Science:Math Exam Resources/Courses/MATH152/April 2013/Question A 06

MATH152 April 2013

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[hide] Question A 06
Is it possible for three vectors in $\mathbb {R} ^{2}$ to be linearly independent? Briefly justify your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Try to formulate this in term of a linear system of equations.

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[show] Solution
Suppose we have three vectors over $\mathbb {R} ^{2}$ say ${\begin{bmatrix}a\\b\end{bmatrix}}\quad {\begin{bmatrix}c\\d\end{bmatrix}}\quad {\begin{bmatrix}e\\f\end{bmatrix}}$ for fixed a,b,c,d,e,f. These vectors are linearly independent if we can write $x{\begin{bmatrix}a\\b\end{bmatrix}}+y{\begin{bmatrix}c\\d\end{bmatrix}}+z{\begin{bmatrix}e\\f\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}$ and have x=y=z=0 be the only solution. We can write this as a system of linear equations given by: ${\begin{aligned}xa+yc+ze&=0\\xb+yd+zf&=0\end{aligned}}$ This is a system of two equations and three unknowns (given by x,y, and z). This always has a nontrivial solution (i.e. for any value of z, I can always find an x and y that solve this) and therefore x=y=z=0 is not the only solution. The answer then is that 3 vectors are always linearly dependent in $\mathbb {R} ^{2}$