# Show that two vectors in plane are linearly independent

Show that two vectors $$\vec a_{1} = (a_{11},a_{12}), \vec a_{2}=(a_{21},a_{22})$$ in plane are linearly independent if and only if $$a_{11}a_{22}-a_{12}a_{21} \neq 0.$$

So here is my textbook's description on linear dependence and linear independence:

In general, given k vectors $$\vec a_{1}, \dots , \vec a_{k}$$, if any one of $$\vec a_{1}, \dots , \vec a_{k}$$ is a linear combination of the other vectors, $$\vec a_{1}, \dots \vec a_{k}$$ are called linearly dependent. If vectors $$\vec a_{1}, \dots , \vec a_{k}$$ are not linearly dependent, then they are called linearly independent.

Using this description, I tried solving the problem but I just keep getting stuck on how the equation $$a_{11}a_{22}-a_{12}a_{21} \neq 0$$ could be related to the linear independence of the two vectors in the question, $$\vec a_{1}$$ and $$\vec a_{2}$$.

• For them to be linearly dependent, then one must be a multiple of the other. – Lord Shark the Unknown Jul 13 at 7:57
• Oh I see. Guess I was stuck with this question because I didn't come up with thinking of it in terms of linear dependence instead of linear independence. Thanks – linearAlg Jul 13 at 8:02

If $$\vec a_1=\lambda\vec a_2$$ or equivalently $$a_{11}=\lambda a_{21}$$ and $$a_{12}=\lambda a_{22}$$ then it is easy to verify that $$a_{11}a_{22}-a_{12}a_{21}=0$$.

Same story if $$\vec a_2=\lambda\vec a_1$$.

If conversely $$a_{11}a_{22}-a_{12}a_{21}=0$$ then $$a_{22}\vec a_1=a_{12}\vec a_2$$.

This gives three options:

• $$a_{22}=a_{21}=0$$ or equivalently $$\vec a_2=0$$ so that $$\vec a_2=0\vec a_1$$
• $$a_{22}\neq 0$$ so that $$\vec a_1=\frac{a_{12}}{a_{22}}\vec a_2$$
• $$a_{12}\neq 0$$ so that $$\vec a_2=\frac{a_{22}}{a_{12}}\vec a_1$$

So in all cases $$\vec a_1$$ and $$\vec a_2$$ appear to be linearly dependent.