but I don't trust that my intuition is right.

share|cite|improve this question
If $c=0$ then you must have $b=0$ and then you must have $a=0$. Hence they are linearly independent. – copper.hat Jun 6 '13 at 1:42
From $c=0$ and $b-2c=0$ you can conclude? And then what about $a$? You were doing fine. The same thing, with less writing, can be done using row reduction. – André Nicolas Jun 6 '13 at 1:42
substitute $c=0$ back into (4) or (5) to show that $b=0$ and then both $b=0$ and $c=0$ into (1) or (2) to show that $a=0$. By definition they are then linearly independent. – Tpofofn Jun 6 '13 at 1:43
@AndréNicolas I'm only starting to learn about matrices now... this was taught in this method so I presume this is how I have to do it on the assignment. – agent154 Jun 6 '13 at 1:50
Yes, at the beginning it makes sense do do things directly from the definition. – André Nicolas Jun 6 '13 at 1:51
up vote 7 down vote accepted

You just stopped too early:

You need to simply substitute $c = 0$ back into the two equations: from equation $(3)$, $c = 0 \implies b = 0$.

With $b = 0, c = 0$ substituted into equation $(1)$ or $(2)$, $b = c = 0 \implies a = 0$. So in the end, since

$$\begin{equation*} a\left[\begin{array}{r}2\\2\\0\end{array}\right]+b\left[\begin{array}{r}1\\-1\\1\end{array}\right]+c\left[\begin{array}{r}4\\2\\-2\end{array}\right]=\left[\begin{array}{r}0\\0\\0\end{array}\right] \end{equation*}\implies a = b = c = 0$$ the vectors are linearly independent.

You could have, similarly, constructed a $3\times 3$ matrix $M$ with the three given vectors as its columns, and computed the determinant of $M$: We know that if $\det M \neq 0$, the given vectors are linearly independent.

$$M = \begin{bmatrix} 2 & 1 & 4 \\ 2 & -1 & 2 \\ 0 & 1 & -2 \end{bmatrix}$$

$$\det M = 12 \neq 0 \implies \;\;\text{linear independence of columns}$$

share|cite|improve this answer

you can take the vectors to form a matrix and check its determinant. If the determinant is non zero, then the vectors are linearly independent. Otherwise, they are linearly dependent.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.